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When the polynomial $f(x)$ is divided by $(x-2)$, the remainder is $-8$. Given that $(x+2)$ is a factor of $f(x)$, find the remainder when $f(x)$ is divided by $(x^2-4)$. I know the answer is $-2x-4$ but I do not know the working behind it?

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    Hint: write $f(x)=(x^2-4)q(x)+r(x)$. where $r(x)=ax+b$. All you need is two values for $r(x)$. – 2017-01-19
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    I actually got up to that step but I do not know how to proceed from there. – 2017-01-19
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    Wait... I think I have solved it... Thank you! – 2017-01-19
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    @lulu That's the hard way. Instead pull out the factor $\,x+2\,$ in common with $f$ and the modulus $\,x^2-4\,$ and then you don't need to solve any equations - see my answer. – 2017-01-19

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An easy way: use $\,{ab\bmod ac = a(b\bmod c)} = $ mod Distributive Law, $ $ i.e.

$\,\ x\!+\!2\mid f\,\Rightarrow\, f\bmod x^2\!-\!4\,$ $=\, (x\!+\!2)\Bigg[\dfrac{f}{x\!+\!2}\bmod x\!-\!\color{#c00}2\Bigg]$ $ =\, \underbrace{(x\!+\!2)\left[\dfrac{f(\color{#c00}{2})}{\color{#c00}2\!+\!2}\right] =\, (x\!+\!2)\left[-2\right]}_{\large f\bmod x-\color{#c00}2\,\ =\,\ f(\color{#c00}{2})\,\,\ =\,\ -8}$

Remark $\ $ As I explain in the linked answer, using this distributive law often greatly simplifies computations by reducing the modulus, e.g. above it reduces the modulus to have degree one, viz. $\,x\!-\!2,\,$ and such mods are easy: $\,g(x)\bmod x\!-2\!\ =\ g(2)\,$ by the Polynomial Remainder Theorem. Further this optimization may eliminate the need to use CRT = Chinese Remiander Theorem, so elminates possibly complicated computation of modular inverses.

See this answer for a more complex and general example of the above method for polynomials.

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    Good point, much easier. – 2017-01-19
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We know that when $f(x)$ is divided by $x-2$ the remainder is $-8$, so $$f(x) = (x-2)q(x) - 8$$ for some polynomial $q(x)$.

We also know that $x+2$ is a factor of of $f(x)$. This means that when we plug in $x=-2$ to $f(x)$ we should get $0$. So $$0 = (-2 -2)q(-2) - 8$$ $$8 = -4q(-2)$$ $$-2 = q(-2)$$ This last result tells us that when $q(x)$ is divided by $x+2$, the remainder is $-2$, so for some polynomial $u(x)$ we have $$q(x)=(x+2)u(x) - 2$$ Combining these results, we now know that $$f(x) = (x-2)((x+2)u(x) - 2) - 8$$ $$f(x) = (x^2-4)u(x) -2(x-2) - 8$$ $$f(x) = (x^2 - 4)u(x) -2x - 4$$ and now you can see the remainder when $f(x)$ is divided by $x^2-4$.