Does the ACT and/or SAT cover polynomial long division (for example: problems in which you might use synthetic division)?
@butterflylemon You’re never required to use synthetic division since you could just use polynomial long division.
Someone else could correct me if I’m wrong (as it’s been 4+ years since I’ve taken the SAT/ACT), but I don’t recall ever having to do polynomial long division on the SAT or ACT. One reason is probably because dividing polynomials takes a bit of time. Another reason is there is usually a simpler approach for a given question.
This is not an SAT or ACT question, but suppose you were asked to find the product of the roots of the polynomial x^3 + 9x^2 + 14x - 24. You could notice that 1 is a root, then divide by x-1 to yield a quadratic, then find the roots of the quadratic. But why waste time doing that?
Long and synthetic division of polynomials are in the Core Curriculum.
Obviously you can always do long instead of synthetic division on the SAT because method is not tested.
I have not yet seen a question where the best solution involves long division of a cubic, for example, but I think it is not impossible that such questions may turn up in the future.
I saw an October PSAT question in the non-calc section something like (numbers changed):
(x^3+4x^2+3x)-(2x^2+9x)=(x+s)(x+t)(x+u) what is s+t+u?
But this is not really factoring a cubic since one of the factors is x. However, the question could be slightly altered for the SAT to make it into a real cubic factoring question.
On the other hand, Question 24 on the calculator section of SAT Official Practice Test 4 reads
f(x)=2x^3+6x^2+4x
g(x)=x^2+3x+2
Which of the following is divisible by 2x+3?
A. f(x)+g(x)
B. f(x)+3g(x) etc.
However, the best method here is probably not long division but…CAS!
quote-(2x^2+9x)=(x+s)(x+t)(x+u) what is s+t+u?
[/quote]
@Plotinus Even if we slightly altered it (E.g. (x^3 + 4x^2 + 3x + 17) - …), I wouldn’t consider this a question where you have to factor the cubic, because you don’t need to factor it.
I agree here, with four answer choices, it would take a lot of effort adding and dividing polynomials. With my non-CAS calculator, I would probably compute f(-3/2) + g(-3/2) etc.
@Plotinus actually nvm what I said, the trick is to notice that f(x) = 2x g(x), and f(x) + 3g(x) = (2x+3)g(x) so the answer is B.
So my best guess is, while there are many problems on polynomials and factoring/roots/etc., I highly doubt any of them will require you to do the divisions or spend a lot of time, since many of these problems can be solved several ways.
@MITer94 If it took YOU two looks to come up with the trick, I rest my case that the easiest solution for most students is going to be CAS:
define f(x) = 2x^3+6x^2+4x ENTER
define g(x)=x^2+3x+2 ENTER
f(x)+ 3g(x) ENTER – The calculator returns (2x+3)((x^2+3x+2)
With CAS and good data entry skills, the problem takes less than 30 sections and no thinking.
Bad problem, bad.
@Plotinus Yeah, I still agree that this isn’t a good problem especially on a multiple choice test where CAS capabilities can be exploited.
The first problem you posted can be solved in the same way (with a CAS calculator) with no thinking provided that it returns the answer as a product of factors.
The first question was in the no calculator section. The second should have been there too, but wasn’t.
Ah okay. Didn’t see the “non-calc” part.