Chapter 11.5, Problem 11AYU

To determine

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

Answer to Problem 11AYU

The expression $\frac{x\left(x-1\right)}{\left(x+4\right)\left(x-3\right)}$ is an improper rational expression. It is expressed as $\frac{x^2-x}{x^2+x-12}=1+\frac{-20}{7\left(x+4\right)}+\frac{6}{7\left(x-3\right)}$ .

Explanation of Solution

Given information:

The expression $\frac{x\left(x-1\right)}{\left(x+4\right)\left(x-3\right)}$ .

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 $\frac{x\left(x-1\right)}{\left(x+4\right)\left(x-3\right)}$ . It is equivalent to $\frac{x^2-x}{x^2+x-12}$ .

Observe that the degree of the numerator of the rational expression is greater than or equal to 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.

  $\begin{array}{l}{x}^2+x-12\begin{array}{c}\hfill 1\\ \hfill \overline{)x^2-x}\end{array}\\ \begin{array}{ccc}& & \begin{array}{ccc}& x^2+x-12& \end{array}\end{array}\\ \begin{array}{ccc}& & \begin{array}{cc}& -2x+12\end{array}\end{array}\end{array}$

Now, since $\text{Dividend}=\text{Quotient}\times \text{Divisor}+\text{Remainder}$ .

Therefore,

  $\frac{x^2-x}{x^2+x-12}=1+\frac{-2x+12}{x^2+x-12}$

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

   $\frac{-2x+12}{x^2+x-12}$ as,

  $\begin{array}{c}\frac{-2x+12}{x^2+x-12}=\frac{A}{\left(x+4\right)}+\frac{B}{\left(x-3\right)}\\ \frac{-2x+12}{x^2+x-12}=\frac{A\left(x-3\right)+B\left(x+4\right)}{\left(x+4\right)\left(x-3\right)}\end{array}$

Therefore, $-2x+12=A\left(x-3\right)+B\left(x+4\right)$

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

  $\begin{array}{c}-2x+12=A\left(x-3\right)+B\left(x+4\right)\\ =Ax-3A+Bx+4B\\ =\left(A+B\right)x-3A+4B\end{array}$

Now, equate the coefficients on both the side of equation to obtain $A+B=-2$ and $-3A+4B=12$ .

Now, solve the two equations involving two variables.

From equation $A+B=-2$ isolate the value of A , $A=-2-B$ .

Now substitute $A=-2-B$ in the equation $-3A+4B=12$ .

  $\begin{array}{c}-3\left(-2-B\right)+4B=12\\ 6+3B+4B=12\\ 7B=6\\ B=\frac{6}{7}\end{array}$

Now substitute $B=\frac{6}{7}$ in the equation $A+B=-2$ .

  $\begin{array}{c}A+\frac{6}{7}=-2\\ A=-\frac{20}{7}\end{array}$

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

  $\frac{-2x+12}{x^2+x-12}=\frac{-20}{7\left(x+4\right)}+\frac{6}{7\left(x-3\right)}$

Therefore, $\frac{x^2-x}{x^2+x-12}=1+\frac{-2x+12}{x^2+x-12}$ is rewritten as,

  $\frac{x^2-x}{x^2+x-12}=1+\frac{-20}{7\left(x+4\right)}+\frac{6}{7\left(x-3\right)}$

Thus, the expression $\frac{x\left(x-1\right)}{\left(x+4\right)\left(x-3\right)}$ is an improper rational expression. It is expressed as $\frac{x^2-x}{x^2+x-12}=1+\frac{-20}{7\left(x+4\right)}+\frac{6}{7\left(x-3\right)}$ .