+99 When a polynomial \(p(x)\) is divided by \(x+1,\) the remainder is \(5\). When \(p(x)\) is divided by \(x+5,\) the remainder is \(-7.\) Find the remainder when \(p(x)\) is divided by \((x+1)(x+5).\)
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!
