+0  
 
0
33
1
avatar+84 

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 \((x+1)(x+5).\)

 Sep 11, 2022
 #1
avatar
+1

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!

 Sep 12, 2022

17 Online Users