# Let h(x)=6x^5-5x^4+4x^3-3x^2-2x+x+7 and m(x)=x^2-1, how do you find the quotient h(x) and m(x)?

Feb 10, 2018

$- 6 {x}^{5} + 5 {x}^{4} - 3 {x}^{3} + 2 {x}^{2} + x - 7$

#### Explanation:

$h \left(x\right) = 6 {x}^{5} - 5 {x}^{4} + 4 {x}^{3} - 3 {x}^{2} - 2 x + x + 7$
$m \left(x\right) = {x}^{2} - 1$
therefore, $h \left(x\right) = \frac{6 {x}^{5} - 5 {x}^{4} + 4 {x}^{3} - 3 {x}^{2} - 2 x + x + 7}{{x}^{2} - 1}$
=$- \left(6 {x}^{5} - 5 {x}^{4} + 3 {x}^{3} - 2 {x}^{2} - x + 1\right)$
=$- 6 {x}^{5} + 5 {x}^{4} - 3 {x}^{3} + 2 {x}^{2} + x + 1$

simplify $\left(- 2 x + x\right)$ and $\left(- 3 {x}^{2} \mathmr{and} {x}^{2}\right)$