How to compute remainder of division of $P(x)$ by $x^2 -3x+2$? 
The remainder of division of $P(x)$ by $x^2−1$ is $2x+1$, and the remainder of division of the same polynomial by $x^2−4$ is $x+4$. Compute the remainder of division of $P(x)$ by $x^2−3x+2$.

I will translate these into math equations
$$P(x) = (x^2-1)Q(x)+ 2x+1$$
$$P(x) = (x^2 -4)R(x)+x+4$$
And let 
$$f(x) = P(x) $$
We're asked to find the remainder when this polynomial is divided by $x^2 -3x+2$. So, there are two equations, which is why I'm confused with what to use in the equation $f(x) = P(x)$. What am I missing here?
Regards
 A: $P(x)=(x^2-1)Q(x)+2x+1;$
$P(x)=(x^2-4)R(x)+x+4;$
$P(x)=(x-1)(x-2)S(x)+ax+b.$
1)$P(1)=2(1)+1=a+b;$
2)$P(2)=2+4=2a+b;$
$3=a+b$; and $6=2a+b;$
$a=3$; $b=0;$
Remainder: $ax+b=3x.$
A: HINT:
$$\frac{P(x)}{x^2-1}=Q(x)+\frac{2x+1}{x^2-1}\implies\frac{P(x)}{x-1}=[(x+1)Q(x)+2]+\frac 3{x-1}\\\frac{P(x)}{x^2-4}=R(x)+\frac{x+4}{x^2-4}\implies\frac{P(x)}{x-2}=[(x+2)R(x)+1]+\frac6{x-2}\\\boxed{\frac{P(x)}{x-2}-\frac{P(x)}{x-1}=\frac{(x-1)P(x)-(x-2)P(x)}{x^2-3x+2}=\frac{P(x)}{x^2-3x+2}}$$
A: You want to find $r(x)$ where the degree of $r$ is less than 2 in the following equation:
$$P(x) = (x^2-3x+2)S(x) + r(x) = (x-1)(x-2)S(x) + r(x)$$
You also have the two formulas you wrote above (representing the polynomial remainders you are given). If you plug in $x=1$ and $x=2$, you have two different values of $x$ that make the $S(x)$ term zero. So, $P(1) = r(1)$ and $P(2) = r(2)$.
From your first equation, you also know:
$$P(1) = (1^2-1)Q(1)+2(1)+1 = 3$$
From your second equation, you have:
$$P(2) = (2^2-4)R(2)+(2)+4 = 6$$
Since $r(x) = ax+b$ (since your quotient has degree 2, your remainder will have a maximum degree of 1)
You have $r(1)=3=a+b$ and $r(2) = 6 = 2a+b$.
This gives $a=3, b=0$.
Plugging in gives:
$$P(x) = (x^2-3x+2)S(x)+3x$$
