1970 AHSME Problems/Problem 11

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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 $\fbox{E}$.

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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