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When $f(x)$ is divided by $x-1$ and $x+2$, the remainders are $4$ and $-2$ respectively. Find the remainder when $f(x)$ is divided by $x^2+x-2$.

Please help. The answer is $2x+2$.

I tried to understand the link between the first sentence and the second sentence but all I could make out was $f(1)=4$ ($4$ being the remainder) and $f(-2)=-2$.

I also broke down $x^2+x-2$ into $(x-1)(x+2)$ but after that, I didn't know how to continue. It seemed like there was a link as both have $x-1$ and $x+2$. I tried to attempt it myself for quite a long time but I couldn't make out anything. I'm not sure what I made out was even on the correct track though.

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Division of a polynomial $A(x)$ by $B(x)$ means finding another two polynomials $Q(x)$ (quotient), $R(x)$ (remainder) with $\deg(R(x))<\deg(B(x))$ ($\deg$ denotes the highest power of $x$ that is present in the polynomial) such that: $$A(x)=B(x)Q(x)+R(x)$$

So by definition: if you divide $f(x)$ by $x-1$ and obtain remainder $4$, this says $f(x)=(x-1)Q(x)+4$ for some polynomial $Q(x)$. This gives $f(1)=4$. Similarly, $f(-2)=-2$.

$$f(x)=(x^2+x-2)Q(x)+ax+b=(x+2)(x-1)Q(x)+ax+b$$

for some $a,b\in\Bbb R$, polynomial $Q(x)$ (we wrote $ax+b$ because as I said $\deg(R(x))<\deg(B(x))$, so the remainder has degree less than $2$). You know $f(1)=4,\,f(-2)=-2$.

$f(1)=a+b=4$ and $f(-2)=-2a+b=-2$. It's a system of two equations with two variables. You should be able to solve it.

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Use the Chinese Remainder theorem (Wikipedia link) in the ring of polynomials.

You'll want to look at the polynomials $g=\frac{1}{3}(x+2)$ and $h=-\frac{1}{3}(x-1)$, which satisfy $$\begin{align*} g&\equiv 1\bmod (x-1) & h&\equiv 0\bmod(x-1)\\ g&\equiv 0\bmod (x+2) & h&\equiv 1\bmod (x+2) \end{align*}$$ Thus, for any $a$ and $b$, $$\begin{align*} ag+bh&\equiv a\bmod (x-1)\\ ag+bh&\equiv b\bmod (x+2) \end{align*}$$ In particular, the polynomial $$k=4g-2h=(\tfrac{4}{3}x+\tfrac{8}{3})+(\tfrac{2}{3}x-\tfrac{2}{3})=2x+2$$ has the property that $$\begin{align*} k&\equiv \hphantom{-}4\bmod (x-1)\\ k&\equiv -2\bmod (x+2) \end{align*}$$ The Chinese remainder theorem then guarantees that $f\equiv k\bmod (x^2+x-2)$ for any $f$ such that $f(1)=4$ and $f(-2)=-2$.

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HINT:

As $x^2+x-2=(x+2)(x-1),$

Let $f(x)=g(x)(x^2+x-2)+A(x+2)+B(x-1)$

$\implies f(x)\equiv A(x+2)\pmod{x-1}\implies A(-1+2)=6$

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