# How do you express (x³ - 7x - 3)/(x² - x -2)  in partial fractions?

Apr 14, 2016

Partial fractions of $\frac{{x}^{3} - 7 x - 3}{{x}^{2} - x - 2}$ are $x + 1 - \frac{1}{x + 1} - \frac{3}{x - 2}$

#### Explanation:

Before we express $\frac{{x}^{3} - 7 x - 3}{{x}^{2} - x - 2}$ we must divide the algebraic expression in numerator by denominator so that only remainder, whose degree is less than that of denominator, is left. Furthe, we should also factorize the denominator.

Dividing $\left({x}^{3} - 7 x - 3\right)$ by $\left({x}^{2} - x - 2\right)$, we get

$\left({x}^{3} - 7 x - 3\right) = \left(x + 1\right) \left({x}^{2} - x - 2\right) - \left(4 x + 1\right)$ and

${x}^{2} - x - 2 = {x}^{2} - 2 x + x - 2 = x \left(x - 2\right) + 1 \left(x - 2\right) = \left(x + 1\right) \left(x - 2\right)$

Hence $\frac{{x}^{3} - 7 x - 3}{{x}^{2} - x - 2} = x + 1 - \frac{4 x + 1}{\left(x + 1\right) \left(x - 2\right)}$

Now let $\frac{4 x + 1}{\left(x + 1\right) \left(x - 2\right)} \Leftrightarrow \frac{A}{x + 1} + \frac{B}{x - 2} = \frac{A \left(x - 2\right) + B \left(x + 1\right)}{\left(x + 1\right) \left(x - 2\right)}$

or (4x+1)/((x+1)(x-2))hArr(Ax-2A+Bx+B)/((x+1)(x-2))=((A+B)x-2A+B)/((x+1)(x-2)

Hence $A + B = 4$ and $B - 2 A = 1$

Now putting $B = 4 - A$ (from first equation) into second, we get

$4 - A - 2 A = 1$ or $3 A = 3$ or $A = 1$ and hence $B = 4 - 1 = 3$ hence

Partial fractions of $\frac{{x}^{3} - 7 x - 3}{{x}^{2} - x - 2}$ are $x + 1 - \frac{1}{x + 1} - \frac{3}{x - 2}$