1961 AHSME Problems/Problem 22

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Problem

If $3x^3-9x^2+kx-12$ is divisible by $x-3$, then it is also divisible by:

$\textbf{(A)}\ 3x^2-x+4\qquad \textbf{(B)}\ 3x^2-4\qquad \textbf{(C)}\ 3x^2+4\qquad \textbf{(D)}\ 3x-4 \qquad \textbf{(E)}\ 3x+4$

Solution

If $3x^3-9x^2+kx-12$ is divisible by $x-3$, then by the Remainder Theorem, plugging in $3$ in the cubic results in $0$. \[3 \cdot 3^3 - 9 \cdot 3^2 + 3k - 12 = 0\] Combine like terms to get \[3k - 12 = 0\] Thus, $k=4$. The cubic is $3x^3-9x^2+4x-12$, and it can be factored (by grouping or synthetic division) into \[(3x^2+4)(x-3)\] Thus, the answer is $\boxed{\textbf{(C)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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