pls help me

Hi.

If we have a polynomial $p(x)$ and we divided it by another polynomial, say, $h(x)$ to give us a "Quotient polynomial" $Q(x)$, and a remainder, $r(x)$. Then we can do the following:

$\dfrac{p(x)}{h(x)}=Q(x)+\dfrac{r(x)}{h(x)}$                   (*)

Now, let's multiply (*) by h(x):

$p(x)=Q(x)h(x)+r(x)$

Theorem:   $deg(r(x)) < deg(h(x))$  (That is, if h(x) is quadratic, then r(x) is at most linear).

Let's now use the givens:

$p(x)=Q_1(x)(x+1)+5 \implies p(-1)=Q_1(-1)(-1+1)+r(-1) \implies p(-1)=r(-1)=5$

Similarly, $p(-5)=r(-5)=-7$

Now, 

$p(x)=Q(x)(x+1)(x+5)+r(x)$, we want to find r(x).

Well, notice: $h(x)=(x+1)(x+5)=x^2+6x+5$; that is, h(x) is a quadratic, so r(x) is at most linear 

Set: $r(x)=Ax+B$ where A and B are constants to be found.

Now we know p(-1)=r(-1)=5, let's substitute that:

$p(-1)=A(-1)+B=5 \implies B-A=5$                         (1)

Moreover, p(-5)=r(-5)=-7:

$p(-5)=-5A+B=-7 \implies B-5A=-7$                    (2)

Subtract equation (2) from (1):

$(B-A)-(B-5A)=5-(-7) \\ \iff 4A=12 \implies A=3 \implies B=8$

So: $r(x)=3x+8$ which is the desired remainder. I hope this helps!