# How do you write the partial fraction decomposition of the rational expression (4x^2+ 21x-22)/((2x-1)(x^2 +x-42))?

Feb 13, 2016

$\frac{14}{55} \left(2 x - 1\right) + \frac{9}{65} \left(x + 7\right) + \frac{248}{143} \left(x - 6\right)$

#### Explanation:

first step is to factor the denominator.

${x}^{2} + x - 42 = \left(x + 7\right) \left(x - 6\right)$

The factors on the denominator are all linear , hence the numerators will be constants.

$\frac{4 {x}^{2} + 21 x - 22}{\left(2 x - 1\right) \left(x + 7\right) \left(x - 6\right)} = \frac{A}{2 x - 1} + \frac{B}{x + 7} + \frac{C}{x - 6}$

multiply through by (2x-1)(x+7)(x-6)

$4 {x}^{2} + 21 x - 22 = A \left(x + 7\right) \left(x - 6\right) + B \left(2 x - 1\right) \left(x - 6\right) + C \left(2 x - 1\right) \left(x + 7\right) \ldots \ldots \ldots . \left(1\right)$

the aim now is to find the values of A , B and C. Note that if x=$\frac{1}{2}$

then the terms with B and C will be zero. If x = -7 ,the terms with A and C will be zero and if x=6 , the terms with A and B will be zero. This is the starting point in finding A , B and C.

let x $= \frac{1}{2} \textcolor{b l a c k}{\text{ in (1)}}$

$- \frac{21}{2} = - \frac{165}{4} \Rightarrow A = \frac{14}{55}$

let x = -7 in (1) : 27 = $195 B \Rightarrow B = \frac{9}{65}$

let x = 6 in (1) : 248 = 143 C$\Rightarrow C = \frac{248}{143}$

 rArr( 4x^2+21x-22)/((2x-1)(x^2+x-42)) = 14/55(2x-1) + 9/65(x+7) + 248/143((x-6)