Chapter 5, Problem 22REAE

Solution

To calculate: The simplified form of $\frac{x^2-13x+40}{x^2-2x-15}÷\left(x^2-5x-24\right)$.

The simplified form is $\frac{1}{\left(x+3\right)\left(x+3\right)}$.

Given information:

The expression is $\frac{x^2-13x+40}{x^2-2x-15}÷\left(x^2-5x-24\right)$.

Calculation:

Given expression is $\frac{x^2-13x+40}{x^2-2x-15}÷\left(x^2-5x-24\right)$.

Multiply by reciprocal.

  $\frac{x^2-13x+40}{x^3-2x-15}\times \frac{1}{x^2-5x-24}$

Factor the numerators and denominators.

  $\begin{array}{l}\frac{x^2-5x-8x+40}{x^2-5x+3x-15}\times \frac{1}{x^2-8x+3x-24}\\ =\frac{x\left(x-5\right)-8\left(x-5\right)}{x\left(x-5\right)+3\left(x-5\right)}\times \frac{1}{x\left(x-8\right)+3\left(x-8\right)}\\ =\frac{\left(x-5\right)\left(x-8\right)}{\left(x-5\right)\left(x+3\right)}\times \frac{1}{\left(x-8\right)\left(x+3\right)}\end{array}$

Multiply the numerators and denominators.

  $\frac{\left(x-5\right)\left(x-8\right)}{\left(x-5\right)\left(x+3\right)\left(x-8\right)\left(x+3\right)}$

Divide out the common factors.

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

Therefore, the simplified form is $\frac{1}{\left(x+3\right)\left(x+3\right)}$.