0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $x^3+3x^2-x-2$? $\endgroup$ – Kieren MacMillan Oct 1 '14 at 2:14
  • $\begingroup$ I could multiply it by x, but I have to use the remainder theorem to solve the question. $\endgroup$ – user3175999 Oct 1 '14 at 2:20
0
$\begingroup$

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 $b$.

$\endgroup$
  • $\begingroup$ so at the end you would have two unknowns and two equations. Then you would isolate a variable in both equations, set the equations equal and then you could solve. $\endgroup$ – user3175999 Oct 3 '14 at 0:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.