Question about the polynomial remainder theorem Given that $f(x) = x^3 - x^2 - ax - b$ has a factor $x - 3$ but leaves a remainder of $13x - 11$ when divided by $x + 4$, find $a$ and $b$.

I get that in order for $x-3$ to be a factor then according to the remainder theorem, to find remainder of $f(x)$ divided by $x-a$, calculate $f(a)$ which in this case $f(3)=0$.

So, $f(3)= 18-3a-b=0$

$3a+b=18$ is the first equation found.

It starts to get tricky when dividing $f(x)$ by $x+4$ and finding values of $a$ and $b$ such that the remainder = $13x-11$ to find our second equation according to the answer , $a=5$ , $b=3$

Could someone please help me with this. This is supposed to be grade 10 math and i've got a degree in mathematics yet I still can't get it.
$g(x)=f(x)-(13x-11)$ has a factor of $x+4$ so 
$$g(-4)=f(-4)-(13(-4)-11)=0$$
$$f(-4)=-64-16+4a-b=13(-4)-11$$
$$
\implies\left\{ 
\begin{array}{c}
4a-y=17 \\ 
3a+b=18  
\end{array}
\right.\implies 
\left\{ 
\begin{array}{c}
a=5 \\ 
b=3 
\end{array}
\right.
$$
$g(x)=f(x)-(13x-11)$ has a factor of $x+4$ so 
$$g(-4)=f(-4)-(13(-4)-11)=0$$
$$f(-4)=-64-16+4a-b=13(-4)-11$$
$$
\implies\left\{ 
\begin{array}{c}
4a-y=17 \\ 
3a+b=18  
\end{array}
\right.\implies 
\left\{ 
\begin{array}{c}
a=5 \\ 
b=3 
\end{array}
\right.
$$