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. Thus,

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

$18+2h=0$

$2h=-18$

$h=-9$

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

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
1959 AHSME
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions