The elephant in the testing room


Hemant Mehta, "The Friendly Atheist", is also a math teacher. This is what he found on one of the tests he was grading this week. The ol' Elephant Excuse. Pretty crafty. So how does a responsible educator of young minds respond to such a stunt? The answer is after the jump ...


If you're going to throw a Hail Mary Pachyderm on your final exam, you damn well best get your artwork correct.


  1. I can’t do that problem because I’ve apparently forgotten how to do algebra. I had a feeling this would happen.

    1. I strongly doubt that this one it real. Both pens are written in identical handwriting. There’s only a first name (and it doesn’t appear that the last name has been blotted out). The scoring system for the exam as written makes no sense either.

    1. @ “That elephant has a tail…”

      I believe an appendage in that location would signify the elephant is male.

  2. X^2 +8x +7 factorises to (x+7)(x+1) so we times the right side top and bottom by x+1, giving ((5x+17)-3(x+1))/(x^2+8x+7). this simplifies to 2x+14/(x^2+8x+7). we know the bottom and top both have a factor of (x+7) so we cancel, giving 2/(x+1)

    I think.

  3. Haha. Excellent excuse. Theres a website of these I think, can’t remember it.

    @joeposts: x^2 + 8x + 7 factorises to (x+1)(x+7). You then convert it to partial fractions.

    1. Well, you could simplify it â€” put it on a common denominator then add up the top. That gives 2(x+7)/(x+1)(x+7), whereupon the (x+7) cancels out, unless x is -7, giving 2/(x+1). After that, it depends on what the beginning of the question said.

      Of course, the other question is whether a student might not be expected to have somewhat better drawing skill and hand-writing at that level.

  4. The problem presented to the student is to simplify the expression, not to “solve” it. The instructor is hoping that the student will recognize that the trinomial that is the denominator of the first algebraic fraction has, as one of its factors, the binomial that is the denominator of the second algebraic fraction. The task of finding a common denominator hinges upon that recognition.

  5. the question may not need “=0” at the end. We don’t know what the question is. It may have been a “simplify this” type of question. in that case, the answer is

  6. Don’t worry about solving for x, just simplify. If I remember what I’m doing it should reduce to (x+7)^2.

  7. I think the idea is to simplify the expression.

    Notice that the denominator of the first term is

    x^2+8x+7 = (x+1)(x+7)

    (x+7) is the common factor, so multiply the term after the minus sign by (x+1)/(x+1) (by 1) so that it can all have the same denominator.

    nominator is now
    (suspicious how that x+7 appeared there)


    canceling x+7 out of the nominator and denominator leaves


    By the way this can’t ever equal zero, hibob42. The elephant is the big divide by zero error you would get if you tried it.

    Also try to avoid x=-1 in the above equation for the same reason.

    I miss algebra.

    1. Nicely done, hhype. Math is fun because through a logical process, you can take something complicated and make it into something much easier to deal with. Very satisfying.

      Its use in physics and engineering is even more fun to me, because you set up a problem that describes something in real life, apply some mathematical rules, and suddenly something pops out that tells you exactly what’s going on. Doing derivations was always the coolest part of physics lecture courses. (Of course labs were more fun.)

      Assume a spherical elephant…

      And sabik — depending on the school, kids can do algebra at relatively young ages. I took algebra at age 12, and I think it’s possible even younger. I’d buy the drawing and handwriting as that of a 12-year-old — and the sense of humor.

    2. try to avoid x=-1 in the above equation

      Also remember to avoid x=-7. You probably shouldn’t cancel out the (x+7) without saying that you have to avoid x=-7 (and possibly dealing with the x=-7 case separately, though in this case that just means noting that there’s a discontinuity).

    3. Technically you are right, x =/= 0 but, if you take a limit of that equation, then as x approaches infinity, the equation approaches zero.

  8. Gah! I got as far as (2x + 14)/((x + 1)(x + 7)) but somehow didn’t see I could factor out the 2, though now it’s painfully obvious. Thanks for the fun guys, it’s been a while.

  9. I would have said “I’ll give you partial credit if you tell me where you get your acid.”

    I did actually write a paper in college on William Blake while on LSD (seemed logical at the time), and it was only slightly less ridiculous than this.

  10. Ahhhh… reminds me of my school days. I once got a single pity point (out of 15) for handing in an otherwise empty Turbo Pascal exam with a drawing of a croissant.

    Mind you, my art work was better.

  11. I’d say I have forgotten all my algebra, but I never really learned it in the first place.

    Somehow I have survived into middle adulthood.

  12. The answer is 2/(x+1). Now to figure out how to show my work!

    [(5x+7)/(x+1)(x+7)]- {3(x+1)/[(x+1)(x+7)]}
    [(5x+7)/(x+1)(x+7)]- {(3x+3)/[(x+1)(x+7)]}
    [(5x+7 -3x-3)]/[(x+1)(x+7)]
    Never had to type out the answers when I had algebra 50yrs ago…oy!

  13. HEY DUDES I FIGURED IT OUT: 2/(x+1). don’t worry everyone, all speculation can be put to rest, because i have solved the question, i am so smrt.

    1. Apparently reading comments before posting is harder than algebra.

      I look forward to the next BB intelligence challenge.

  14. I’m in the same boat as Arborman, apparently.

    Hmm, if you can’t at least make a reasonable pass at algebra I don’t think you an really claim to have achieved adulthood.

  15. I feel like everyone doing the math is getting an A+ at missing the point. Bonus points for not knowing fun when it stares at them from an elephant drawing.

  16. Okay… this teacher teaches at my high school, Neuqua Valley High School in Naperville IL. It’s a small world after all.

  17. As much as I find this to be fairly funny, as a teacher I’m also a wee bit concerned. It is a pretty serious ethics breach to post a student’s work online without their permission. At any rate, each jurisdiction has its own rules, and I sincerely hope that Hemant Mehta isn’t violating her code of conduct. Some of you may laugh, but it is amazing what teachers can get in trouble for these days.


      Not to mention copyright infringement…

  18. I hope nobody gets in trouble!!! besides the name is not visible on the test.
    Writing the name of the school in the comments is not that discreet though.
    At any rate, a big elephant would have fit in my test yesterday better than one answer I made up!
    Ciao Eva

  19. The answer is 2/(x+1) EXCEPT at x=-7, where it is undefined.

    Remember, when you start from


    and cancel the (x+7) in the numerator and denominator, you’re possibly dividing the numerator and denominator by 0, effectively trying to cancel out a 0/0. This happens for the above expression when x+7=0, or x=-7.

  20. @#55 but at x=-7 you are dividing 0 by 0, so why would it be undefined?
    Sure, x/0 is undefined (infinity), but x=0 in this case.
    Long time no algebra for me, but seems ok to divide 0 by any number, znd getting 0
    what is the rule?

  21. I swear I had a physics teacher in Wisconsin who would give you +1 for any cutely drawn animal. I ended up with a d- both semesters and the ability to draw a variety of cute bunny rabbits.

  22. I will never solve a problem like this. I’d pay people like you to do it for me then on sell the answer to others for an administration fee.

  23. Whats hilarious is that this is obviously a JOKE and you guys either solve the math prolem or point out other errors in logic or credibility. EVERY SINGLE ONE OF YOU ABOVE ME IS A BLATANTLY JOYLESS NERD!!!!!!!
    I thought it was cute. The truth never stands in the way of a good story.

  24. I repeated my year for this equation: x^5+x+1

    After in my university my teacher canÂ´t do this too…

  25. @#57: 0/0 is undefined — or more specifically, indeterminate. You mentioned the “rule” that zero divided by anything is zero; but we also know the rule that anything divided by zero is undefined. And don’t forget that anything divided by itself is one. So should 0/0 be undefined, zero, or one?

    To resolve this, we must have an explicit definition of division; usually, we take it to be the inverse of multiplication. “6/2 = ?” is essentially “? x 2 = 6”; the number in question is obviously 3. Now, 0/3 = 0 since 0 is the number we multiply by 3 to get 0, but 3/0 is undefined, since there’s no number that works here: any number times 0 is 0, not 3! But if we look at 0/0, we have the question of what number times 0 gives 0 — and that, clearly, could be anything. Unless we have proper context (see l’Hospital’s Rule), we cannot determine a value for 0/0.

    (My apologies if this has already been addressed; I don’t know how long the moderation lag may be.)

  26. if Hemant Mehta fails at teaching a student how to solve something so remarkably easy, tail less elephant or not, he should resign as math teacher. Maybe he’d do well as teacher of Natural Sciences…

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