Chapter 11.5, Problem 9AYU

To determine

To express:The expression $\frac{5x^3+2x-1}{x^2-4}$ as the sum of a polynomial and a proper rational expression if it is an improper fraction.

Answer to Problem 9AYU

The expression $\frac{5x^3+2x-1}{x^2-4}$ is an improper rational expression. It is expressed as $\frac{5x^3+2x-1}{x^2-4}=5x+\frac{43}{4\left(x-2\right)}+\frac{45}{4\left(x+2\right)}$ .

Explanation of Solution

Given information:

The expression $\frac{5x^3+2x-1}{x^2-4}$ .

Formula used:

Improper rational expression of the form $\frac{P}{Q}$ where Q has non repeated linear factors of the form $\left(x-a_1\right)\left(x-a_2\right)\cdots \left(x-a_n\right)$ then $\frac{P}{Q}$ is expressed as $\frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_n}{x-a_n}$ .

Calculation:

Consider the expression $\frac{5x^3+2x-1}{x^2-4}$ .

Observe that the degree of the numerator of the rational expression is greater than the denominator so the provided expression is improper.

When a polynomial is divided by its factor then dividend is the product of divisor and quotient increased by remainder.

Apply the method of long division.

  $\begin{array}{l}{x}^2-4\begin{array}{c}\hfill 5x\\ \hfill \overline{)5x^3+2x-1}\end{array}\\ \begin{array}{ccc}& & 5x^3-20x\end{array}\\ \begin{array}{ccc}& & \begin{array}{cc}& 22x-1\end{array}\end{array}\end{array}$

Now, since $\text{Dividend}=\text{Quotient}\times \text{Divisor}+\text{Remainder}$ . Here, dividend is $5x^3+2x-1$ quotient is $5x$ , divisor is $x^2-4$ and remainder is $22x-1$ .

Therefore,

  $\frac{5x^3+2x-1}{x^2-4}=5x+\frac{22x-1}{x^2-4}$

Factors of the denominator are of the form,

  $\begin{array}{c}{x}^2-4=x^2-2^2\\ =\left(x-2\right)\left(x+2\right)\end{array}$ .

Recall that improper rational expression of the form $\frac{P}{Q}$ where Q has non repeated linear factors of the form $\left(x-a_1\right)\left(x-a_2\right)\cdots \left(x-a_n\right)$ then $\frac{P}{Q}$ is expressed as $\frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_n}{x-a_n}$ .

Rewrite the expression $\frac{22x-1}{x^2-4}$ as,

  $\begin{array}{c}\frac{22x-1}{x^2-4}=\frac{A}{\left(x-2\right)}+\frac{B}{\left(x+2\right)}\\ \frac{22x-1}{x^2-4}=\frac{A\left(x+2\right)+B\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\\ \frac{22x-1}{x^2-4}=\frac{A\left(x+2\right)+B\left(x-2\right)}{x^2-4}\end{array}$

Therefore, $22x-1=A\left(x+2\right)+B\left(x-2\right)$

Now, solve the above expression for values of A and B .

  $\begin{array}{c}22x-1=A\left(x+2\right)+B\left(x-2\right)\\ 22x-1=Ax+2A+Bx-2B\\ 22x-1=\left(A+B\right)x+2A-2B\end{array}$

Now, equate the coefficients on both the side of equation to obtain $A+B=22$ and $2A-2B=-1$ .

Now, solve the two equations involving two variables.

From equation $A+B=22$ isolate the value of A , $A=22-B$ .

Now substitute $A=22-B$ in the equation $2A-2B=-1$ .

  $\begin{array}{c}2\left(22-B\right)-2B=-1\\ 44-2B-2B=-1\\ -4B=-45\\ B=\frac{45}{4}\end{array}$

Now substitute $B=\frac{45}{4}$ in the equation $A+B=22$ .

  $\begin{array}{c}A+\frac{45}{4}=22\\ A=22-\frac{45}{4}\\ A=\frac{43}{4}\end{array}$

Now, substitute the values of A and B in the partial fraction decomposition.

  $\frac{22x-1}{x^2-4}=\frac{43}{4\left(x-2\right)}+\frac{45}{4\left(x+2\right)}$

Therefore, $\frac{5x^3+2x-1}{x^2-4}=5x+\frac{22x-1}{x^2-4}$ is rewritten as,

  $\frac{5x^3+2x-1}{x^2-4}=5x+\frac{43}{4\left(x-2\right)}+\frac{45}{4\left(x+2\right)}$

Thus, the expression $\frac{5x^3+2x-1}{x^2-4}$ is an improper rational expression. It is expressed as $\frac{5x^3+2x-1}{x^2-4}=5x+\frac{43}{4\left(x-2\right)}+\frac{45}{4\left(x+2\right)}$ .