Can anyone help out with a SAT math question from the new format. It’s in practice test 4 calculator section number 25:
F(x)= 2x^3+6x^2+4x
G(x)= x^2+3x+2
The polynomials f(x) and g(x) are defined above. Which of the following polynomials is divisible by 2x+3
A) h(x)= f(x) + g(x)
B) p(x)= f(x) + 3g(x)
C) r(x)= 2f(x) + 3g(x)
D) s(x)= 3f(x) + 2g(x)
The explanation in the book made sense but I was wondering if anyone knew a simpler way to solve it. Thanks
Since the polynomial we seek is divisible by 2x+3 you know that x=-1.5 is a zero of that polynomial. Find f(-1.5) and g(-1.5) and then look to see which combination of f and g gives a value of zero.
Hah! I just saw that @DrSteve already took care of this in another thread…
I think it’s more interesting that we’re both spending Saturday night answering SAT questions. 
You could also choose a value for x( like x=2) and evaluate f and g at 2 and check the answer choices to see if it’s divisible by 2x + 3 (in this case, if it’s divisible by 2(2) +3=7)
@AlphaDragon that doesn’t always work in general, mostly since “divisibility” (when referring to integers) doesn’t mean the same thing as divisibility (when referring to polynomials).
Formally, a polynomial P(x) is divisible by a polynomial Q(x) iff there exists a polynomial R(x) such that P(x) = Q(x)*R(x), where the coefficients of P, Q, R range over some ring (oftentimes the integers, but we can also use the set of reals).
For example, it is invalid to state that p(x) = f(x) + 3g(x) is divisible by x+4 since when x = 1, p(1) = 30 and x+4 = 5, and 30 is divisible by 5.
However, it is true that if a polynomial P(x) is divisible by Q(x) over the integers, then if d is an integer such that Q(d) ≠ 0, then the integer P(d) is divisible by the integer Q(d).
So are you saying that @AlphaDragon 's method can generate false-positives? Cause that can happen any time you use the make-up-numbers approach. When it does, you just try again with a new number. I like alpha dragon 's solution.
@pckeller I think I meant to say, from a purely logical point of view (ignoring answer choices and everything), the implication doesn’t always hold.
From an SAT perspective, I agree that AlphaDragon’s solution is good, provided that you find an integer that works for exactly one and fails for the other three. I forgot about the whole “false-positives” thing.
The only possible complication is if they were to throw some question involving divisibility over the reals (instead of over integers), for example, one that involved knowing that 2x^2 + 3x + 1 was divisible by 2x + 2 over the reals, since any integer you try won’t work. But I don’t really see the SAT doing something like that.