Find the cubic polynomial given linear reminders after division by quadratic polynomials? 
A cubic polynomial gives remainders $(13x-2)$ and $(-1-7x)$ when divide by
  $x^2-x-3$ and $x^2-2x+5$ respectively. Find the polynomial.

I have written this as:
$P(x)=(x^2-x-3)Q(x)+(13x-2)$
$P(x)=(x^2-2x+5)G(x)+(-1-7x)$
and
$P(x)=ax^3+bx^2+cx+d$
The first method I thought about is factoring $(x^2-x-3)$ and $(x^2-2x+5)$ so that I could find roots and thereby $P(root)$ would equal the reminder for the given value of root. As the two quadratic equations would give four roots i.e. four values I can substitute, I would be able to calculate all the four constant $a$, $b$, $c$ and $d$. However, the first polynomial has very ugly roots and the second one has no real roots so I guess I should find another way. What do you suggest?

Using the Extended Euclidean Algorithm as implemented in this answer, we get
$$
\begin{array}{r}
&&1&x+6&(x-8)/53\\\hline
1&0&1&-x-6&(x^2-2x+5)/53\\
0&1&-1&x+7&(-x^2+x+3)/53\\
x^2-x-3&x^2-2x+5&x-8&53&0\\
\end{array}
$$
That is,
$$
(x+7)(x^2-2x+5)-(x+6)(x^2-x-3)=53\tag{1}
$$
We can now use the Chinese Remainder Theorem. $(1)$ tells us that
$$
\frac{x+7}{53}\,(x^2-2x+5)\equiv
\left\{\begin{array}{}
0&\pmod{x^2-2x+5}\\
1&\pmod{x^2-x-3}
\end{array}\right.\tag{2}
$$
$$
-\frac{x+6}{53}\,(x^2-x-3)\equiv
\left\{\begin{array}{}
1&\pmod{x^2-2x+5}\\
0&\pmod{x^2-x-3}
\end{array}\right.\tag{3}
$$
Add $13x-2$ times $(2)$ and $-1-7x$ times $(3)$ to get
$$
\frac1{53}\left(20x^4+99x^3-185x^2+338x-88\right)\tag{4}
$$
taking the remainder of $(4)$ mod $(x^2-x-3)(x^2-2x+5)$ gives
$$
3x^3-5x^2+6x+4\tag{5}
$$
