1960 AHSME Problems/Problem 1

Problem

If $2$ is a solution (root) of $x^3+hx+10=0$ , then $h$ equals:

$\textbf{(A) }10\qquad \textbf{(B) }9 \qquad \textbf{(C) }2\qquad \textbf{(D) }-2\qquad \textbf{(E) }-9$

Substitute $2$ for $x$ . We are given that this equation is true. Thus,

$2^3+2h+10 =0$

$18+2h=0$

$2h=-18$

$h=-9$

Thus, the answer is $\boxed{\textbf{(E) }-9}$ .

1960 AHSC (Problems • Answer Key • Resources)
Preceded by 1959 AHSME	Followed by Problem 2
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