1. ## Remainder theorem problem

Im currently stuck on this question:

When P(z) is divided by z+1 the remainder is -8 and when divided by z-3 the remainder is 4. Find the remainder when P(z) is divided by (z-3)(z+1)

I know that P(-1) = -8 and P(3) = 4 but i have no clue what to do from here.

Thanks for any help

2. We have $P(z)=(z+1)Q(z)-8.$

Let $Q(z)=(z-3)R(z)+k.$

Thus $4=P(3)=4Q(3)-8=4k-8$ $\implies$ $k=3.$

Hence $P(z)=(z+1)(z-3)R(z)+3(z+1)-8=(z+1)(z-3)R(z)+3z-5$ so the remainder when $P(z)$ is divided by $(z+1)(z-3)$ is $3z-5.$

3. Im not sure what youve done here

Thus

And whats R(z)?

4. I am basically using the result that given any polynomials $f(x),g(x)$ with $g(x)\not\equiv0,$ there exist polynomials $q(x),r(x)$ with either $r(x)\equiv0$ or $\deg r(x)<\deg g(x)$ such that $f(x)=q(x)g(x)+r(x).$ (This applies to polynomials over a field such as the field of rationals, reals or complex numbers.) The polynomial $r(x)$ is the remainder when $f(x)$ is divided by $g(x).$

Actually, I didn’t do a great job in my post above. Let me re-do the proof in a clearer way.

Given polynomials $P(z)$ and $(z-1)(z+3)$ we have that there exist a polynomial $R(z)$ and constants $k,h$ such that

$P(z)\ =\ (z-1)(z+3)R(z)+kx+h$

As $P(-1)=-8$ and $P(3)=4$ we have $-k+h=-8$ and $3k+h=4.$ Solving the two simultaneous equations should yield $k=3,\,h=-5.$ Hence the remainder is $3x-5.$

5. Ahh i see :d

Thanks very much for your help