1988 AHSME Problems/Problem 5

Problem

If $b$ and $c$ are constants and $(x + 2)(x + b) = x^2 + cx + 6$ , then $c$ is

$\textbf{(A)}\ -5\qquad \textbf{(B)}\ -3\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 5$

Solution

We first start out by expanding the left side of the equation, $(x+2)(x+b)=x^{2}+bx+2x+2b=x^2+(2+b)x+2b=x^2+cx+6$ . We know the constant terms have to be equal so we have $2b=6$ , so $b=3$ . Plugging $b=3$ back in yields $x^2+(2+3)x+6=x^2+cx+6$ . Thus, $c=5 \implies \boxed{\text{E}}$ .

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1988 AHSME Problems/Problem 5

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