Find the remainder when $f(x)$ is divided by $x^2+x-2$ 
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. 
 A: 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.
A: 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 $f(-2)=-2$.
A: HINT:
As $x^2+x-2=(x+2)(x-1),$
Let $f(x)=g(x)(x^2+x-2)+A(x+2)+B(x-1)$
$\implies f(x)\equiv A(x+2)\pmod{x-1}\implies A(-1+2)=6$
