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Difference between revisions of "1976 AHSME Problems/Problem 19"

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We know that <math>p(1)=3</math> and <math>p(3)=5</math>. We write <math>p(x)</math> as <math>p(x)=q(x)(x-1)(x-3)+r(x)</math>, where <math>r(x)=ax+b</math>. Plugging in <math>x=1</math>, we get <math>a+b=3</math>. Plugging in <math>x=3</math>, we know <math>3a+b=5</math>. We have a systems of equations, where we can solve that <math>a=1</math> and <math>b=2</math>. So, our answer is <math>1(x)+(2)=x+2\Rightarrow \textbf{(B)}</math>. ~MathJams
 
We know that <math>p(1)=3</math> and <math>p(3)=5</math>. We write <math>p(x)</math> as <math>p(x)=q(x)(x-1)(x-3)+r(x)</math>, where <math>r(x)=ax+b</math>. Plugging in <math>x=1</math>, we get <math>a+b=3</math>. Plugging in <math>x=3</math>, we know <math>3a+b=5</math>. We have a systems of equations, where we can solve that <math>a=1</math> and <math>b=2</math>. So, our answer is <math>1(x)+(2)=x+2\Rightarrow \textbf{(B)}</math>. ~MathJams
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==1976 AHSME Problems==
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{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}}

Latest revision as of 19:50, 12 July 2020

A polynomial $p(x)$ has remainder three when divided by $x-1$ and remainder five when divided by $x-3$. The remainder when $p(x)$ is divided by $(x-1)(x-3)$ is

$\textbf{(A) }x-2\qquad \textbf{(B) }x+2\qquad \textbf{(C) }2\qquad \textbf{(D) }8\qquad \textbf{(E) }15$


Solution

We know that $p(1)=3$ and $p(3)=5$. We write $p(x)$ as $p(x)=q(x)(x-1)(x-3)+r(x)$, where $r(x)=ax+b$. Plugging in $x=1$, we get $a+b=3$. Plugging in $x=3$, we know $3a+b=5$. We have a systems of equations, where we can solve that $a=1$ and $b=2$. So, our answer is $1(x)+(2)=x+2\Rightarrow \textbf{(B)}$. ~MathJams

1976 AHSME Problems

1976 AHSME (Problems • Answer Key • Resources)
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