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

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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>.
 
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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==See Also==
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{{AHSME 50p box|year=1953|num-b=35|num-a=37}}
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{{MAA Notice}}

Latest revision as of 00:41, 4 February 2020

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 $\boxed{\textbf{(C) }36}$.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 35 		Followed by
Problem 37
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions


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