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

Mar 22, 2018

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

#### Explanation:

As we have ${x}^{2} \left(x + 2\right)$ in the denominator, we can have partial fractions as

$\frac{4 x + 4}{{x}^{2} \left(x + 2\right)} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x + 2}$

or $4 x + 4 = A x \left(x + 2\right) + B \left(x + 2\right) + C {x}^{2}$ ..............(P)

If we put $x = 0$ in (P), we get $2 B = 4$ i.e. $B = 2$

and if we put $x = - 2$, then $4 C = - 4$ or $C = - 1$

Comparing coefficient of ${x}^{2}$ in (P), we get

$A + C = 0$ i.e. $A = - C = 1$

Hence, partial fractions are

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

Mar 22, 2018

Partial fraction : $\frac{4 x + 4}{{x}^{2} \left(x + 2\right)} = \frac{1}{x} + \frac{2}{x} ^ 2 - \frac{1}{x + 2}$

#### Explanation:

Let $\frac{4 x + 4}{{x}^{2} \left(x + 2\right)} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x + 2}$ or

$\frac{4 x + 4}{{x}^{2} \left(x + 2\right)} = \frac{A x \left(x + 2\right) + B \left(x + 2\right) + C {x}^{2}}{{x}^{2} \left(x + 2\right)}$ or

$A x \left(x + 2\right) + B \left(x + 2\right) + C {x}^{2} = 4 x + 4$ or

${x}^{2} \left(A + C\right) + x \left(2 A + B\right) + 2 B = 4 x + 4$

Equating the co-efficient of ${x}^{2}$ in both sides we get, $A + C = 0$

Equating the co-efficient of $x$ in both sides we get, $2 A + B = 4$

Equating constant term in both sides we get, $2 B = 4 \mathmr{and} B = 2$

B=2 :. 2A+2=4 :. 2A = 2 :. A=1; A+C=0

$\therefore C = - A \therefore C = - 1$ Hence partial fraction is

$\frac{4 x + 4}{{x}^{2} \left(x + 2\right)} = \frac{1}{x} + \frac{2}{x} ^ 2 - \frac{1}{x + 2}$ [Ans]