# How do you express 1/((x+6)(x^2+3)) in partial fractions?

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{\frac{1}{39}}{x + 6} + \frac{- \frac{1}{39} x + \frac{2}{13}}{{x}^{2} + 3}$

#### Explanation:

We start setting up the equation with the unknown constants A, B, C

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{A}{x + 6} + \frac{B x + C}{{x}^{2} + 3}$

After setting this right like this, convert it to one fraction
using the LCD$= \left(x + 6\right) \left({x}^{2} + 3\right)$

so that

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{A \left({x}^{2} + 3\right) + \left(B x + C\right) \left(x + 6\right)}{\left(x + 6\right) \left({x}^{2} + 3\right)}$

simplify

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{A {x}^{2} + 3 A + \left(B {x}^{2} + C x + 6 B x + 6 C\right)}{\left(x + 6\right) \left({x}^{2} + 3\right)}$

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{A {x}^{2} + 3 A + B {x}^{2} + C x + 6 B x + 6 C}{\left(x + 6\right) \left({x}^{2} + 3\right)}$

Rearrange according to degree from highest to lowest

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{A {x}^{2} + B {x}^{2} + 6 B x + C x + 3 A + 6 C}{\left(x + 6\right) \left({x}^{2} + 3\right)}$

Determine the appropriate numerical coefficients

$\frac{0 \cdot {x}^{2} + 0 \cdot x + 1 \cdot {x}^{0}}{\left(x + 6\right) \left({x}^{2} + 3\right)}$

$= \frac{\left(A + B\right) {x}^{2} + \left(6 B + C\right) x + \left(3 A + 6 C\right) {x}^{0}}{\left(x + 6\right) \left({x}^{2} + 3\right)}$

We can now obtain the equations to solve for A, B, C

$A + B = 0 \text{ }$first equation
$6 B + C = 0 \text{ }$second equation
$3 A + 6 C = 1 \text{ }$third equation

Simultaneous solution of these three equations result to

$A = \frac{1}{39}$ and $B = - \frac{1}{39}$ and $C = \frac{2}{13}$

so that our final answer is

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{\frac{1}{39}}{x + 6} + \frac{- \frac{1}{39} x + \frac{2}{13}}{{x}^{2} + 3}$

God bless ....I hope the explanation is useful.