A rational function is given in this problem and it is formed by the involvement of a quadratic expression $x^2+13x+15$ and two linear expressions in one variable $2x+3$ and $x+3$.

$\dfrac{x^2+13x+15}{(2x+3)(x+3)^2}$

In this rational expression,

- $x^2+13x+15$ is the expression in the numerator.
- In the denominator, the linear expression $x+3$ is multiplied by itself and their product is expressed as square of $x+3$ in exponential notation.
- Finally, the repeated linear factor $x+3$ is multiplied by the linear expression $2x+3$.

The given rational function consists of repeated linear factor has to decompose into the partial fractions in this problem. So, let us learn how to decompose the rational expression into partial fractions.

The degree of the algebraic expression $x^2+13x+15$ in the numerator is $2$. In the denominator, the square of the linear expression $x+3$ can be expanded as a quadratic expression. The multiplication of the quadratic expression with the linear factor $2x+3$ forms a cubic expression. Therefore, the degree of the expression in the denominator is $3$.

$\dfrac{x^2+13x+15}{(2x+3)(x+3)^2}$

The degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator. Hence, the above rational function is called the proper rational expression. In this proper rational function, one linear factor is not repeated but another linear factor is repeated twice. So, it is a proper rational function that consists of both repeated and non-repeated linear factors. Therefore, the given rational function should be decomposed by combining the decompositions of both repeated and non-repeated proper rational function.

$\implies$ $\dfrac{x^2+13x+15}{(2x+3)(x+3)^2}$ $\,=\,$ $\dfrac{A}{2x+3}$ $+$ $\dfrac{B}{x+3}$ $+$ $\dfrac{C}{(x+3)^2}$

Due to the involvement of linear expressions in one variable in denominator, the expression in each numerator of partial fraction is a constant. Hence, the constants are $A.$ $B$ and $C$ are written in numerator of the partial fractions.

Now, add the fractional functions as per the addition of the fractions.

$\implies$ $\dfrac{x^2+13x+15}{(2x+3)(x+3)^2}$ $\,=\,$ $\dfrac{A \times (x+3)^2+B \times (2x+3) \times (x+3)+C \times (2x+3)}{(2x+3)(x+3)^2}$

$\implies$ $\dfrac{x^2+13x+15}{(2x+3)(x+3)^2}$ $\,=\,$ $\dfrac{A(x+3)^2+B(2x+3)(x+3)+C(2x+3)}{(2x+3)(x+3)^2}$

$\implies$ $\dfrac{x^2+13x+15}{\cancel{(2x+3)(x+3)^2}}$ $\,=\,$ $\dfrac{A \times (x+3)^2+B \times (2x+3)(x+3)+C \times (2x+3)}{\cancel{(2x+3)(x+3)^2}}$

$\implies$ $x^2$ $+$ $13x$ $+$ $15$ $\,=\,$ $A(x+3)^2$ $+$ $B(2x+3)(x+3)$ $+$ $C(2x+3)$

Therefore, the given proper rational function is simplified as the following equation.

$\,\,\,\therefore\,\,\,\,\,\,$ $A(x+3)^2$ $+$ $B(2x+3)(x+3)$ $+$ $C(2x+3)$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

Look at the first fraction in the first term on the right-hand side of the equation. In this term, $A$ is a constant and $2x+3$ is a linear expression in one variable.

$\dfrac{x^2+13x+15}{(2x+3)(x+3)^2}$ $\,=\,$ $\dfrac{A}{2x+3}$ $+$ $\dfrac{B}{x+3}$ $+$ $\dfrac{C}{(x+3)^2}$

For calculating the value of $A$, make the linear expression $2x+3$ is equal to zero.

$\implies$ $2x+3 \,=\, 0$

$\implies$ $2x \,=\, -3$

$\implies$ $x \,=\, \dfrac{-3}{2}$

$\,\,\,\therefore\,\,\,\,\,\,$ $x \,=\, -\dfrac{3}{2}$

Now, substitute the value of $x$ in the following equation to find the value of $A$.

$A(x+3)^2$ $+$ $B(2x+3)(x+3)$ $+$ $C(2x+3)$ $\,=\,$ $x^2+13x+15$

$\implies$ $A\Bigg(\bigg(-\dfrac{3}{2}\bigg)+3\Bigg)^2$ $+$ $B\Bigg(2\bigg(-\dfrac{3}{2}\bigg)+3\Bigg)\Bigg(\bigg(-\dfrac{3}{2}\bigg)+3\Bigg)$ $+$ $C\Bigg(2\bigg(-\dfrac{3}{2}\bigg)+3\Bigg)$ $\,=\,$ $\bigg(-\dfrac{3}{2}\bigg)^2+13\bigg(-\dfrac{3}{2}\bigg)+15$

$\implies$ $A\bigg(-\dfrac{3}{2}+3\bigg)^2$ $+$ $B\bigg(-\dfrac{2 \times 3}{2}+3\bigg)\bigg(-\dfrac{3}{2}+3\bigg)$ $+$ $C\bigg(-\dfrac{2 \times 3}{2}+3\bigg)$ $\,=\,$ $\dfrac{9}{4}-\dfrac{13 \times 3}{2}+15$

$\implies$ $A\bigg(\dfrac{-3+3 \times 2}{2}\bigg)^2$ $+$ $B\bigg(-\dfrac{6}{2}+3\bigg)\bigg(\dfrac{-3+3 \times 2}{2}\bigg)$ $+$ $C\bigg(-\dfrac{\cancel{2} \times 3}{\cancel{2}}+3\bigg)$ $\,=\,$ $\dfrac{9}{4}-\dfrac{39}{2}+15$

$\implies$ $A\bigg(\dfrac{-3+6}{2}\bigg)^2$ $+$ $B\bigg(-\dfrac{\cancel{6}}{\cancel{2}}+3\bigg)\bigg(\dfrac{-3+6}{2}\bigg)$ $+$ $C(-3+3)$ $\,=\,$ $\dfrac{9 \times 1-39 \times 2+15 \times 4}{4}$

$\implies$ $A\bigg(\dfrac{3}{2}\bigg)^2$ $+$ $B(-3+3)\bigg(\dfrac{3}{2}\bigg)$ $+$ $C(0)$ $\,=\,$ $\dfrac{9-78+60}{4}$

$\implies$ $A\bigg(\dfrac{9}{4}\bigg)$ $+$ $B(0)\bigg(\dfrac{3}{2}\bigg)$ $+$ $C(0)$ $\,=\,$ $\dfrac{-9}{4}$

$\implies$ $A \times \dfrac{9}{4}$ $+$ $B \times 0 \times \dfrac{3}{2}$ $+$ $C \times 0$ $\,=\,$ $-\dfrac{9}{4}$

$\implies$ $A \times \dfrac{9}{4}$ $\,=\,$ $-\dfrac{9}{4}$

$\implies$ $A$ $\,=\,$ $-\dfrac{9}{4} \times \dfrac{4}{9}$

$\implies$ $A$ $\,=\,$ $-\dfrac{9 \times 4}{4 \times 9}$

$\implies$ $A$ $\,=\,$ $-\dfrac{\cancel{9} \times \cancel{4}}{\cancel{4} \times \cancel{9}}$

$\,\,\,\therefore\,\,\,\,\,\,$ $A \,=\, -1$

Look at the denominators of the second and third terms on the right hand side of the equation. They both consist $x+3$ as the expression in the denominator. However, it is possible to find the value of $C$ due to the repeated linear factor.

$\dfrac{x^2+13x+15}{(2x+3)(x+3)^2}$ $\,=\,$ $\dfrac{A}{2x+3}$ $+$ $\dfrac{B}{x+3}$ $+$ $\dfrac{C}{(x+3)^2}$

For finding the value of $C$, make the linear expression $x+3$ equals to zero.

$\implies$ $x+3 \,=\, 0$

$\,\,\,\therefore\,\,\,\,\,\,$ $x \,=\, -3$

Now, substitute $x \,=\, -3$ on both sides of the below equation and find the value of $C$ by simplification.

$A(x+3)^2$ $+$ $B(2x+3)(x+3)$ $+$ $C(2x+3)$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

$\implies$ $A(-3+3)^2$ $+$ $B\big(2(-3)+3\big)(-3+3)$ $+$ $C\big(2(-3)+3\big)$ $\,=\,$ $(-3)^2$ $+$ $13(-3)$ $+$ $15$

$\implies$ $A(0)^2$ $+$ $B(-6+3)(0)$ $+$ $C(-6+3)$ $\,=\,$ $9$ $-$ $39$ $+$ $15$

$\implies$ $A(0)$ $+$ $B(-3)(0)$ $+$ $C(-3)$ $\,=\,$ $-15$

$\implies$ $A \times 0$ $+$ $B \times (-3) \times 0$ $+$ $C \times (-3)$ $\,=\,$ $-15$

$\implies$ $0$ $+$ $0$ $+$ $(-3) \times C$ $\,=\,$ $-15$

$\implies$ $(-3) \times C$ $\,=\,$ $-15$

$\implies$ $C$ $\,=\,$ $\dfrac{-15}{-3}$

$\implies$ $C$ $\,=\,$ $\dfrac{\cancel{-15}}{\cancel{-3}}$

$\,\,\,\therefore\,\,\,\,\,\,$ $C \,=\, 5$

The values of both $A$ and $C$ are calculated by substituting the suitable values in the simplified equation but the value of $B$ cannot be evaluated in the same way. Hence, we choose another method to find the value of $B$ in this step.

$\implies$ $A(x+3)^2$ $+$ $B(2x+3)(x+3)$ $+$ $C(2x+3)$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

On the left-hand side of the equation, expand $x+3$ whole square as per square of sum of two terms formula in the first term, get the product of linear factors $2x+3$ and $x+3$ as per the multiplication of algebraic expressions in the second term.

$\implies$ $A(x^2+2 \times 3 \times x+3^2)$ $+$ $B\big(2x \times (x+3)+3 \times (x+3)\big)$ $+$ $C(2x+3)$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

$\implies$ $A(x^2+6x+9)$ $+$ $B(2x \times x+2x \times 3+3 \times x+3 \times 3)$ $+$ $C(2x+3)$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

$\implies$ $A(x^2+6x+9)$ $+$ $B(2x^2+6x+3x+9)$ $+$ $C(2x+3)$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

$\implies$ $A(x^2+6x+9)$ $+$ $B(2x^2+9x+9)$ $+$ $C(2x+3)$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

$\implies$ $A \times (x^2+6x+9)$ $+$ $B \times (2x^2+9x+9)$ $+$ $C(2x+3)$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

Now, distribute each constant over the addition of the terms by using the distributive rule of the multiplication over addition.

$\implies$ $A \times x^2$ $+$ $A \times 6x$ $+$ $A \times 9$ $+$ $B \times 2x^2$ $+$ $B \times 9x$ $+$ $B \times 9$ $+$ $C \times 2x$ $+$ $C \times 3$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

Now, write the expression on the left-hand side of the equation in descending order.

$\implies$ $Ax^2$ $+$ $6Ax$ $+$ $9A$ $+$ $2Bx^2$ $+$ $9Bx$ $+$ $9B$ $+$ $2Cx$ $+$ $3C$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

$\implies$ $Ax^2$ $+$ $2Bx^2$ $+$ $6Ax$ $+$ $9Bx$ $+$ $2Cx$ $+$ $9A$ $+$ $9B$ $+$ $3C$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

$\implies$ $(A+2B)x^2$ $+$ $(6A+9B+2C)x$ $+$ $(9A+9B+3C)$ $\,=\,$ $x^2$ $+$ $13x$ $+$ $15$

Compare the literal coefficients of $x^2$ on both sides of the equation to find the value of $B$.

$\implies$ $A+2B \,=\, 1$

The value of $A$ is evaluated and it is equal to $-1$. Now, substitute the value of $A$ in the above equation for finding the value of $B$.

$\implies$ $-1+2B \,=\, 1$

$\implies$ $2B \,=\, 1+1$

$\implies$ $2B \,=\, 2$

$\implies$ $B \,=\, \dfrac{2}{2}$

$\implies$ $B \,=\, \dfrac{\cancel{2}}{\cancel{2}}$

$\implies$ $B \,=\, \dfrac{2}{2}$

$\,\,\,\therefore\,\,\,\,\,\,$ $B \,=\, 1$

The given proper rational function is decomposed as the sum of the partial fractions.

$\dfrac{x^2+13x+15}{(2x+3)(x+3)^2}$ $\,=\,$ $\dfrac{A}{2x+3}$ $+$ $\dfrac{B}{x+3}$ $+$ $\dfrac{C}{(x+3)^2}$

We have evaluated that $A \,=\, -1,$ $B \,=\, 1$ and $C \,=\, 5$. Now, substitute them in the above equation.

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{x^2+13x+15}{(2x+3)(x+3)^2}$ $\,=\,$ $\dfrac{-1}{2x+3}$ $+$ $\dfrac{1}{x+3}$ $+$ $\dfrac{5}{(x+3)^2}$

Latest Math Topics

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved