The expression $4x^3+21x-6$ has the same remainder when divided by $x-a$ or by $x+b$, where $a$ not equals to $b$. Find the value of $a^2+b^2-ab$. I have the answer but I don't know how to work through the question.
Find the value after Remainder Theorem
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$\begingroup$
polynomials
real-numbers
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2I think it should be $a$ not equals to $-b$. – 2017-01-19
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0Or precisely $|a|\ne|b|$ – 2017-01-19
2 Answers
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Let $f(x)=4x^3+21x-6$
$\displaystyle f(a)=f(-b) \implies 4(a^3+b^3)+21(a+b)=0 \implies a^2-ab+b^2=\frac{-21}{4}$
Since $f(x)$ is a polynomial and it's remainder is $f(a)$ when divided by $x-a$
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Given same remainder, by Polynomial Remainder Theorem we have,
$\displaystyle\ \ \ \ \ \ f(a)=f(-b)$
$\displaystyle\Rightarrow 4a^3+21a-6=-4b^3-21b-6$
$\displaystyle\Rightarrow 4(a^3+b^3)+21(a+b)=0$
$\displaystyle\Rightarrow \frac{(a^3+b^3)}{(a+b)}=a^2+b^2-ab=\frac{-21}{4}$
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1We should assume that $a+b \not=0$ – 2017-01-19