# How do you write the partial fraction decomposition of the rational expression (7x-2)/((x-3)^2(x+1))?

Jul 31, 2017

The answer is $= \frac{- \frac{9}{64}}{x - 3} + \frac{\frac{9}{16}}{x - 3} ^ 2 + \frac{\frac{19}{4}}{x - 3} ^ 3 + \frac{\frac{9}{64}}{x + 1}$

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{7 x - 2}{{\left(x - 3\right)}^{2} \left(x + 1\right)} = \frac{A}{x - 3} + \frac{B}{x - 3} ^ 2 + \frac{C}{x - 3} ^ 3 + \frac{D}{x + 1}$

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

The denominators are the same, we compare the numerators

$\left(7 x - 2\right) = A {\left(x - 3\right)}^{2} \left(x + 1\right) + B \left(x - 3\right) \left(x + 1\right) + C \left(x + 1\right) + D {\left(x - 3\right)}^{3}$

Let $x = 3$, $\implies$, $19 = 4 C$, $\implies$, $C = \frac{19}{4}$

Let $x = - 1$, $\implies$, $- 9 = - 64 D$, $\implies$, $D = \frac{9}{64}$

Coefficients of ${x}^{3}$,

$0 = A + D$, $\implies$, $A = - D = - \frac{9}{64}$

Coefficients of ${x}^{2}$

$0 = - 5 A + B - 9 D$

$\implies$, $B = 5 A + 9 D = - \frac{45}{64} + \frac{81}{64} = \frac{36}{64} = \frac{9}{16}$

Therefore,

$\frac{7 x - 2}{{\left(x - 3\right)}^{2} \left(x + 1\right)} = \frac{- \frac{9}{64}}{x - 3} + \frac{\frac{9}{16}}{x - 3} ^ 2 + \frac{\frac{19}{4}}{x - 3} ^ 3 + \frac{\frac{9}{64}}{x + 1}$

Jul 31, 2017

(7x-2)/((x-3)^2(x+1)) = color(blue)(9/(16(x-3)) + 19/(4(x-3)^2) - 9/(16(x+1))

#### Explanation:

We recognize that there will be three different fractions with denominators of

• $x - 3$

• ${\left(x - 3\right)}^{2}$

• $x + 1$

So we set it up as

$\frac{7 x - 2}{{\left(x - 3\right)}^{2} \left(x + 1\right)} = \frac{A}{x - 3} + \frac{B}{{\left(x - 3\right)}^{2}} + \frac{C}{x + 1}$

Multiply both sides by the quantity ${\left(x - 3\right)}^{2} \left(x + 1\right)$:

$7 x - 2 = A \left(x - 3\right) \left(x + 1\right) + B \left(x + 1\right) + C {\left(x - 3\right)}^{2}$

Expand the terms:

$7 x - 2 = A \left({x}^{2} - 2 x + 3\right) + B \left(x + 1\right) + C \left({x}^{2} - 6 x + 9\right)$

Collect the terms by power of $x$:

$7 x - 2 = {x}^{2} \left(A + C\right) + x \left(- 2 A + B - 6 C\right) + - 3 A + B + 9 C$

Equate the coefficients on both sides (there is no ${x}^{2}$ term on left hand side, so that we set equal to $0$:

$0 = A + C$

$7 = - 2 A + B - 6 C$

$- 2 = - 3 A + B + 9 C$

Now we set up an augmented matrix (sorry for the poor quality):

$\left(\begin{matrix}1 \textcolor{w h i t e}{a a a} & 0 \textcolor{w h i t e}{a a a} & 1 \textcolor{w h i t e}{a a a} | - 2 \\ - 2 \textcolor{w h i t e}{a a a} & 1 & - 6 \textcolor{w h i t e}{} | 7 \\ - 3 \textcolor{w h i t e}{a a a} & 1 \textcolor{w h i t e}{a a a} & 9 \textcolor{w h i t e}{a a a} | - 2\end{matrix}\right)$

In order to avoid mayhem, I won't work out the solving of the matrix (maybe you already know how, if not there's plenty of other sources, maybe some of Socratic!), but the solutions are

$A = \frac{9}{16}$

$B = \frac{19}{4}$

$C = - \frac{9}{16}$

Plugging these into the original equation, we have

(7x-2)/((x-3)^2(x+1)) = color(blue)(ul(9/(16(x-3)) + 19/(4(x-3)^2) - 9/(16(x+1))