1953 AHSME Problems/Problem 36

Problem

Determine $m$ so that $4x^2-6x+m$ is divisible by $x-3$ . The obtained value, $m$ , is an exact divisor of:

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 64$

Solution

Since the given expression is a quadratic, the factored form would be $(x-3)(4x+y)$ , where $y$ is a value such that $-12x+yx=-6x$ and $-3(y)=m$ . The only number that fits the first equation is $y=6$ , so $m=-18$ . The only choice that is a multiple of 18 is $\boxed{\textbf{(C) }36}$ .

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1953 AHSME Problems/Problem 36

Problem

Solution