# Quotient + remainder I need help I would have 20 points

## Answers

We want to compute[tex]\dfrac{-5x^4+4x^3-20x^2+15x-16}{-x^2-3}[/tex]The goal is to write it in the following form:[tex]-5x^4+4x^3-20x^2+15x-16=Q(x)(-x^2-3)+R(x)[/tex]First step: How many "copies" of [tex]-x^2[/tex] does [tex]-5x^4[/tex] contain? We can write [tex]-5x^4=(-x^2)(5x^2)[/tex], so the answer is [tex]5x^2[/tex] copies.If we distribute [tex]5x^2[/tex] to [tex]-x^2-3[/tex], we get[tex]5x^2(-x^2-3)=-5x^4-15x^2[/tex]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[tex](-5x^4+4x^3-20x^2+15x-16)-(-5x^4-15x^2)=4x^3-5x^2+15x-16[/tex]Second step: How many copies of [tex]-x^2[/tex] can we find in [tex]4x^3[/tex]? Well, [tex]4x^3=(-4x)(-x^2)[/tex], so the answer is [tex]-4x[/tex]. Then[tex](5x^2-4x)(-x^2-3)=-5x^4+4x^3-15x^2+12x[/tex]but this still doesn't match the original numerator. We still have to deal with[tex](-5x^4+4x^3-20x^2+15x-16)-(5x^2-4x)(-x^2-3)=-5x^2+3x-16[/tex]Third step: How many copies of [tex]-x^2[/tex] can we get out of [tex]-5x^2[/tex]? [tex]-5x^2=5(-x^2)[/tex]. Then[tex](5x^2-4x+5)(-x^2-3)=-5x^4+4x^3-20x^2+12x-15[/tex]This also doesn't match the original numerator. We have a difference between what we have and what we want of[tex](-5x^4+4x^3-20x^2+15x-16)-(5x^2-4x+5)(-x^2-3)=3x-1[/tex]But we also can't divide [tex]3x[/tex] into chunks of [tex]-x^2[/tex], and we can't continue the division algorithm.Our final answer would be[tex]\dfrac{-5x^4+4x^3-20x^2+15x-16}{-x^2-3}=5x^2-4x+5+\dfrac{3x-1}{-x^2-3}[/tex]For the second problem, you should find[tex]\dfrac{-15x^3-13x+4}{5x^2+2}=-3x+\dfrac{-7x+4}{5x^2+2}[/tex]