Question

If (ax+2)(bx+7) = 15x^2 + cx +14 for all values of x, and a+b =8, what are two possible values of C ?

Answer 1

$(ax+2)(bx+7)=abx^2+(7a+2b)x+14=15x^2+cx+14$

For the two quadratics to match, the coefficients of corresponding terms must be equal, so that

$\begin{cases}ab=15\\7a+2b=c\end{cases}$

From the given options, we are only allowing $c$ to be an integer, which means $a,b$ must also be integers. Then we have eight possible choices for $a,b$:

1, 15

3, 5

5, 3

15, 1

plus the same combinations with negative signs on both numbers. But since the choices for $c$ are also positive, we can ignore the negative versions.

If $(a,b)=(1,15)$, then $7a+2b=37$.

If $(a,b)=(3,5)$, then $7a+2b=31$.

If $(a,b)=(5,3)$, then $7a+2b=41$.

If $(a,b)=(15,1)$, then $7a+2b=107$.

So the answer is C.