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Difference between revisions of "1953 AHSME Problems/Problem 36"

(Created page with "==Problem== Determine <math>m</math> so that <math>4x^2-6x+m</math> is divisible by <math>x-3</math>. The obtained value, <math>m</math>, is an exact divisor of: <math>\te...")
 
(→‎Solution)
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==Solution==
 
==Solution==
  
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Since the given expression is a quadratic, the factored form would be <math>(x-3)(4x+y)</math>, where <math>y</math> is a value such that <math>-12x+yx=-6x</math> and <math>-3(y)=m</math>. The only number that fits the first equation is <math>y=6</math>, so <math>m=18</math>. The only choice that is a multiple of 18 is <math>\boxed{\textbf{(C) }36}</math>.
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Since the given expression is a quadratic, the factored form would be <math>(x-3)(4x+y)</math>, where <math>y</math> is a value such that <math>-12x+yx=-6x</math> and <math>-3(y)=m</math>. The only number that fits the first equation is <math>y=6</math>, so <math>m=-18</math>. The only choice that is a multiple of 18 is <math>\boxed{\textbf{(C) }36}</math>.

Revision as of 12:56, 18 February 2019

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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