Chapter 1.4, Problem 28E

To determine

To calculate: The product of the rational expression,

  $\frac{x^2-x-6}{x^2+2x}\cdot \frac{x^3+x^2}{x^2-2x-3}$

Answer to Problem 28E

The product of rational expression is $\frac{\left(x+3\right)x}{\left(x+2\right)}$

Explanation of Solution

Given information:

The expression is given as:

  $\frac{x^2-x-6}{x^2+2x}\cdot \frac{x^3+x^2}{x^2-2x-3}$

Formula used:

For the rational expression:

Fractions property:

  $\frac{A}{B}\cdot \frac{C}{D}=\frac{AC}{BD}$

Factoring trinomials: The factor of rational expression which contain three terms is of the from $x^2+bx+c$ ,

  $\left(x+r\right)\left(x+s\right)=x^2+\left(r+s\right)x+rs$

Choose the values of $r$ and $s$ which satisfied these equations $r+s=b$ and $rs=c$

Calculation:

Consider the, rational expression

  $\frac{x^2-x-6}{x^2+2x}\cdot \frac{x^3+x^2}{x^2-2x-3}$

Use the fraction property for multiplying rational expression,

  $\frac{A}{B}\cdot \frac{C}{D}=\frac{AC}{BD}$

  $\frac{\left(x^2-x-6\right)\left(x^3+x^2\right)}{\left(x^2+2x\right)\left(x^2-2x-3\right)}$

Factor the first term in numerator byFactoring trinomials rule and common $x^2$ from the second term,

  $\begin{array}{l}\frac{\left(x^2-3x+2x-6\right)x^2\left(x+1\right)}{\left(x^2+2x\right)\left(x^2-2x-3\right)}\\ \frac{\left(x+3\right)\left(x-3\right)x^2\left(x+1\right)}{\left(x^2+2x\right)\left(x^2-2x-3\right)}\end{array}$

Factor the second term in denominator and common $x$ from the first term,

  $\begin{array}{l}\frac{\left(x+3\right)\left(x-3\right)x^2\left(x+1\right)}{x\left(x+2\right)\left(x^2-3x+1x-3\right)}\\ \frac{\left(x+3\right)\left(x-3\right)x^2\left(x+1\right)}{x\left(x+2\right)\left(x-3\right)\left(x+1\right)}\end{array}$

Cancel common factors from the numerators and denominator,

  $\frac{\left(x+3\right)x}{\left(x+2\right)}$

Thus, the product of rational expression is $\frac{\left(x+3\right)x}{\left(x+2\right)}$