1961 AHSME Problems/Problem 22

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 $\boxed{\textbf{(C)}}$ .

Video Solution

https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=2515 - AMBRIGGS

1961 AHSC (Problems • Answer Key • Resources)
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1961 AHSME Problems/Problem 22

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