Divide (x^5+2x^3-x-8)/(x^2-2x+1)

$\frac{x^{5} + 2 x^{3} - x - 8}{x^{2} - 2 x + 1}$

Step 1

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$\frac{x^{5} + 2 x^{3} - x - 8}{x^{2} - 2 x + 1^{2}}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$\frac{x^{5} + 2 x^{3} - x - 8}{x^{2} - 2 \cdot x \cdot 1 + 1^{2}}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

$\frac{x^{5} + 2 x^{3} - x - 8}{{(x - 1)}^{2}}$

Step 2

Split the fraction $\frac{x^{5} + 2 x^{3} - x - 8}{{(x - 1)}^{2}}$ into two fractions.

$\frac{x^{5} + 2 x^{3} - x}{{(x - 1)}^{2}} + \frac{- 8}{{(x - 1)}^{2}}$

Step 3

Split the fraction $\frac{x^{5} + 2 x^{3} - x}{{(x - 1)}^{2}}$ into two fractions.

$\frac{x^{5} + 2 x^{3}}{{(x - 1)}^{2}} + \frac{- x}{{(x - 1)}^{2}} + \frac{- 8}{{(x - 1)}^{2}}$

Step 4

Split the fraction $\frac{x^{5} + 2 x^{3}}{{(x - 1)}^{2}}$ into two fractions.

$\frac{x^{5}}{{(x - 1)}^{2}} + \frac{2 x^{3}}{{(x - 1)}^{2}} + \frac{- x}{{(x - 1)}^{2}} + \frac{- 8}{{(x - 1)}^{2}}$

Step 5

Move the negative in front of the fraction.

$\frac{x^{5}}{{(x - 1)}^{2}} + \frac{2 x^{3}}{{(x - 1)}^{2}} - \frac{x}{{(x - 1)}^{2}} + \frac{- 8}{{(x - 1)}^{2}}$

Step 6

Move the negative in front of the fraction.

$\frac{x^{5}}{{(x - 1)}^{2}} + \frac{2 x^{3}}{{(x - 1)}^{2}} - \frac{x}{{(x - 1)}^{2}} - \frac{8}{{(x - 1)}^{2}}$