Chapter 11.3, Problem 67MR

To determine

To divide:

  $\frac{x^2+9x+20}{x^2+5x-24}÷\frac{x^2+15x+56}{x^2+x-12}$

Answer to Problem 67MR

  $\frac{x^2+9x+20}{x^2+5x-24}÷\frac{x^2+15x+56}{x^2+x-12}=\frac{{\left(x+4\right)}^2\left(x+5\right)}{{\left(x+8\right)}^2\left(x+7\right)}$

Explanation of Solution

Given:

  $\frac{x^2+9x+20}{x^2+5x-24}÷\frac{x^2+15x+56}{x^2+x-12}$

Concept Used:

• To multiply/divide rational expressions first simplify numerator and denominator of each rational expression and then proceed to find the operation by canceling the common terms of numerator and denominator.
• $\frac{P\left(x\right)}{Q\left(x\right)}÷\frac{S\left(x\right)}{R\left(x\right)}=\frac{P\left(x\right)}{Q\left(x\right)}\cdot \frac{R\left(x\right)}{S\left(x\right)}$
• $\frac{P\left(x\right)}{Q\left(x\right)}\cdot \frac{R\left(x\right)}{S\left(x\right)}=\frac{P\left(x\right)R\left(x\right)}{Q\left(x\right)S\left(x\right)}$

Calculation:

In order to divide:

  $\frac{x^2+9x+20}{x^2+5x-24}÷\frac{x^2+15x+56}{x^2+x-12}$

First, rewrite it as

  $\frac{x^2+9x+20}{x^2+5x-24}÷\frac{x^2+15x+56}{x^2+x-12}=\frac{x^2+9x+20}{x^2+5x-24}\cdot \frac{x^2+x-12}{x^2+15x+56}$

Now, simplify the numerator and denominator of each rational expression of given multiplication, as

  $\begin{array}{c}\frac{x^2+9x+20}{x^2+5x-24}\cdot \frac{x^2+x-12}{x^2+15x+56}=\frac{x^2+4x+5x+20}{x^2+8x-3x-24}\cdot \frac{x^2+4x-3x-12}{x^2+8x+7x+56}\\ =\frac{x\left(x+4\right)+5\left(x+4\right)}{x\left(x+8\right)-3\left(x+8\right)}\cdot \frac{x\left(x+4\right)-3\left(x+4\right)}{x\left(x+8\right)+7\left(x+8\right)}\\ =\frac{\left(x+4\right)\left(x+5\right)}{\left(x+8\right)\left(x-3\right)}\cdot \frac{\left(x+4\right)\left(x-3\right)}{\left(x+8\right)\left(x+7\right)}\end{array}$

Now, multiplying numerator and denominator of rational expressions, as

  $\begin{array}{c}\frac{\left(x+4\right)\left(x+5\right)}{\left(x+8\right)\left(x-3\right)}\cdot \frac{\left(x+4\right)\left(x-3\right)}{\left(x+8\right)\left(x+7\right)}=\frac{\left(x+4\right)\left(x+5\right)\left(x+4\right)\left(x-3\right)}{\left(x+8\right)\left(x-3\right)\left(x+8\right)\left(x+7\right)}\\ =\frac{\left(x+4\right)\left(x+5\right)\left(x+4\right)}{\left(x+8\right)\left(x+8\right)\left(x+7\right)},\left(\begin{array}{l}\text{cancelling the }\\ \text{common factor }\left(x-3\right)\end{array}\right)\\ =\frac{{\left(x+4\right)}^2\left(x+5\right)}{{\left(x+8\right)}^2\left(x+7\right)}\end{array}$

Thus,

  $\frac{x^2+9x+20}{x^2+5x-24}÷\frac{x^2+15x+56}{x^2+x-12}=\frac{{\left(x+4\right)}^2\left(x+5\right)}{{\left(x+8\right)}^2\left(x+7\right)}$