# How do you divide (7x^3 - x^2 – 5x – 6 )/((x + 1) )?

Feb 8, 2018

$7 {x}^{2} - 8 x + 3 - \frac{9}{x + 1}$

#### Explanation:

I don't like to divide using long polynomial division.

${P}_{x} = \frac{7 {x}^{3} - {x}^{2} - 5 x - 6}{x + 1}$

Lets graph this:
graph{(7x^3-x^2-5x-6)/(x+1) [-25.93, 30.5, -15.17, 13.03]}

Let $u = x + 1 \implies u - 1 = x$.

${P}_{u} = \frac{7 {\left(u - 1\right)}^{3} - {\left(u - 1\right)}^{2} - 5 \left(u - 1\right) - 6}{u}$
${P}_{u} = \frac{7 \left({u}^{3} - 3 {u}^{2} + 3 u - 1\right) - \left({u}^{2} - 2 u + 1\right) - 5 u + 5 - 6}{u}$
${P}_{u} = \frac{7 {u}^{3} - 21 {u}^{2} + 21 u - 7 - {u}^{2} + 2 u - 1 - 5 u - 1}{u}$
${P}_{u} = \frac{7 {u}^{3} - 22 {u}^{2} + 18 u - 9}{u}$
${P}_{u} = 7 {u}^{2} - 22 u + 18 - \frac{9}{u}$

Going back to $x$-world

${P}_{x} = 7 {\left(x + 1\right)}^{2} - 22 \left(x + 1\right) + 18 - \frac{9}{x + 1}$
${P}_{x} = 7 \left({x}^{2} + 2 x + 1\right) - 22 x - 22 + 18 - \frac{9}{x + 1}$
${P}_{x} = 7 {x}^{2} + 14 x + 7 - 22 x - 4 - \frac{9}{x + 1}$
${P}_{x} = 7 {x}^{2} - 8 x + 3 - \frac{9}{x + 1}$

Compare with the graph here:
graph{(7x^2 - 8x + 3 - 9/(x+1)) [-25.93, 30.5, -15.17, 13.03]}