74 Partial Fractions
Learning Objectives
In this section, you will:
- Decompose [latex]\frac{P\left(x\right)}{Q\left(x\right)}[/latex], where [latex]Q\left(x\right)[/latex] has only nonrepeated linear factors.
- Decompose [latex]\frac{P\left(x\right)}{Q\left(x\right)}[/latex], where [latex]Q\left(x\right)[/latex] has repeated linear factors.
- Decompose [latex]\frac{P\left(x\right)}{Q\left(x\right)}[/latex], where [latex]Q\left(x\right)[/latex] has a nonrepeated irreducible quadratic factor.
- Decompose [latex]\frac{P\left(x\right)}{Q\left(x\right)}[/latex], where [latex]Q\left(x\right)[/latex] has a repeated irreducible quadratic factor.
Earlier in this chapter, we studied systems of two equations in two variables, systems of three equations in three variables, and nonlinear systems. Here we introduce another way that systems of equations can be utilized—the decomposition of rational expressions.
Fractions can be complicated; adding a variable in the denominator makes them even more so. The methods studied in this section will help simplify the concept of a rational expression.
Decomposing[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}\,[/latex]Where Q(x) Has Only Nonrepeated Linear Factors
Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at partial fraction decomposition, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified rational expression to the original expressions, called the partial fractions.
For example, suppose we add the following fractions:
We would first need to find a common denominator,[latex]\,\left(x+2\right)\left(x-3\right).[/latex]
Next, we would write each expression with this common denominator and find the sum of the terms.
Partial fraction decomposition is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.
We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator.
When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of[latex]\,{x}^{2}-x-6\,[/latex]are[latex]\,\left(x-3\right)\left(x+2\right),\,[/latex]the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition.
Partial Fraction Decomposition of[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)\,[/latex]Has Nonrepeated Linear Factors
The partial fraction decomposition of[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}\,[/latex]when[latex]\,Q\left(x\right)\,[/latex]has nonrepeated linear factors and the degree of[latex]\,P\left(x\right)\,[/latex]is less than the degree of[latex]\,Q\left(x\right)\,[/latex]is
Given a rational expression with distinct linear factors in the denominator, decompose it.
- Use a variable for the original numerators, usually[latex]\,A,B[/latex]or[latex]\,C,\,[/latex]depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use[latex]\,{A}_{n}\,[/latex]for each numerator
[latex]\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\frac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\cdots \text{+}\frac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}[/latex]
- Multiply both sides of the equation by the common denominator to eliminate fractions.
- Expand the right side of the equation and collect like terms.
- Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
Decomposing a Rational Function with Distinct Linear Factors
Decompose the given rational expression with distinct linear factors.
[hidden-answer a=”156294″]
We will separate the denominator factors and give each numerator a symbolic label, like[latex]\,A,B\text{\hspace{0.17em},}[/latex]or[latex]\,C.[/latex]
Multiply both sides of the equation by the common denominator to eliminate the fractions:
The resulting equation is
Expand the right side of the equation and collect like terms.
Set up a system of equations associating corresponding coefficients.
Add the two equations and solve for[latex]\,B.[/latex]
Substitute[latex]\,B=1\,[/latex]into one of the original equations in the system.
Thus, the partial fraction decomposition is
Another method to use to solve for[latex]\,A\,[/latex]or[latex]\,B\,[/latex]is by considering the equation that resulted from eliminating the fractions and substituting a value for[latex]\,x\,[/latex]that will make either the A– or B-term equal 0. If we let[latex]\,x=1,\,[/latex]the
Next, either substitute[latex]\,B=1\,[/latex]into the equation and solve for[latex]\,A,\,[/latex]or make the B-term 0 by substituting[latex]\,x=-2\,[/latex]into the equation.
We obtain the same values for[latex]\,A\,[/latex]and[latex]\,B\,[/latex]using either method, so the decompositions are the same using either method.
Although this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the Heaviside method, named after Charles Heaviside, a pioneer in the study of electronics.
[/hidden-answer]
Try It
Find the partial fraction decomposition of the following expression.
[hidden-answer a=”fs-id1165137605239″]
[latex]\frac{3}{x-3}-\frac{2}{x-2}[/latex]
[/hidden-answer]
Decomposing[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}\,[/latex]Where Q(x) Has Repeated Linear Factors
Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.
Partial Fraction Decomposition of[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)\,[/latex]Has Repeated Linear Factors
The partial fraction decomposition of[latex]\,\frac{P\left(x\right)}{Q\left(x\right)},\,[/latex]when[latex]\,Q\left(x\right)\,[/latex]has a repeated linear factor occurring[latex]\,n\,[/latex]times and the degree of[latex]\,P\left(x\right)\,[/latex]is less than the degree of[latex]\,Q\left(x\right),\,[/latex]is
Write the denominator powers in increasing order.
Given a rational expression with repeated linear factors, decompose it.
- Use a variable like[latex]\,A,B,\,[/latex]or[latex]\,C\,[/latex]for the numerators and account for increasing powers of the denominators.
[latex]\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left(ax+b\right)}+\frac{{A}_{2}}{{\left(ax+b\right)}^{2}}+ \text{. }\text{. }\text{. + }\frac{{A}_{n}}{{\left(ax+b\right)}^{n}}[/latex]
- Multiply both sides of the equation by the common denominator to eliminate fractions.
- Expand the right side of the equation and collect like terms.
- Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
Decomposing with Repeated Linear Factors
Decompose the given rational expression with repeated linear factors.
[hidden-answer a=”fs-id1165137645114″]
The denominator factors are[latex]\,x{\left(x-2\right)}^{2}.\,[/latex]To allow for the repeated factor of[latex]\,\left(x-2\right),\,[/latex]the decomposition will include three denominators:[latex]\,x,\left(x-2\right),\,[/latex]and[latex]\,{\left(x-2\right)}^{2}.\,[/latex]Thus,
Next, we multiply both sides by the common denominator.
On the right side of the equation, we expand and collect like terms.
Next, we compare the coefficients of both sides. This will give the system of equations in three variables:
Solving for[latex]\,A[/latex], we have
Substitute[latex]\,A=1\,[/latex]into equation (1).
Then, to solve for[latex]\,C,\,[/latex]substitute the values for[latex]\,A\,[/latex]and[latex]\,B\,[/latex]into equation (2).
Thus,
Try It
Find the partial fraction decomposition of the expression with repeated linear factors.
[hidden-answer a=”fs-id1165133047556″]
[latex]\frac{6}{x-1}-\frac{5}{{\left(x-1\right)}^{2}}[/latex]
[/hidden-answer]
Decomposing[latex]\,\frac{P\left(x\right)}{Q\left(x\right)},\,[/latex]Where Q(x) Has a Nonrepeated Irreducible Quadratic Factor
So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators[latex]\,A,B,\,[/latex]or[latex]\,C\,[/latex]representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as[latex]\,Ax+B,Bx+C,\,[/latex]etc.
Decomposition of[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)\,[/latex]Has a Nonrepeated Irreducible Quadratic Factor
The partial fraction decomposition of[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}\,[/latex]such that[latex]\,Q\left(x\right)\,[/latex]has a nonrepeated irreducible quadratic factor and the degree of[latex]\,P\left(x\right)\,[/latex]is less than the degree of[latex]\,Q\left(x\right)\,[/latex]is written as
The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator:[latex]\,A,B,C,\,[/latex]and so on.
Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it.
- Use variables such as[latex]\,A,B,\,[/latex]or[latex]\,C\,[/latex]for the constant numerators over linear factors, and linear expressions such as[latex]\,{A}_{1}x+{B}_{1},{A}_{2}x+{B}_{2},\,[/latex]etc., for the numerators of each quadratic factor in the denominator.
[latex]\frac{P\left(x\right)}{Q\left(x\right)}=\frac{A}{ax+b}+\frac{{A}_{1}x+{B}_{1}}{\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\right)}+\frac{{A}_{2}x+{B}_{2}}{\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\right)}+\cdot \cdot \cdot +\frac{{A}_{n}x+{B}_{n}}{\left({a}_{n}{x}^{2}+{b}_{n}x+{c}_{n}\right)}[/latex]
- Multiply both sides of the equation by the common denominator to eliminate fractions.
- Expand the right side of the equation and collect like terms.
- Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
Decomposing[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}\,[/latex]When Q(x) Contains a Nonrepeated Irreducible Quadratic Factor
Find a partial fraction decomposition of the given expression.
[hidden-answer a=”fs-id1165133088179″]
We have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus,
We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator.
Notice we could easily solve for[latex]\,A\,[/latex]by choosing a value for[latex]\,x\,[/latex]that will make the[latex]\,Bx+C\,[/latex]term equal 0. Let[latex]x=-3\,[/latex]and substitute it into the equation.
Now that we know the value of[latex]\,A,\,[/latex]substitute it back into the equation. Then expand the right side and collect like terms.
Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations.
Solve for[latex]\,B\,[/latex]using equation (1) and solve for[latex]\,C\,[/latex]using equation (3).
Thus, the partial fraction decomposition of the expression is
Could we have just set up a system of equations to solve (Figure)?
Yes, we could have solved it by setting up a system of equations without solving for[latex]\,A\,[/latex]first. The expansion on the right would be:
So the system of equations would be:
Try It
Find the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic factor.
[hidden-answer a=”fs-id1165134042251″]
[latex]\frac{3}{x-1}+\frac{2x-4}{{x}^{2}+1}[/latex]
[/hidden-answer]
Decomposing[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}\,[/latex]When Q(x) Has a Repeated Irreducible Quadratic Factor
Now that we can decompose a simplified rational expression with an irreducible quadratic factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers.
Decomposition of[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}\,[/latex]When Q(x) Has a Repeated Irreducible Quadratic Factor
The partial fraction decomposition of[latex]\,\frac{P\left(x\right)}{Q\left(x\right)},\,[/latex]when[latex]\,Q\left(x\right)\,[/latex]has a repeated irreducible quadratic factor and the degree of[latex]\,P\left(x\right)\,[/latex]is less than the degree of[latex]\,Q\left(x\right),\,[/latex]is
Write the denominators in increasing powers.
Given a rational expression that has a repeated irreducible factor, decompose it.
- Use variables like[latex]\,A,B,\,[/latex]or[latex]\,C\,[/latex]for the constant numerators over linear factors, and linear expressions such as[latex]\,{A}_{1}x+{B}_{1},{A}_{2}x+{B}_{2},\,[/latex]etc., for the numerators of each quadratic factor in the denominator written in increasing powers, such as
[latex]\frac{P\left(x\right)}{Q\left(x\right)}=\frac{A}{ax+b}+\frac{{A}_{1}x+{B}_{1}}{\left(a{x}^{2}+bx+c\right)}+\frac{{A}_{2}x+{B}_{2}}{{\left(a{x}^{2}+bx+c\right)}^{2}}+\cdots +\text{ }\frac{{A}_{n}+{B}_{n}}{{\left(a{x}^{2}+bx+c\right)}^{n}}[/latex]
- Multiply both sides of the equation by the common denominator to eliminate fractions.
- Expand the right side of the equation and collect like terms.
- Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
Decomposing a Rational Function with a Repeated Irreducible Quadratic Factor in the Denominator
Decompose the given expression that has a repeated irreducible factor in the denominator.
[hidden-answer a=”fs-id1165134041975″]
The factors of the denominator are[latex]\,x,\left({x}^{2}+1\right),\,[/latex]and[latex]\,{\left({x}^{2}+1\right)}^{2}.\,[/latex]Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form[latex]\,Ax+B.\,[/latex]So, let’s begin the decomposition.
We eliminate the denominators by multiplying each term by[latex]\,x{\left({x}^{2}+1\right)}^{2}.\,[/latex]Thus,
Expand the right side.
Now we will collect like terms.
Set up the system of equations matching corresponding coefficients on each side of the equal sign.
We can use substitution from this point. Substitute[latex]\,A=1\,[/latex]into the first equation.
Substitute[latex]\,A=1\,[/latex]and[latex]\,B=0\,[/latex]into the third equation.
Substitute[latex]\,C=1\,[/latex]into the fourth equation.
Now we have solved for all of the unknowns on the right side of the equal sign. We have[latex]\,A=1,\,[/latex][latex]B=0,\,[/latex][latex]C=1,\,[/latex][latex]D=-1,\,[/latex]and[latex]\,E=-2.\,[/latex]We can write the decomposition as follows:
Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.
[hidden-answer a=”fs-id1165131801235″]
[latex]\frac{x-2}{{x}^{2}-2x+3}+\frac{2x+1}{{\left({x}^{2}-2x+3\right)}^{2}}[/latex]
[/hidden-answer]
Access these online resources for additional instruction and practice with partial fractions.
Key Concepts
- Decompose[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}\,[/latex]by writing the partial fractions as[latex]\,\frac{A}{{a}_{1}x+{b}_{1}}+\frac{B}{{a}_{2}x+{b}_{2}}.\,[/latex]Solve by clearing the fractions, expanding the right side, collecting like terms, and setting corresponding coefficients equal to each other, then setting up and solving a system of equations. See (Figure).
- The decomposition of[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}\,[/latex]with repeated linear factors must account for the factors of the denominator in increasing powers. See (Figure).
- The decomposition of[latex]\,\frac{P\left(x\right)}{Q\left(x\right)}\,[/latex]with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor, as in[latex]\,\frac{A}{x}+\frac{Bx+C}{\left(a{x}^{2}+bx+c\right)}.\,[/latex]See (Figure).
- In the decomposition of[latex]\,\frac{P\left(x\right)}{Q\left(x\right)},\,[/latex]where[latex]\,Q\left(x\right)\,[/latex]has a repeated irreducible quadratic factor, when the irreducible quadratic factors are repeated, powers of the denominator factors must be represented in increasing powers as
[latex]\frac{Ax+B}{\left(a{x}^{2}+bx+c\right)}+\frac{{A}_{2}x+{B}_{2}}{{\left(a{x}^{2}+bx+c\right)}^{2}}+\cdots \text{+}\frac{{A}_{n}x+{B}_{n}}{{\left(a{x}^{2}+bx+c\right)}^{n}}.[/latex]
See (Figure).
Section Exercises
Verbal
Can any quotient of polynomials be decomposed into at least two partial fractions? If so, explain why, and if not, give an example of such a fraction
[hidden-answer a=”fs-id1165137410930″]
No, a quotient of polynomials can only be decomposed if the denominator can be factored. For example,[latex]\,\frac{1}{{x}^{2}+1}\,[/latex]cannot be decomposed because the denominator cannot be factored.
[/hidden-answer]
Can you explain why a partial fraction decomposition is unique? (Hint: Think about it as a system of equations.)
Can you explain how to verify a partial fraction decomposition graphically?
[hidden-answer a=”fs-id1165131974776″]
Graph both sides and ensure they are equal.
[/hidden-answer]
You are unsure if you correctly decomposed the partial fraction correctly. Explain how you could double-check your answer.
Once you have a system of equations generated by the partial fraction decomposition, can you explain another method to solve it? For example if you had[latex]\,\frac{7x+13}{3{x}^{2}+8x+15}=\frac{A}{x+1}+\frac{B}{3x+5},\,[/latex]we eventually simplify to[latex]\,7x+13=A\left(3x+5\right)+B\left(x+1\right).\,[/latex]Explain how you could intelligently choose an[latex]\,x[/latex]-value that will eliminate either[latex]\,A\,[/latex]or[latex]\,B\,[/latex]and solve for[latex]\,A\,[/latex]and[latex]\,B.[/latex]
[hidden-answer a=”fs-id1165137451391″]
If we choose[latex]\,x=-1,\,[/latex]then the B-term disappears, letting us immediately know that[latex]\,A=3.\,[/latex]We could alternatively plug in[latex]\,x=-\frac{5}{3},\,[/latex]giving us a B-value of[latex]\,-2.[/latex]
[/hidden-answer]
Algebraic
For the following exercises, find the decomposition of the partial fraction for the nonrepeating linear factors.
[latex]\frac{5x+16}{{x}^{2}+10x+24}[/latex]
[latex]\frac{3x-79}{{x}^{2}-5x-24}[/latex]
[hidden-answer a=”fs-id1165137668085″]
[latex]\frac{8}{x+3}-\frac{5}{x-8}[/latex]
[/hidden-answer]
[latex]\frac{10x+47}{{x}^{2}+7x+10}[/latex]
[hidden-answer a=”fs-id1165134342483″]
[latex]\frac{1}{x+5}+\frac{9}{x+2}[/latex]
[/hidden-answer]
[latex]\frac{x}{6{x}^{2}+25x+25}[/latex]
[latex]\frac{32x-11}{20{x}^{2}-13x+2}[/latex]
[hidden-answer a=”fs-id1165133026294″]
[latex]\frac{3}{5x-2}+\frac{4}{4x-1}[/latex]
[/hidden-answer]
[latex]\frac{x+1}{{x}^{2}+7x+10}[/latex]
[latex]\frac{5x}{{x}^{2}-9}[/latex]
[hidden-answer a=”fs-id1165132976175″]
[latex]\frac{5}{2\left(x+3\right)}+\frac{5}{2\left(x-3\right)}[/latex]
[/hidden-answer]
[latex]\frac{10x}{{x}^{2}-25}[/latex]
[latex]\frac{6x}{{x}^{2}-4}[/latex]
[hidden-answer a=”fs-id1165134575087″]
[latex]\frac{3}{x+2}+\frac{3}{x-2}[/latex]
[/hidden-answer]
[latex]\frac{2x-3}{{x}^{2}-6x+5}[/latex]
[latex]\frac{4x-1}{{x}^{2}-x-6}[/latex]
[hidden-answer a=”fs-id1165133035991″]
[latex]\frac{9}{5\left(x+2\right)}+\frac{11}{5\left(x-3\right)}[/latex]
[/hidden-answer]
[latex]\frac{4x+3}{{x}^{2}+8x+15}[/latex]
[latex]\frac{3x-1}{{x}^{2}-5x+6}[/latex]
[hidden-answer a=”fs-id1165135196639″]
[latex]\frac{8}{x-3}-\frac{5}{x-2}[/latex]
[/hidden-answer]
For the following exercises, find the decomposition of the partial fraction for the repeating linear factors.
[latex]\frac{-5x-19}{{\left(x+4\right)}^{2}}[/latex]
[latex]\frac{x}{{\left(x-2\right)}^{2}}[/latex]
[hidden-answer a=”fs-id1165135680115″]
[latex]\frac{1}{x-2}+\frac{2}{{\left(x-2\right)}^{2}}[/latex]
[/hidden-answer]
[latex]\frac{7x+14}{{\left(x+3\right)}^{2}}[/latex]
[latex]\frac{-24x-27}{{\left(4x+5\right)}^{2}}[/latex]
[hidden-answer a=”fs-id1165135466377″]
[latex]-\frac{6}{4x+5}+\frac{3}{{\left(4x+5\right)}^{2}}[/latex]
[/hidden-answer]
[latex]\frac{-24x-27}{{\left(6x-7\right)}^{2}}[/latex]
[latex]\frac{5-x}{{\left(x-7\right)}^{2}}[/latex]
[hidden-answer a=”fs-id1165135543287″]
[latex]-\frac{1}{x-7}-\frac{2}{{\left(x-7\right)}^{2}}[/latex]
[/hidden-answer]
[latex]\frac{5x+14}{2{x}^{2}+12x+18}[/latex]
[latex]\frac{5{x}^{2}+20x+8}{2x{\left(x+1\right)}^{2}}[/latex]
[hidden-answer a=”fs-id1165137583307″]
[latex]\frac{4}{x}-\frac{3}{2\left(x+1\right)}+\frac{7}{2{\left(x+1\right)}^{2}}[/latex]
[/hidden-answer]
[latex]\frac{4{x}^{2}+55x+25}{5x{\left(3x+5\right)}^{2}}[/latex]
[latex]\frac{54{x}^{3}+127{x}^{2}+80x+16}{2{x}^{2}{\left(3x+2\right)}^{2}}[/latex]
[hidden-answer a=”fs-id1165131862958″]
[latex]\frac{4}{x}+\frac{2}{{x}^{2}}-\frac{3}{3x+2}+\frac{7}{2{\left(3x+2\right)}^{2}}[/latex]
[/hidden-answer]
[latex]\frac{{x}^{3}-5{x}^{2}+12x+144}{{x}^{2}\left({x}^{2}+12x+36\right)}[/latex]
For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor.
[latex]\frac{4{x}^{2}+6x+11}{\left(x+2\right)\left({x}^{2}+x+3\right)}[/latex]
[hidden-answer a=”fs-id1165134279720″]
[latex]\frac{x+1}{{x}^{2}+x+3}+\frac{3}{x+2}[/latex]
[/hidden-answer]
[latex]\frac{-2{x}^{2}+10x+4}{\left(x-1\right)\left({x}^{2}+3x+8\right)}[/latex]
[hidden-answer a=”fs-id1165137442876″]
[latex]\frac{4-3x}{{x}^{2}+3x+8}+\frac{1}{x-1}[/latex]
[/hidden-answer]
[latex]\frac{{x}^{2}+3x+1}{\left(x+1\right)\left({x}^{2}+5x-2\right)}[/latex]
[latex]\frac{4{x}^{2}+17x-1}{\left(x+3\right)\left({x}^{2}+6x+1\right)}[/latex]
[hidden-answer a=”fs-id1165137863885″]
[latex]\frac{2x-1}{{x}^{2}+6x+1}+\frac{2}{x+3}[/latex]
[/hidden-answer]
[latex]\frac{4{x}^{2}}{\left(x+5\right)\left({x}^{2}+7x-5\right)}[/latex]
[latex]\frac{4{x}^{2}+5x+3}{{x}^{3}-1}[/latex]
[hidden-answer a=”fs-id1165135543055″]
[latex]\frac{1}{{x}^{2}+x+1}+\frac{4}{x-1}[/latex]
[/hidden-answer]
[latex]\frac{-5{x}^{2}+18x-4}{{x}^{3}+8}[/latex]
[latex]\frac{3{x}^{2}-7x+33}{{x}^{3}+27}[/latex]
[hidden-answer a=”fs-id1165137786222″]
[latex]\frac{2}{{x}^{2}-3x+9}+\frac{3}{x+3}[/latex]
[/hidden-answer]
[latex]\frac{{x}^{2}+2x+40}{{x}^{3}-125}[/latex]
[latex]\frac{4{x}^{2}+4x+12}{8{x}^{3}-27}[/latex]
[hidden-answer a=”fs-id1165134081096″]
[latex]-\frac{1}{4{x}^{2}+6x+9}+\frac{1}{2x-3}[/latex]
[/hidden-answer]
[latex]\frac{-50{x}^{2}+5x-3}{125{x}^{3}-1}[/latex]
[latex]\frac{-2{x}^{3}-30{x}^{2}+36x+216}{{x}^{4}+216x}[/latex]
[hidden-answer a=”fs-id1165134134227″]
[latex]\frac{1}{x}+\frac{1}{x+6}-\frac{4x}{{x}^{2}-6x+36}[/latex]
[/hidden-answer]
For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor.
[latex]\frac{3{x}^{3}+2{x}^{2}+14x+15}{{\left({x}^{2}+4\right)}^{2}}[/latex]
[latex]\frac{{x}^{3}+6{x}^{2}+5x+9}{{\left({x}^{2}+1\right)}^{2}}[/latex]
[hidden-answer a=”fs-id1165135160221″]
[latex]\frac{x+6}{{x}^{2}+1}+\frac{4x+3}{{\left({x}^{2}+1\right)}^{2}}[/latex]
[/hidden-answer]
[latex]\frac{{x}^{3}-{x}^{2}+x-1}{{\left({x}^{2}-3\right)}^{2}}[/latex]
[latex]\frac{{x}^{2}+5x+5}{{\left(x+2\right)}^{2}}[/latex]
[hidden-answer a=”fs-id1165137733720″]
[latex]\frac{x+1}{x+2}+\frac{2x+3}{{\left(x+2\right)}^{2}}[/latex]
[/hidden-answer]
[latex]\frac{{x}^{3}+2{x}^{2}+4x}{{\left({x}^{2}+2x+9\right)}^{2}}[/latex]
[latex]\frac{{x}^{2}+25}{{\left({x}^{2}+3x+25\right)}^{2}}[/latex]
[hidden-answer a=”fs-id1165137430918″]
[latex]\frac{1}{{x}^{2}+3x+25}-\frac{3x}{{\left({x}^{2}+3x+25\right)}^{2}}[/latex]
[/hidden-answer]
[latex]\frac{2{x}^{3}+11x+7x+70}{{\left(2{x}^{2}+x+14\right)}^{2}}[/latex]
[latex]\frac{5x+2}{x{\left({x}^{2}+4\right)}^{2}}[/latex]
[hidden-answer a=”fs-id1165133112786″]
[latex]\frac{1}{8x}-\frac{x}{8\left({x}^{2}+4\right)}+\frac{10-x}{2{\left({x}^{2}+4\right)}^{2}}[/latex]
[/hidden-answer]
[latex]\frac{2x-9}{{\left({x}^{2}-x\right)}^{2}}[/latex]
[hidden-answer a=”fs-id1165137431316″]
[latex]-\frac{16}{x}-\frac{9}{{x}^{2}}+\frac{16}{x-1}-\frac{7}{{\left(x-1\right)}^{2}}[/latex]
[/hidden-answer]
[latex]\frac{5{x}^{3}-2x+1}{{\left({x}^{2}+2x\right)}^{2}}[/latex]
Extensions
For the following exercises, find the partial fraction expansion.
[latex]\frac{{x}^{2}+4}{{\left(x+1\right)}^{3}}[/latex]
[hidden-answer a=”fs-id1165134201601″]
[latex]\frac{1}{x+1}-\frac{2}{{\left(x+1\right)}^{2}}+\frac{5}{{\left(x+1\right)}^{3}}[/latex]
[/hidden-answer]
[latex]\frac{{x}^{3}-4{x}^{2}+5x+4}{{\left(x-2\right)}^{3}}[/latex]
For the following exercises, perform the operation and then find the partial fraction decomposition.
[latex]\frac{7}{x+8}+\frac{5}{x-2}-\frac{x-1}{{x}^{2}-6x-16}[/latex]
[hidden-answer a=”fs-id1165137550537″]
[latex]\frac{5}{x-2}-\frac{3}{10\left(x+2\right)}+\frac{7}{x+8}-\frac{7}{10\left(x-8\right)}[/latex]
[/hidden-answer]
[latex]\frac{1}{x-4}-\frac{3}{x+6}-\frac{2x+7}{{x}^{2}+2x-24}[/latex]
[latex]\frac{2x}{{x}^{2}-16}-\frac{1-2x}{{x}^{2}+6x+8}-\frac{x-5}{{x}^{2}-4x}[/latex]
[hidden-answer a=”fs-id1165137767858″]
[latex]-\frac{5}{4x}-\frac{5}{2\left(x+2\right)}+\frac{11}{2\left(x+4\right)}+\frac{5}{4\left(x+4\right)}[/latex]
[/hidden-answer]
Glossary
- partial fractions
- the individual fractions that make up the sum or difference of a rational expression before combining them into a simplified rational expression
- partial fraction decomposition
- the process of returning a simplified rational expression to its original form, a sum or difference of simpler rational expressions


5 Culture
Module Learning Objectives (MLO)
In this chapter, you will
- MLO 5.1 Identify foundational concepts and theories related to culture, identity, and dress. [CLO 1]
- MLO 5.2 Identify the factors that influence different cultural aspects of dress. [CLO 1]
- MLO 5.3 Identify the different cultural perspectives an individual can take to approach understanding the dress and identity of others. [CLO 1]
- MLO 5.4 Explain the role of dress in identity development. [CLO 1]
- MLO 5.5 Examine how marginalized communities in the United States use dress and appearance to negotiate their identities. [CLO 2]
- MLO 5.6 Deconstruct your own perspectives and approach to understanding the dress of others. [CLO 4]
What Is Culture?
Culture refers to aspects of human-made elements, including tools, dress, and media, in addition to values, attitudes, and norms. Dress is a significant part of almost every culture.
As more cultures have cross-cultural contact, people from begin to change aspects of their cultures by incorporating aspects of different cultures they come into contact with. This includes changes in dress. This process of cultural change is often referred to as cultural authentication. It should be noted though, that there is a long history of forced assimilation, especially for Native or Indigenous communities in North America (Little, 2018). That is, Native communities were forced to assimilate into European culture; Native people were not interested in incorporating European cultural elements.
Culture vs. Cultured
All people have culture. Culture is not something held only by society elites, such as only the wealthiest, most educated, or most sophisticated.
Culture is a system of learned behavior patterns that are characteristic of the members of a society (Hoebel, 1958). Note the emphasis on the learning of behaviors or ways of doing things. Culture is learned, and individuals learn culture through an ongoing process of socialization. Parents, families, schools, peers, and workplaces, for example, all socialize individuals to ways of doing things. We find differences across cultures in dress, language, food preferences, and other behaviors in part because these are learned behaviors—not knowledge that is innate, instinctual, or determined by genetic programming.
Cultural patterns are characteristic behaviors and often include a complex array of choices. A culture may afford more than one way of doing the same thing. Hence, diversity in behaviors may be found in some aspects of any culture
Culture is a complex whole that includes knowledge, belief, art, morals, law, customs, and any other capabilities and habits that members of a society acquire (Linton, 1936). Linton emphasized that culture is a complex whole—a network of behaviors related to all aspects of life. Dress is shaped by and reflects many characteristics in any culture, so dress is a complex map of cultural characteristics.
Culture includes both abstract and concrete components (Scupin, 1998).
Abstract components include the meanings of symbols, events, activities, or actions and how the meanings are created and selected. Concrete components include the forms of action, behavior, event, activity, or artifact. Dress may be a concrete object, such as a shirt, a pair of shoes, or a hairstyle, but those artifacts develop meanings in a culture. The fashion process, situations in which we use dress, and groups associated with wearing of types of dress all bring meaning to those artifacts.
Culture is mentifacts, what people know; sociofacts, what people do; and artifacts, what people make (Spradley, 1972):
Mentifacts: Ideas, Ideals, Values, Knowledge, and How We Know
Culture shapes how people think about things. Mentifacts are the ideas, values, and knowledge that shape how we see a culture and how we know or recognize patterns. This includes stereotypes that are held about groups of people who look certain ways. Note that “how we know” also refers to educational and media systems. Fashion magazines are part of these media systems; so are television and the Internet.
How people in a culture think and what they value are often reflected in dress. For example, in the United States, where we value the freedom to consume and material plenty, consumers tend to prefer large wardrobes (Sproles & Burns, 1994). In some European countries, the average consumer has a comparatively small wardrobe that often includes several high-quality, expensive items. These few items are worn over and over during the season in which they are fashionable.
Sociofacts: Characteristics of Social Organizations and How People Organize Themselves.
Sociofacts reflect how people behave in groups and social interactions. For example, police officers wear uniforms to indicate their occupations and rank within the police force. A store like Target has a dress code—red shirts and khaki pants—to help customers identify employees. Many individuals dress up or “clean up” when going to dinner at someone’s house; this small act shows respect to the hosts and indicates participation in a social event.
Sometimes patterns of dress in the larger society reflect how people organize themselves. In the United States today, socioeconomic status is usually only vaguely communicated in our clothing. One hundred years ago, the social class of everyone passing us on the street would have been made clear by their dress.
Artifacts: Things People Make and Tools and Processes for Making Them
Dress such as clothing, makeup, tattoos, and shoes are artifacts made by people. Artifacts reflect multiple aspects of a culture, such as mentifacts, sociofacts, and the technological knowledge of a culture that shapes manufacturing processes and materials used. For example, Gore-Tex fabric, now used in high-performance sportswear, was invented through the NASA space program.
Technological advancements and economic reality may exclude some clothing options. In ready-to-wear clothing that most people in the United States buy today, only simple seams, darts, and a few gathers comprise construction. Knits and pandex rather than intricate construction details accomplish, a tight fit. During the 1940s and earlier, however, ordinary clothing often had many tucks, darts, godets, and complex seams, for example. The handwork and sewing skill required for those designs is too expensive to produce today in assembly-line factories. Complex fit through construction also requires customized fitting that is too expensive, time-consuming, and difficult for most of us. We save that expense for business suits and special-occasion garments such as wedding dresses.
Society vs. Culture
Anthropologists disagree on how to distinguish the terms culture and society. We will use the terms fairly interchangeably. One definition of society is “a group of people living and working together in a systematic way” (Mead, 1934). An important implication of this definition is that society requires people to coordinate their actions with each other. Each individual cannot haphazardly do their own thing with no concern for others. With no traffic laws, for example, we would run into each other fairly frequently. Indeed, with no coordination of human effort, automobiles and roads would never have been invented. Dress would have no meaning; fashions and traditions in dress would not exist. We would have no idea who anyone might be on first meeting of them. Dress is a product of systematic human interaction, and it helps us to coordinate our interactions with others.
What Factors Influence Types of Dress Worn in a Culture?
Ruth Benedict (1959) drew an “arc of human potential” to indicate that every culture makes choices from among a wide array of possibilities for any form of behavior. Each culture, then, chooses different language sounds, foods, dress materials and designs, and other behaviors.
In any culture, the following factors shape choices for dress and other behaviors:
- climate and natural resources
- religion, ideology, ritual
- technology
- culture contact and diffusion of ideas
- social and political organization
- history
- aesthetic rules.
This short film features the Korowai tribe, whose members had no contact with anyone outside of their community until the 1970s. This isolation affected their dress in numerous ways.
To access a transcript for the video above, download this file: The Tree House People Korowai Video Transcript [DOC]
Principles of Cultural Perspective
Taking a cultural perspective on dress throughout the semester requires that we adopt specific some ways of thinking about people and the world:
Holistic approach
The meaning of dress can be understood only through the study of all aspects of a culture. Dress does not mean one single thing at a time. Many meanings and aspects of a culture are embedded in any example of dress. And sometimes these meanings are difficult to read. For example, why is it that in the United States, where we value individuality so much, so many students on college campuses wear jeans and T-shirts to classes?
Cultural relativism
Seek to understand dress as it has meaning to a society “insider.” What dress worn in nations outside your home country means to you, a visitor or tourist, is not necessarily related to what it means to people in that culture. We need to examine the characteristics of a culture and talk to people within that culture to find out what their dress means. Something as simple as color may have very different meanings in dress in different cultures. For example, red is a common funeral color in Zambia, Africa, and a wedding dress color in traditional China. What would it mean if someone wore a bright red dress to a close relative’s funeral or a bride wore a bright red dress to her wedding in the United States today?
Ethnocentrism
Ethnocentrism is judgment of people of other cultures by one’s own cultural standards and beliefs. Dress that is different from our own can be challenging to accept or appreciate.
Ethnocentricity is hard to avoid, for example, when members of other cultures or some groups within our own culture do things to the body that are harmful to the individuals. Acting to end these practices may be seen as important for humanity. It is equally important to recognize, however, that these practices may have deep meanings and roots in religious values or beauty standards previously considered good or “healthful.” Simply marching in and making such practices illegal will not necessarily end them. We must take great care when we attempt to change deeply held beliefs about these practices. Change is likely to be slow.
Sometimes it can be quite difficult to avoid ethnocentrism. Here are some examples that individuals may or may not view as harmful:
- corset wearing
- foot binding
- tanning
- web sites promoting anorexia
- female circumcision
- little people lengthening their legs
- embracing one’s transgender identity
- women exposing their breasts in public
- steroid use
- scarification, or the carving of permanent designs in the skin
- practicing bondage and discipline, dominance and submission, sadochism and masochism (BDSM) or kink such as wearing dog collars, chastity belts, or other restrictive garments.
Video Example:
Watch this short video about BDSM dress, identity, and women’s empowerment. The research presentation is titled “Paddles, Strap-Ons, Latex, and Leather: Negotiations of BDSM Women’s Dress, Embodiment, and Bodies in Motion through Spatial-Temporal Dynamics.”
The video highlights how these women embrace their BDSM identity when others sometimes view these practices as harmful.
Cultural Appropriation
Another term related to changing cultural aesthetics or norms is cultural appropriation. The concept of cultural appropriation is highly debated. For example, in a video that went viral, Amandla Stenberg discusses her opinions about white people adopting Black hairstyles (2015). Stenberg discusses how the adoption of hairstyles such as braids or cornrows by white people is wrong because when Black people wear these styles, they are viewed as “thugs” or “gangsters,” yet when white people adopt these styles they can be seen as “cool” or “edgy,” which reinforces longstanding racial hierarchies and stereotypes of white people having power. There are also arguments such as appreciation versus appropriation, where individuals will argue that their adoption of a particular style or aesthetic is not wrong; they merely appreciate that part of another culture (Brucculieri, 2018).
In the following film, about 28 minutes long, Denise Green of Cornell University delivers a compelling lecture on cultural appropriation and the Cowichan sweater (CTV News, 2015).
To access a transcript for the video above, download this file: What is the cost of not fitting in Video Transcript [DOC]
In this short film, Bethany Yellowtail, an Indigenous designer, discusses her experiences as a fashion designer and cultural appropriation.
To access a transcript for the video above, download this file: Indigenous Cultural Appropriation in Fashion with Bethany Yellowtail and James St. James Video Transcript [DOC]
Required reading
References
Benedict, R. (1959). Patterns of culture. Boston: Houston Mifflin.
Hoebel, E. A. (1958). Man in the primitive world: An introduction to anthropology (2nd edition). New York: McGraw-Hill.
Linton, R. (1936). The study of man: An introduction. New York: D. Appleton-Century.
Mead, G.H. (1934). Mind, self, and society (Ed. by Charles W. Morris). Chicago: University of Chicago Press.
Scupin, R. (1998). Cultural anthropology: A global perspective (3rd ed.). Upper Saddle River, NJ: Prentice-Hall.
Spradley, J. P. (1972). Culture and cognition: Rules, maps, and plans. San Francisco: Chandler Publishing.
Sproles, G. B., & Burns, L. D. (1994). Changing appearances. New York: Fairchild.
Culture Case Study
Step One: Become familiar with the case study.
- The case study attached below is a Word document and can be downloaded. It includes the task, evaluation, and template for the case study:
Step Two: Submit your complete assignment on Canvas.
- Format your document.
- Remember to check the submission against the rubric.


Checking Other Sources
Now watch this video:
This step may sometimes be the last one you do, but it could also be the first. The News tab is better than the main Google search for this step because it only shows real news sources. While not every source that’s included is perfectly reliable, they are all news outlets that really exist. Here is an example of how this works:
You can also use our custom news search, tiny,cc.news-search, which searches Canadian and international sources of reliable news.
By taking this step, you can be sure you get the whole story. Remember, all sources make mistakes sometimes, but reliable ones will correct them.
Looking at other sources can help you find out if the first place you saw something might have been leaving something out. This is also a good way of discovering any possible bias that might exist in any one source
You can also use this step to find out whether something agrees with what most experts on that topic think – what’s called the consensus view. While it’s generally good reporting to give both sides of a story, including views that experts agree aren’t right can result in spreading misinformation.
Wikipedia is a good place to find the consensus on a topic because each Wikipedia article represents the consensus of its editors. Just remember to watch for warning banners that suggest a specific article might not be reliable.
You can use our custom search https://mediasmarts.ca/break-fake/science-search to find the consensus on specialist topics like science and medicine.
If you want to know if another specialist source is reliable, Verify the Source.


States of Consciousness
Putting It Together: States of Consciousness
Learning Objectives
In this module, you learned to:
- describe consciousness and biological rhythms
- describe what happens to the brain and body during sleep
- explain how drugs affect consciousness
Remember that psychology is the study of the mind and behavior. Consider how your behavior is affected by varying states of consciousness—what happens if you don’t get enough sleep? How does your concentration and mood change? What happens to your conscious awareness when you take strong medications or drugs? In this module, you learned that varying states of consciousness, whether it be tiredness from jet lag, deep sleep, daydreaming, alcohol or drug use, hypnosis, or meditation, can all change physiological components in the mind as well as behavior. While we have control over some of these altered states, like meditation, there are sometimes variables outside of our control that lead to things like sleep disorders. Drug use often begins as a personal choice or something prescribed by a doctor, but many drugs have addictive qualities that make them hard to put down.
Substance abuse disorders can develop out of drug use, and those with drug addictions struggle with physical and psychological dependence on the drug. This is an especially important issue for us today because death by drug overdoses has been dramatically increasing over the past few years. Take a look at the interactive below to see the increasing prevalence of drug overdoses.
link tO Learning
Visit the NYTimes article, “You Draw It: Just How Bad Is the Drug Overdose Epidemic?” to take a look at current trends and the alarming amount of deaths caused by drug overdoses.
As you read in the article above, in 2015 there were 52,000 American deaths from all drug overdoses. Two thirds of them, 33,000, were from opioids, compared to 16,000 in 2010 and 4,000 in 1999. Death from opioid drug overdoses were nearly equal to the number of deaths from car crashes, with deaths from heroin alone accounting for more deaths than from gun violence. In 2016, deaths from overdoses increased over the previous year by 26% in Connecticut, 35% in Delaware, 39% in Maine, and 62% in Maryland. Nearly half of all opioid overdose deaths involve a prescription opioid.
The governor of Maryland declared a State of Emergency in March 2017 to combat the epidemic and CDC director Thomas Frieden has said that “America is awash in opioids; urgent action is critical.”
With this sudden, extreme increase in drug overdoses, psychologists and psychiatrists will continue to play an important role in researching, educating, and preventing substance abuse disorders. You can read more about what researchers are doing at the Addiction Connection or even find some practical tips on ways to prevent a drug overdose in this article from Psychology Today.
Licenses and Attributions (Click to expand)
CC licensed content, Original
- Provided by: Lumen Learning. License: CC BY: Attribution
CC licensed content, Shared previously
- Opioid Epidemic. Provided by: Wikipedia. Located at: https://en.wikipedia.org/wiki/Opioid_epidemic. License: CC BY-SA: Attribution-ShareAlike
- Generic temazepam 10 mg tablets. Provided by: Wikipedia. Located at: https://en.wikipedia.org/wiki/Temazepam#/media/File:Temazepam_10mg_tablets-1.jpg. License: CC BY-SA: Attribution-ShareAlike


44 Solder your own Circuits
Solderless breadboards are easy and fast while you are testing ideas and building a first prototype, but their failings become apparent and annoying very quickly. Jumper wire connections pull out easily. The fairly high and variable resistances of the pin-in-breadboard connections are a problem in sensitive analog circuits like bridges. Switch up to something better sooner, rather than later for best results.
Soldered Prototyping Boards
Your circuits will provide much better connections with lower resistances if you solder them up on a printed circuit board (PCB). You can start easily by using prototyping boards like this shield for the UNO, or various wings for the Feather with your components and hookup wire to make permanent connections. Permaproto boards make the transition for a solderless breadboard to a first prototype easy by maintaining the same geometry. Once you are ready, you can create your own designs with the connections built right into the board for production and high reliability.
PCB Design and Layout Software
http://fritzing.org/home/ provides an application for drawing user friendly schematics and layout drawings for soldered printed circuits or jumpers and breadboards. It is widely used in the maker community and provides the basis for many illustrations in this book.
Eagle
Download and install Eagle for free and follow the tutorials at Autodesk, Sparkfun or Adafruit
I installed the application in the default Applications directory, but have been putting libraries and script files in my Documents/eagle directory so they should be available even if I am using a different computer. (The Documents directory is replicated across all my machines through iCloud.)
KiCad
Try open source KiCad that enjoys corporate support from both Arduino and Raspberry Pi.
PCB Fabrication Options
https://pcbshopper.com/ allows you to review multiple options for fabrication and assembly.


8 Putting It Together
During earlier eras, the definition of art was aligned with craftsmanship and guilds, but as societies changed, so, too, did the meaning and purpose of art. Over time, art evolved beyond practical and religious functions and became an autonomous expression of the artist’s creative process and of the surrounding culture.
Aesthetics is concerned with how we perceptually engage in the changing and complex concepts of beauty and the sublime (Ocvirk, 6).
Exploring the definition of art is an act of critical thinking. Critical thinking is creative thinking, and the critical-thinking process often requires a belief in the question, rather than an expectation of hard truths or answers. Through active questioning, exploration, and trial and error, we uncover multiple valid perspectives.
Consider the example of the nkisi figures introduced at the beginning of this module. Recall how that misunderstanding of visual culture was representative of the larger confrontation and oppression of African societies by Europeans. Consider also, in the final example of video games, how the introduction of new media keeps alive the ongoing debate about what art is.
Works Cited
Ocvirk, Stinson, Wigg, Bone, Cayton. Art Fundamentals, Theory and Practice, 12 Edition. New York: McGraw Hill, 2013. Print.
Sayre, Henry. A World of Art, Sixth edition. Boston: Prentice Hall, 2010. Print.


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Introduction
1. What is Open Education?
2. Why OER?
3. Who Writes OER?
4. How to Go Open
5. OER Supports
6. Self-Check Quiz
7. Reflection and Wrap Up
8. How can OER advocate Anti-Racism?
9. Defining Open Pedagogy
10. Examples of Open Pedagogy
11. Open Pedagogy and Anti-Racism
12. Reflection and Wrap Up
13. Defining Anti-racism
14. How to Be an Anti-Racist
15. What is Anti-Racist Pedagogy?
16. Culturally Responsive Pedagogy
17. Self-Check Quiz
18. Reflection and Wrap Up
19. Anti-Racist OER Examples
20. Faculty Perspectives
21. Ways to Incorporate Anti-Racism in your Subject Area
22. Action Plan
23. Reflection
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17.6 Precipitation and Dissolution
Learning Objectives
- Write chemical equations and equilibrium expressions representing solubility equilibria
- Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations
The preservation of medical laboratory blood samples, mining of sea water for magnesium, formulation of over-the-counter medicines such as Milk of Magnesia and antacids, and treating the presence of hard water in your home’s water supply are just a few of the many tasks that involve controlling the equilibrium between a slightly soluble ionic solid and an aqueous solution of its ions.
In some cases, we want to prevent dissolution from occurring. Tooth decay, for example, occurs when the calcium hydroxylapatite, which has the formula Ca5(PO4)3(OH), in our teeth dissolves. The dissolution process is aided when bacteria in our mouths feast on the sugars in our diets to produce lactic acid, which reacts with the hydroxide ions in the calcium hydroxylapatite. Preventing the dissolution prevents the decay. On the other hand, sometimes we want a substance to dissolve. We want the calcium carbonate in a chewable antacid to dissolve because the CO32- ions produced in this process help soothe an upset stomach.
In this section, we will find out how we can control the dissolution of a slightly soluble ionic solid by the application of Le Châtelier’s principle. We will also learn how to use the equilibrium constant of the reaction to determine the concentration of ions present in a solution.
The Solubility Product Constant
Silver chloride is what’s known as a sparingly soluble ionic solid (Figure 17.6a). Recall from the solubility rules in an earlier chapter that halides of Ag+ are not normally soluble. However, when we add an excess of solid AgCl to water, it dissolves to a small extent and produces a mixture consisting of a very dilute solution of Ag+ and Cl– ions in equilibrium with undissolved silver chloride:
This equilibrium, like other equilibria, is dynamic; some of the solid AgCl continues to dissolve, but at the same time, Ag+ and Cl– ions in the solution combine to produce an equal amount of the solid. At equilibrium, the opposing processes have equal rates.
The equilibrium constant for the equilibrium between a slightly soluble ionic solid and a solution of its ions is called the solubility product (Ksp) of the solid. Recall from the chapter on solutions and colloids that we use an ion’s concentration as an approximation of its activity in a dilute solution. For silver chloride, at equilibrium:
When looking at dissolution reactions such as this, the solid is listed as a reactant, whereas the ions are listed as products. The solubility product constant, as with every equilibrium constant expression, is written as the product of the concentrations of each of the ions, raised to the power of their stoichiometric coefficients. Here, the solubility product constant is equal to Ag+ and Cl– when a solution of silver chloride is in equilibrium with undissolved AgCl. There is no denominator representing the reactants in this equilibrium expression since the reactant is a pure solid; therefore [AgCl] does not appear in the expression for Ksp.
Some common solubility products are listed in Table 17.6a according to their Ksp values. Each of these equilibrium constants is much smaller than 1 because the compounds listed are only slightly soluble. A small Ksp represents a system in which the equilibrium lies to the left, so that relatively few hydrated ions would be present in a saturated solution.
| Substance | Ksp at 25 °C |
|---|---|
| CuCl | 1.2 × 10–6 |
| CuBr | 6.27 × 10–9 |
| AgI | 1.5 × 10–16 |
| PbS | 7 × 10–29 |
| Al(OH)3 | 2 × 10–32 |
| Fe(OH)3 | 4 × 10–38 |
Example 17.6a
Writing Equations and Solubility Products
Write the ionic equation for the dissolution and the solubility product expression for each of the following slightly soluble ionic compounds:
- AgI, silver iodide, a solid with antiseptic properties
- CaCO3, calcium carbonate, the active ingredient in many over-the-counter chewable antacids
- Mg(OH)2, magnesium hydroxide, the active ingredient in Milk of Magnesia
- Mg(NH4)PO4, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium
- Ca5(PO4)3OH, the mineral apatite, a source of phosphate for fertilizers
(Hint: When determining how to break (d) and (e) up into ions, refer to the list of polyatomic ions in the section on chemical nomenclature.)
Solution
- [latex]\text{AgI}(s)\;{\rightleftharpoons}\;\text{Ag}^{+}(aq)\;+\;\text{I}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ag}^{+}][\text{I}^{-}][/latex]
- [latex]\text{CaCO}_3(s)\;{\rightleftharpoons}\;\text{Ca}^{2+}(aq)\;+\;\text{CO}_3^{\;\;2-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ca}^{2+}][\text{CO}_3^{\;\;2-}][/latex]
- [latex]\text{Mg(OH)}_2(s)\;{\rightleftharpoons}\;\text{Mg}^{2+}(aq)\;+\;2\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^{-}]^2[/latex]
- [latex]\text{Mg(NH}_4)\text{PO}_4(s)\;{\rightleftharpoons}\;\text{Mg}^{2+}(aq)\;+\;\text{NH}_4^{\;\;+}(aq)\;+\;\text{PO}_4^{\;\;3-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Mg}^{2+}][\text{NH}_4^{\;\;+}][\text{PO}_4^{\;\;3-}][/latex]
- [latex]\text{Ca}_5(\text{PO}_4)3\text{OH}(s)\;{\rightleftharpoons}\;5\text{Ca}^{2+}(aq)\;+\;3\text{PO}_4^{\;\;3-}(aq)\;+\;\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ca}^{2+}]^5[\text{PO}_4^{\;\;3-}]^3[\text{OH}^{-}][/latex]
Exercise 17.6a
Write the ionic equation for the dissolution and the solubility product for each of the following slightly soluble compounds:
- BaSO4
- Ag2SO4
- Al(OH)3
- Pb(OH)Cl
Check Your Answer[1]
Now we will extend the discussion of Ksp and show how the solubility product constant is determined from the solubility of its ions, as well as how Ksp can be used to determine the molar solubility of a substance.
Ksp and Solubility
Recall that the definition of solubility is the maximum possible concentration of a solute in a solution at a given temperature and pressure. We can determine the solubility product of a slightly soluble solid from that measure of its solubility at a given temperature and pressure, provided that the only significant reaction that occurs when the solid dissolves is its dissociation into solvated ions, that is, the only equilibrium involved is:
In this case, we calculate the solubility product by taking the solid’s solubility expressed in units of moles per litre (mol/L), known as its molar solubility.
Example 17.6b
Calculation of Ksp from Equilibrium Concentrations
Fluorite, CaF2, is a slightly soluble solid that dissolves according to the equation:
The concentration of Ca2+ in a saturated solution of CaF2 is 2.15 × 10–4M; therefore, that of F– is 4.30 × 10–4M, that is, twice the concentration of Ca2+. What is the solubility product of fluorite?
Solution
First, write out the Ksp expression, then substitute in concentrations and solve for Ksp:
A saturated solution is a solution at equilibrium with the solid. Thus:
As with other equilibrium constants, we do not include units with Ksp.
Exercise 17.6b
Check Your Learning Exercise (Text Version)
In a saturated solution that is in contact with solid Mg(OH)2, the concentration of Mg2+ is 1.31 × 10–4M. What is the solubility product for Mg(OH)2?
MgOH2(s) ↔ Mg2+(aq) + 2OH–(aq)
- 1.31
- 4.99 x 10-6
- 1.31 x 10-4
- 8.99 x 10-12
Check Your Answer[2]
Source: “Exercise 17.6b” is adapted from “Example 15.1-2” in General Chemistry 1 & 2, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson, licensed under CC BY 4.0.
Example 17.6c
Determination of Molar Solubility from Ksp
The Ksp of copper(I) bromide, CuBr, is 6.3 × 10–9. Calculate the molar solubility of copper bromide.
Solution
The solubility product constant of copper(I) bromide is 6.3 × 10–9.
The reaction is:
First, write out the solubility product equilibrium constant expression:
Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the CuBr column empty as it is a solid and does not contribute to the Ksp:
At equilibrium:
Therefore, the molar solubility of CuBr is 7.9 × 10–5M.
Exercise 17.6c
The Ksp of AgI is 1.5 × 10–16. Calculate the molar solubility of silver iodide.
Check Your Answer[3]
Example 17.6d
Determination of Molar Solubility from Ksp, Part II
The Ksp of calcium hydroxide, Ca(OH)2, is 1.3 × 10–6. Calculate the molar solubility of calcium hydroxide.
Solution
The solubility product constant of calcium hydroxide is 1.3 × 10–6.
The reaction is:
First, write out the solubility product equilibrium constant expression:
Create an ICE table, leaving the Ca(OH)2 column empty as it is a solid and does not contribute to the Ksp:
At equilibrium:
Therefore, the molar solubility of Ca(OH)2 is 1.3 × 10–2 M.
Using Barium Sulfate for Medical Imaging
Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the Ksp of barium sulfate is 1.1 × 10–10, very little of it dissolves as it coats the lining of the patient’s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure 17.6b).
Further diagnostic testing can be done using barium sulfate and fluoroscopy. In fluoroscopy, a continuous X-ray is passed through the body so the doctor can monitor, on a TV or computer screen, the barium sulfate’s movement as it passes through the digestive tract. Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn’s disease, and ulcers in addition to other conditions.
Visit Barium X-Rays (Upper and Lower GI) at Johns Hopkins Medicine for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose.
Predicting Precipitation
The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:
We can establish this equilibrium either by adding solid calcium carbonate to water or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. If we add calcium carbonate to water, the solid will dissolve until the concentrations are such that the value of the reaction quotient (Q = [Ca2+][CO32-]) is equal to the solubility product (Ksp = 8.7 × 10–9). If we mix a solution of calcium nitrate, which contains Ca2+ ions, with a solution of sodium carbonate, which contains CO32- ions, the slightly soluble ionic solid CaCO3 will precipitate, provided that the concentrations of Ca2+ and CO32- ions are such that Q is greater than Ksp for the mixture. The reaction shifts to the left and the concentrations of the ions are reduced by formation of the solid until the value of Q equals Ksp. A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that Q is less than Ksp, then the solution is not saturated and no precipitate will form.
We can compare numerical values of Q with Ksp to predict whether precipitation will occur, as Example 5 shows. (Note: Since all forms of equilibrium constants are temperature dependent, we will assume a room temperature environment going forward in this chapter unless a different temperature value is explicitly specified.)
Example 17.6e
Precipitation of Mg(OH)2
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of lime, Ca(OH)2, a readily available inexpensive source of OH– ion:
The concentration of Mg2+(aq) in sea water is 0.0537 M. Will Mg(OH)2 precipitate when enough Ca(OH)2 is added to give a [OH–] of 0.0010 M?
Solution
This problem asks whether the reaction:
shifts to the left and forms solid Mg(OH)2 when [Mg2+] = 0.0537 M and [OH–] = 0.0010 M. The reaction shifts to the left if Q is greater than Ksp. Calculation of the reaction quotient under these conditions is shown here:
Because Q is greater than Ksp (Q = 5.4 × 10–8 is larger than Ksp = 8.9 × 10–12), we can expect the reaction to shift to the left and form solid magnesium hydroxide. Mg(OH)2(s) forms until the concentrations of magnesium ion and hydroxide ion are reduced sufficiently so that the value of Q is equal to Ksp.
Example 17.6f
Precipitation of AgCl upon Mixing Solutions
Does silver chloride precipitate when equal volumes of a 2.0 × 10–4–M solution of AgNO3 and a 2.0 × 10–4–M solution of NaCl are mixed?
(Note: The solution also contains Na+ and NO3– ions, but when referring to solubility rules, one can see that sodium nitrate is very soluble and cannot form a precipitate.)
Solution
The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:
The solubility product is 1.6 × 10–10 (see Appendix K).
AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO3 and NaCl is greater than Ksp. The volume doubles when we mix equal volumes of AgNO3 and NaCl solutions, so each concentration is reduced to half its initial value. Consequently, immediately upon mixing, [Ag+] and [Cl–] are both equal to:
The reaction quotient, Q, is momentarily greater than Ksp for AgCl, so a supersaturated solution is formed:
Since supersaturated solutions are unstable, AgCl will precipitate from the mixture until the solution returns to equilibrium, with Q equal to Ksp.
Exercise 17.6d
Will KClO4 precipitate when 20 mL of a 0.050-M solution of K+ is added to 80 mL of a 0.50-M solution of ClO4–? (Remember to calculate the new concentration of each ion after mixing the solutions before plugging into the reaction quotient expression.)
Check Your Answer[4]
In the previous two examples, we have seen that Mg(OH)2 or AgCl precipitate when Q is greater than Ksp. In general, when a solution of a soluble salt of the Mm+ ion is mixed with a solution of a soluble salt of the Xn– ion, the solid, MpXq precipitates if the value of Q for the mixture of Mm+ and Xn– is greater than Ksp for MpXq. Thus, if we know the concentration of one of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will assume that precipitation begins when the reaction quotient becomes equal to the solubility product constant.
Example 17.6g
Precipitation of Calcium Oxalate
Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, C2O42-, for this purpose (Figure 17.6c). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC2O4·H2O (which also contains water bound in the solid). The concentration of Ca2+ in a sample of blood serum is 2.2 × 10–3M. What concentration of C2O42- ion must be established before CaC2O4·H2O begins to precipitate?
Solution
The equilibrium expression is:
For this reaction:
(see Appendix K)
CaC2O4 does not appear in this expression because it is a solid. Water does not appear because it is the solvent.
Solid CaC2O4 does not begin to form until Q equals Ksp. Because we know Ksp and [Ca2+], we can solve for the concentration of C2O42- that is necessary to produce the first trace of solid:
A concentration of C2O42- = 8.9 × 10–6 M is necessary to initiate the precipitation of CaC2O4 under these conditions.
Exercise 17.6e
If a solution contains 0.0020 mol of CrO42– per litre, what concentration of Ag+ ion must be reached by adding solid AgNO3 before Ag2CrO4 begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.
Check Your Answer[5]
It is sometimes useful to know the concentration of an ion that remains in solution after precipitation. We can use the solubility product for this calculation too: If we know the value of Ksp and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same type as that in Example 17.6g—calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation.
Example 17.6h
Concentrations Following Precipitation
Clothing washed in water that has a manganese [Mn2+(aq)] concentration exceeding 0.1 mg/L (1.8 × 10–6M) may be stained by the manganese upon oxidation, but the amount of Mn2+ in the water can be reduced by adding a base. If a person doing laundry wishes to add a buffer to keep the pH high enough to precipitate the manganese as the hydroxide, Mn(OH)2, what pH is required to keep [Mn2+] equal to 1.8 × 10–6M?
Solution
The dissolution of Mn(OH)2 is described by the equation:
We need to calculate the concentration of OH– when the concentration of Mn2+ is 1.8 × 10–6M. From that, we calculate the pH. At equilibrium:
or
so
Now we calculate the pH from the pOH:
[latex]\text{pH} = 14.00\;-\;\text{pOH} = 14.00\;-\;3.80 = 10.52[/latex]
If the person doing laundry adds a base, such as the sodium silicate (Na4SiO4) in some detergents, to the wash water until the pH is raised to 10.52, the manganese ion will be reduced to a concentration of 1.8 × 10–6M; at that concentration or less, the ion will not stain clothing.
Exercise 17.6f
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of Ca(OH)2. The concentration of Mg2+(aq) in sea water is 5.37 × 10–2M. Calculate the pH at which [Mg2+] is diminished to 1.0 × 10–5M by the addition of Ca(OH)2.
Check Your Answer[6]
Due to their light sensitivity, mixtures of silver halides are used in fibre optics for medical lasers, in photochromic eyeglass lenses (glass lenses that automatically darken when exposed to sunlight), and—before the advent of digital photography—in photographic film. Even though AgCl (Ksp = 1.6 × 10–10), AgBr (Ksp = 5.0 × 10–13), and AgI (Ksp = 1.5 × 10–16) are each quite insoluble, we cannot prepare a homogeneous solid mixture of them by adding Ag+ to a solution of Cl–, Br–, and I–; essentially all of the AgI will precipitate before any of the other solid halides form because of its smaller value for Ksp. However, we can prepare a homogeneous mixture of the solids by slowly adding a solution of Cl–, Br–, and I– to a solution of Ag+.
When two anions form slightly soluble compounds with the same cation, or when two cations form slightly soluble compounds with the same anion, the less soluble compound (usually, the compound with the smaller Ksp) generally precipitates first when we add a precipitating agent to a solution containing both anions (or both cations). When the Ksp values of the two compounds differ by two orders of magnitude or more (e.g., 10–2 vs. 10–4), almost all of the less soluble compound precipitates before any of the more soluble one does. This is an example of selective precipitation, where a reagent is added to a solution of dissolved ions causing one of the ions to precipitate out before the rest.
The Role of Precipitation in Wastewater Treatment
Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Figure 17.6d). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions (PO42-) are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.
One common way to remove phosphates from water is by the addition of calcium hydroxide, known as lime, Ca(OH)2. The lime is converted into calcium carbonate, a strong base, in the water. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca5(PO4)3(OH), which then precipitates out of the solution:
The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO2 in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.
View Phosphorus Removal – Wastewater for more information on how phosphorus is removed from wastewater.
Common Ion Effect
As we saw when we discussed buffer solutions, the hydronium ion concentration of an aqueous solution of acetic acid decreases when the strong electrolyte sodium acetate, NaCH3CO2, is added. We can explain this effect using Le Châtelier’s principle. The addition of acetate ions causes the equilibrium to shift to the left, decreasing the concentration of H3O+ to compensate for the increased acetate ion concentration. This increases the concentration of CH3CO2H:
Because sodium acetate and acetic acid have the acetate ion in common, the influence on the equilibrium is called the common ion effect.
The common ion effect can also have a direct effect on solubility equilibria. Suppose we are looking at the reaction where silver iodide is dissolved:
If we were to add potassium iodide (KI) to this solution, we would be adding a substance that shares a common ion with silver iodide. Le Châtelier’s principle tells us that when a change is made to a system at equilibrium, the reaction will shift to counteract that change. In this example, there would be an excess of iodide ions, so the reaction would shift toward the left, causing more silver iodide to precipitate out of solution.
Example 17.6i
Common Ion Effect
Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr2). The Ksp of CdS is 1.0 × 10–28.
Solution
The first thing you should notice is that the cadmium sulfide is dissolved in a solution that contains cadmium ions. We need to use an ICE table to set up this problem and include the CdBr2 concentration as a contributor of cadmium ions:
We can solve this equation using the quadratic formula, but we can also make an assumption to make this calculation much simpler. Since the Ksp value is so small compared with the cadmium concentration, we can assume that the change between the initial concentration and the equilibrium concentration is negligible, so that 0.010 + x ~ 0.010. Going back to our Ksp expression, we would now get:
Therefore, the molar solubility of CdS in this solution is 1.0 × 10–26M.
Exercise 17.6g
Calculate the molar solubility of aluminum hydroxide, Al(OH)3, in a 0.015-M solution of aluminum nitrate, Al(NO3)3. The Ksp of Al(OH)3 is 2 × 10–32.
Check Your Answer[7]
Key Equations
- [latex]\text{M}_p\text{X}_q(s)\;{\leftrightharpoons}\;p\text{M}^{\text{m}+}(aq)\;+\;q\text{X}^{\text{n}-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{M}^{\text{m}+}]^p[\text{X}^{\text{n}-}]^q[/latex]
Attribution & References
-
- [latex]\text{BaSO}_4(s)\;{\rightleftharpoons}\;\text{Ba}^{2+}(aq)\;+\;\text{SO}_4^{\;\;2-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ba}^{2+}][\text{SO}_4^{\;\;2-}][/latex];
- [latex]\text{Ag}_2\text{SO}_4(s)\;{\rightleftharpoons}\;2\text{Ag}^{+}(aq)\;+\;\text{SO}_4^{\;\;2-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ag}^{+}]^2[\text{SO}_4^{\;\;2-}][/latex];
- [latex]\text{Al(OH)}_3(s)\;{\rightleftharpoons}\;\text{Al}^{2+}(aq)\;+\;3\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Al}^{3+}][\text{OH}^{-}]^3[/latex];
- [latex]\text{Pb(OH)Cl}(s)\;{\rightleftharpoons}\;\text{Pb}^{2+}(aq)\;+\;\text{OH}^{-}(aq)\;+\;\text{Cl}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Pb}^{2+}][\text{OH}^{-}][\text{Cl}^{-}][/latex]
- d) 8.99 x 10-12 ↵
- 1.2 × 10–8 M ↵
- No, Q = 4.0 × 10–3, which is less than Ksp = 1.05 × 10–2 ↵
- 4.5 × 10–9M ↵
- 10.97 ↵
- 1 × 10–10M ↵
equilibrium constant for the dissolution of a slightly soluble electrolyte
solubility of a compound expressed in units of moles per liter (mol/L)
process in which ions are separated using differences in their solubility with a given precipitating reagent
effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base