# How do you complete the following division: (x^4 + x^3 - 5x^2 + 26x - 21)/(x^2 + 3x - 4)?

Dec 11, 2016

${x}^{2} + 2 x + 5 + \frac{13}{5 \left(x + 4\right)} + \frac{2}{5 \left(x - 1\right)}$

#### Explanation:

Divide the numerator by the denominator using long division. So, $\frac{{x}^{4} + {x}^{3} - 5 {x}^{2} + 26 x - 21}{{x}^{2} + 3 x - 4} = {x}^{2} + 2 x + 5 + \frac{3 x - 1}{{x}^{2} + 3 x - 4}$.

We can now start the actual partial fraction decomposition process.

${x}^{2} + 3 x - 4$ can be factored as $\left(x + 4\right) \left(x - 1\right)$.

$\frac{A}{x + 4} + \frac{B}{x - 1} = \frac{3 x - 1}{\left(x + 4\right) \left(x - 1\right)}$

$A \left(x - 1\right) + B \left(x + 4\right) = 3 x - 1$

$A x - A + B x + 4 B = 3 x - 1$

$\left(A + B\right) x + \left(4 B - A\right) = 3 x - 1$

We now write a system of equations:

$\left\{\begin{matrix}A + B = 3 \\ 4 B - A = - 1\end{matrix}\right.$

Solve:

$B = 3 - A \to 4 \left(3 - A\right) - A = - 1$

$12 - 4 A - A = - 1$

$- 5 A = - 13$

$A = \frac{13}{5}$

$\frac{13}{5} + B = 3$

$B = \frac{2}{5}$

Therefore, the partial fraction decomposition of $\frac{{x}^{4} + {x}^{3} - 5 {x}^{2} + 26 x - 21}{{x}^{2} + 3 x - 4}$ is ${x}^{2} + 2 x + 5 + \frac{13}{5 \left(x + 4\right)} + \frac{2}{5 \left(x - 1\right)}$.

Hopefully this helps!