# How do you write the partial fraction decomposition of the rational expression (7 x^2 - 7 x+12)/(x^4+x^2-2)?

$\frac{7 {x}^{2} - 7 x + 12}{{x}^{4} + {x}^{2} - 2} = \frac{\frac{7}{3} x + \frac{2}{3}}{{x}^{2} + 2} + \frac{- \frac{13}{3}}{x + 1} + \frac{2}{x - 1}$

#### Explanation:

From the given, we start from the factoring part of the denominator

$\frac{7 {x}^{2} - 7 x + 12}{{x}^{4} + {x}^{2} - 2} = \frac{7 {x}^{2} - 7 x + 12}{\left({x}^{2} + 2\right) \left({x}^{2} - 1\right)} = \frac{7 {x}^{2} - 7 x + 12}{\left({x}^{2} + 2\right) \left(x + 1\right) \left(x - 1\right)}$

so that

$\frac{7 {x}^{2} - 7 x + 12}{{x}^{4} + {x}^{2} - 2} = \frac{7 {x}^{2} - 7 x + 12}{\left({x}^{2} + 2\right) \left(x + 1\right) \left(x - 1\right)}$

and

$\frac{7 {x}^{2} - 7 x + 12}{\left({x}^{2} + 2\right) \left(x + 1\right) \left(x - 1\right)} = \frac{A x + B}{{x}^{2} + 2} + \frac{C}{x + 1} + \frac{D}{x - 1}$

Let us now form the equations using the LCD we continue

$\frac{7 {x}^{2} - 7 x + 12}{\left({x}^{2} + 2\right) \left(x + 1\right) \left(x - 1\right)} =$

$\frac{\left(A x + B\right) \left({x}^{2} - 1\right) + C \left({x}^{2} + 2\right) \left(x - 1\right) + D \left({x}^{2} + 2\right) \left(x + 1\right)}{\left({x}^{2} + 2\right) \left(x + 1\right) \left(x - 1\right)}$

Expand and then simplify

$\frac{A {x}^{3} - A x + B {x}^{2} - B + C {x}^{3} - C {x}^{2} + 2 C x - 2 C + D {x}^{3} + D {x}^{2} + 2 D x + 2 D}{\left({x}^{2} + 2\right) \left(x + 1\right) \left(x - 1\right)}$

$\left(A + C + D\right) {x}^{3} + \left(B - C + D\right) {x}^{2} + \left(- A + 2 C + 2 D\right) {x}^{1} + \left(- B - 2 C + 2 D\right) {x}^{0}$

Equate now the numerical coefficients of the numerators of left and right sides of the equation

$A + C + D = 0 \text{ " }$first equation
$B - C + D = 7 \text{ " }$second equation
$- A + 2 C + 2 D = - 7 \text{ " }$third equation
$- B - 2 C + 2 D = 12 \text{ " }$fourth equation

4 equations with 4 unknowns, solve the unknowns A,B,C,D by elimination of variables

by adding first and third equations the result will be
$3 C + 3 D = - 7 \text{ " }$fifth equation

by adding second and fourth equations the result will be
$- 3 C + 3 D = 19 \text{ " }$sixth equation

by adding the fifth and sixth equations, the result will be
$6 D = 12$

$D = 2$
it follows
$C = - \frac{13}{3}$
Solve A using first equation

$A = - C - D = - \left(- \frac{13}{3}\right) - 2 = \frac{+ 13 - 6}{3} = \frac{7}{3}$
$A = \frac{7}{3}$

Solve B using second equation

$B = C - D + 7 = - \frac{13}{3} - 2 + 7 = - \frac{13}{3} + 5 = \frac{2}{3}$

$B = \frac{2}{3}$

$\frac{7 {x}^{2} - 7 x + 12}{{x}^{4} + {x}^{2} - 2} = \frac{\frac{7}{3} x + \frac{2}{3}}{{x}^{2} + 2} + \frac{- \frac{13}{3}}{x + 1} + \frac{2}{x - 1}$