Chapter 1.4, Problem 26E

To determine

To calculate: The product of the rational expression,

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

Answer to Problem 26E

The product of rational expression is $-\frac{\left(x-1\right)}{\left(x+1\right)}$

Explanation of Solution

Given information:

The expression is given as:

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

Formula used:

For the rational expression:

Fractions property:

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

Factoring trinomials: The factor of rationalexpression 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

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

Use the fraction property,

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

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

Factor the both numerator and denominator terms,

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

Common minus from the numerator,

  $-\frac{\left(x+3\right)\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)\left(x+3\right)}$

Cancel common factors from the numerator and denominator,

  $-\frac{\left(x-1\right)}{\left(x+1\right)}$

Thus, the product of rational expression is $-\frac{\left(x-1\right)}{\left(x+1\right)}$