# How do you express (-7x^2-15x-8)/((x+3)(x^2+4)) in partial fractions?

Oct 7, 2016

$\frac{- 7 {x}^{2} - 15 x - 8}{\left(x + 3\right) \left({x}^{2} + 4\right)} = \frac{- 2}{x + 3} + \frac{- 5 x}{{x}^{2} + 4}$

#### Explanation:

$1.$ First, we factorise the denominator, but since this is done, we skip to the next step: Writing a partial fraction for each factor:

$\frac{- 7 {x}^{2} - 15 x - 8}{\left(x + 3\right) \left({x}^{2} + 4\right)} = \frac{a}{x + 3} + \frac{b x + c}{{x}^{2} + 4}$

$2.$ Multiply out the factors to remove the fractions:

$- 7 {x}^{2} - 15 x - 8 = a \left({x}^{2} + 4\right) + \left(b x + c\right) \left(x + 3\right)$

$3.$ Solve the equation for the constants:

$- 7 {x}^{2} - 15 x - 8 = \left(- 7 x - 8\right) \left(x + 1\right)$
$\left(- 7 x - 8\right) \left(x + 1\right) = a \left({x}^{2} + 4\right) + \left(b x + c\right) \left(x + 3\right)$

"a") If we set $x = - 3$, then we can solve for $a$ because $b x + c = 0$

$\left(- 7 \cdot \left(- 3\right) - 8\right) \left(- 3 + 1\right) = a \left({\left(- 3\right)}^{2} + 4\right)$

$- 26 = 13 a$

$a = - 2$

$\left(- 7 x - 8\right) \left(x + 1\right) = - 2 \left({x}^{2} + 4\right) + \left(b x + c\right) \left(x + 3\right)$

"b") If we set $x = 0$, we can solve for $c$ because $b x = 0$

$\left(- 8\right) \left(+ 1\right) = - 2 \left(+ 4\right) + c \left(3\right)$

$- 8 = - 8 + 3 c$

$3 c = 0 , c = 0$

$\left(- 7 x - 8\right) \left(x + 1\right) = - 2 \left({x}^{2} + 4\right) + \left(b x\right) \left(x + 3\right)$

"c") If we set $x = 1$, we can now solve for $b$ relatively simply

$\left(- 7 - 8\right) \left(1 + 1\right) = - 2 \left(1 + 4\right) + b \left(1 + 3\right)$

$- 30 = - 10 + 4 b$

$- 20 = 4 b$

$b = - 5$

$4.$Now we substitute the values back into the partial fractions:

$\frac{- 7 {x}^{2} - 15 x - 8}{\left(x + 3\right) \left({x}^{2} + 4\right)} = \frac{- 2}{x + 3} + \frac{- 5 x}{{x}^{2} + 4}$