1970 AHSME Problems/Problem 11

Problem

If two factors of $2x^3-hx+k$ are $x+2$ and $x-1$ , the value of $|2h-3k|$ is

$\text{(A) } 4\quad \text{(B) } 3\quad \text{(C) } 2\quad \text{(D) } 1\quad \text{(E) } 0$

Solution

From the Remainder Theorem, we have $2(-2)^3 - h(-2) + k = 0$ and $2(1)^3 - h(1) + k = 0$ . Simplifying both of those equations gives $-16 + 2h + k = 0$ and $2 - h + k = 0$ . Since $k = 16 - 2h$ and $k = h - 2$ , we set those equal to get:

$16 - 2h = h - 2$

$3h = 18$

$h = 6$

This gives $k = 4$ when substituting back into either of the two equations, and $|2h - 3k| = 0$ , which is answer $\fbox{E}$ .

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1970 AHSME Problems/Problem 11

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