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question

What is the remainder when $x^{32}+71x-62$ is divided by $15x^{2}-1$?

My thoughts:

I know that if it was to be divided by something like $3x^2-x$, the remainder would be 1 power less than the divisor. But in this scenario we have a gap between the powers, to clarify, it is $15x^2$ vs -1 and one has a power of two while the other has a power of 0 on the variable $x$.

thus I do not know how to apply the remainder theorem on this problem and would request help

My question is that are you able to set the remainder to be (ax+b)?

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    Assuming you mean as polynomials over the rational numbers? – 2017-01-24
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    Remainder theorem. – 2017-01-24

2 Answers 2

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$$x^{32}+71x-62=(15x^2-1)q(x)+ax+b$$

Now make $x=\frac{1}{\sqrt{15}}$ and $x=-\frac{1}{\sqrt{15}}$ and find $a$ and $b$, and the remainder will be:

$$r(x)=ax+b$$

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    My question is that are you able to set the remainder to be (ax+b) – 2017-01-24
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    Hi @JohnRawls, in fact we are! You already know that remainder must have a degree less than $2$ and then we can write it as $ax+b$. If that is not the case, after solve the problem like I did, you will naturally find $a=0$ if that is the case. Is it clear? – 2017-01-24
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Since $x^2\equiv \frac{-1}{15}\pmod{15x^2-1}$, you have that:

$$x^{32}\equiv \left(\frac{-1}{15}\right)^{16}=\frac{1}{15^{16}}\pmod{15x^2-1}$$

So:

$$x^{32}+71x-62\equiv 71x-\left(62-\frac{1}{15^{16}}\right)\pmod{15x^2-1}$$

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    How did $15^{-16}$ turn into $15^{-8}$? – 2017-01-24
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    Fixed. Stupid error. – 2017-01-24