# A Question Regarding Remainder Theorem

What is the remainder when $x^3 + 3x^2 - x - 2$ is divided by $(x+3)(x+5)$? You have to solve this using the remainder theorem, which states: If $f(x)$ is divided by $(x-p)$, giving a quotient $g(x)$ and a remainder $r$ then $r = f(p)$.

I expanded $(x+3)(x+5)$ to try and find $p$. $$(x+3)(x+5) = x^2 + 8x + 15 = 8x + x^2 + 15$$ $$p = -\frac{x^2+15}{8}$$

When I tried to solve for $f(p)$, I got a term of order six, which is $x^6$, I knew this is wrong. Can someone point out my mistake and give me the right answer.

• You've got $x^2+8x+15$, and you're trying to divide $x^3+3x^2-x-2$ by it. One great way to do that is to eliminate the largest power of $x$ first, then try to eliminate the second largest, etc. How might you change $x^2+8x+15$ so that it could be used to eliminate the $x^3$ from $x^3+3x^2-x-2$? – Kieren MacMillan Oct 1 '14 at 2:14
• I could multiply it by x, but I have to use the remainder theorem to solve the question. – user3175999 Oct 1 '14 at 2:20

Since the divisor has degree $2$, the remainder has degree $1$. Write it as $r(x)=ax+b$: $$x^3 + 3x^2 - x - 2 = q(x)(x+3)(x+5)+ax+b$$ Plug $x=-3$ and $x=-5$ to find $a$ and $b$.