Question

Quotient + remainder I need help I would have 20 points

Answer 1

We want to compute

$\dfrac{-5x^4+4x^3-20x^2+15x-16}{-x^2-3}$

The goal is to write it in the following form:

$-5x^4+4x^3-20x^2+15x-16=Q(x)(-x^2-3)+R(x)$

First step: How many "copies" of $-x^2$ does $-5x^4$ contain? We can write $-5x^4=(-x^2)(5x^2)$, so the answer is $5x^2$ copies.

If we distribute $5x^2$ to $-x^2-3$, we get

$5x^2(-x^2-3)=-5x^4-15x^2$

If we were done, then this product would have matched the numerator at the start. But it doesn't; there are still some terms that need to vanish. In particular, we still have to account for

$(-5x^4+4x^3-20x^2+15x-16)-(-5x^4-15x^2)=4x^3-5x^2+15x-16$

Second step: How many copies of $-x^2$ can we find in $4x^3$? Well, $4x^3=(-4x)(-x^2)$, so the answer is $-4x$. Then

$(5x^2-4x)(-x^2-3)=-5x^4+4x^3-15x^2+12x$

but this still doesn't match the original numerator. We still have to deal with

$(-5x^4+4x^3-20x^2+15x-16)-(5x^2-4x)(-x^2-3)=-5x^2+3x-16$

Third step: How many copies of $-x^2$ can we get out of $-5x^2$? $-5x^2=5(-x^2)$. Then

$(5x^2-4x+5)(-x^2-3)=-5x^4+4x^3-20x^2+12x-15$

This also doesn't match the original numerator. We have a difference between what we have and what we want of

$(-5x^4+4x^3-20x^2+15x-16)-(5x^2-4x+5)(-x^2-3)=3x-1$

But we also can't divide $3x$ into chunks of $-x^2$, and we can't continue the division algorithm.

Our final answer would be

$\dfrac{-5x^4+4x^3-20x^2+15x-16}{-x^2-3}=5x^2-4x+5+\dfrac{3x-1}{-x^2-3}$

For the second problem, you should find

$\dfrac{-15x^3-13x+4}{5x^2+2}=-3x+\dfrac{-7x+4}{5x^2+2}$