# How do you write the partial fraction decomposition of the rational expression (4x^2+3)/((x-5)^3)?

Jul 31, 2018

$\frac{4 {x}^{2} + 3}{x - 5} ^ 3 = \frac{4}{x - 5} + \frac{40}{x - 5} ^ 2 + \frac{103}{x - 5} ^ 3$

#### Explanation:

Let ,

(4x^2+3)/(x-5)^3=A/(x-5)+B/(x-5)^2+C/(x-5)^3 ...tocolor(red)((M)

$\therefore 4 {x}^{2} + 3 = A {\left(x - 5\right)}^{2} + B \left(x - 5\right) + C$

$\therefore 4 {x}^{2} + 3 = A \left({x}^{2} - 10 x + 25\right) + B \left(x - 5\right) + C$

$\therefore 4 {x}^{2} + 3 = A {x}^{2} - 10 A x + 25 A + B x - 5 B + C$

$\therefore 4 {x}^{2} + 0 \cdot x + 3 = A {x}^{2} + x \left(B - 10 A\right) + \left(25 A - 5 B + C\right)$

Comparing coefficients of ${x}^{2} , x \mathmr{and}$ free term,we get

color(blue)(A=4......................to(1)

$B - 10 A = 0. \ldots \ldots . . , . \to \left(2\right)$

$25 A - 5 B + C = 3. . \to \left(3\right)$

Substitute color(blue)(A=4 into $\left(2\right)$

:.B-10(4)=0=>color(blue)(B=40

Again subst. $A = 4 \mathmr{and} B = 40$ into $\left(3\right)$

$\therefore 25 \left(4\right) - 5 \left(40\right) + C = 3$

:.C=3-100+200=>color(blue)(C=103

Hence, from $\textcolor{red}{\left(M\right)}$,we have

$\frac{4 {x}^{2} + 3}{x - 5} ^ 3 = \frac{\textcolor{b l u e}{4}}{x - 5} + \frac{\textcolor{b l u e}{40}}{x - 5} ^ 2 + \frac{\textcolor{b l u e}{103}}{x - 5} ^ 3$