Chapter 11, Problem 24CT

To determine

The partial fraction decomposition of the rational expression $\frac{3x+7}{{\left(x+3\right)}^2}$.

Answer to Problem 24CT

Solution:

The partial fraction decomposition of the rational expression $\frac{3x+7}{{\left(x+3\right)}^2}$ is $\frac{3x+7}{{\left(x+3\right)}^2}=\frac{3}{x+3}+\frac{-2}{{\left(x+3\right)}^2}$.

Explanation of Solution

Given information:

The rational expression, $\frac{3x+7}{{\left(x+3\right)}^2}$.

Explanation:

Here denominator is ${\left(x+3\right)}^2$.

Thus, the denominator has a repeated linear factor $x+3$.

Therefore, the partial factors of $\frac{3x+7}{{\left(x+3\right)}^2}$ takes form as $\frac{A}{x+3}+\frac{B}{{\left(x+3\right)}^2}$.

Then, $\frac{3x+7}{{\left(x+3\right)}^2}=\frac{A}{x+3}+\frac{B}{{\left(x+3\right)}^2}$ where $A,$ and $B$ are to be determined.

Multiply both sides of equation by ${\left(x+3\right)}^2$.

$3x+7=A\left(x+3\right)+B$

By simplifying,

$\Rightarrow 3x+7=Ax+3A+B$

Combine the like terms,

$\Rightarrow 3x+7=Ax+\left(3A+B\right)$

By equating coefficients of $x$ and constants,

$A=3$ ……. (1)

$3A+B=7$……. (2)

Substitute $A=3$ in equation (2),

$\Rightarrow 3\left(3\right)+B=7$

$\Rightarrow B=7-9=-2$

Thus, $A=3,B=-2$.

Hence, $\frac{3x+7}{{\left(x+3\right)}^2}=\frac{3}{x+3}+\frac{-2}{{\left(x+3\right)}^2}$.