Difference between revisions of "1960 AHSME Problems/Problem 1"

Problem

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

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

Solution

Substitute $2$ for $x$. We are given that this equation is true. Solving for $h$ gives $h=-9$. The answer is $\boxed{\textbf{(E)}}$.