Chapter 6.2, Problem 57WE

To determine

To calculate: The simplified form of the expression $\frac{x^3+3x^2-4x-12}{2x^2-18}\cdot \frac{x^3-3x^2+3x-9}{3x^3-12x}$.

Answer to Problem 57WE

The simplified form of the expression $\frac{x^3+3x^2-4x-12}{2x^2-18}\cdot \frac{x^3-3x^2+3x-9}{3x^3-12x}$ is $\frac{x^2+3}{6x}$.

Explanation of Solution

Given information:

The expression $\frac{x^3+3x^2-4x-12}{2x^2-18}\cdot \frac{x^3-3x^2+3x-9}{3x^3-12x}$.

Formula used:

Fractions are multiplied as $\frac{a}{b}\cdot \frac{c}{d}=\frac{ab}{cd}$.

To factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

Calculation:

Consider the expression $\frac{x^3+3x^2-4x-12}{2x^2-18}\cdot \frac{x^3-3x^2+3x-9}{3x^3-12x}$.

Take the common terms from the numerator of both the terms,

  $\begin{array}{c}\frac{x^3+3x^2-4x-12}{2x^2-18}\cdot \frac{x^3-3x^2+3x-9}{3x^3-12x}=\frac{x^2\left(x+3\right)-4\left(x+3\right)}{2x^2-18}\cdot \frac{x^2\left(x-3\right)+3\left(x-3\right)}{3x^3-12x}\\ =\frac{\left(x+3\right)\left(x^2-4\right)}{2x^2-18}\cdot \frac{\left(x-3\right)\left(x^2+3\right)}{3x^3-12x}\end{array}$

Take the common terms from the denominator of both the terms,

  $\begin{array}{c}\frac{x^3+3x^2-4x-12}{2x^2-18}\cdot \frac{x^3-3x^2+3x-9}{3x^3-12x}=\frac{x^2\left(x+3\right)-4\left(x+3\right)}{2x^2-18}\cdot \frac{x^2\left(x-3\right)+3\left(x-3\right)}{3x^3-12x}\\ =\frac{\left(x+3\right)\left(x^2-4\right)}{2\left(x^2-9\right)}\cdot \frac{\left(x-3\right)\left(x^2+3\right)}{3x\left(x^2-4\right)}\end{array}$

Recall that square of difference of two numbers a and bis ${\left(a-b\right)}^2=a^2+b^2-2ab$.

Therefore,

  $\begin{array}{c}\frac{x^3+3x^2-4x-12}{2x^2-18}\cdot \frac{x^3-3x^2+3x-9}{3x^3-12x}=\frac{x^2\left(x+3\right)-4\left(x+3\right)}{2x^2-18}\cdot \frac{x^2\left(x-3\right)+3\left(x-3\right)}{3x^3-12x}\\ =\frac{\left(x+3\right)\left(x^2-4\right)}{2\left(x^2-9\right)}\cdot \frac{\left(x-3\right)\left(x^2+3\right)}{3x\left(x^2-4\right)}\\ =\frac{\left(x+3\right)\left(x^2-4\right)}{2\left(x^2-3^2\right)}\cdot \frac{\left(x-3\right)\left(x^2+3\right)}{3x\left(x^2-4\right)}\\ =\frac{\left(x+3\right)\left(x^2-4\right)}{2\left(x-3\right)\left(x+3\right)}\cdot \frac{\left(x-3\right)\left(x^2+3\right)}{3x\left(x^2-4\right)}\end{array}$

Strike off the similar terms,

  $\begin{array}{c}\frac{x^3+3x^2-4x-12}{2x^2-18}\cdot \frac{x^3-3x^2+3x-9}{3x^3-12x}=\frac{\left(x+3\right)\left(x^2-4\right)}{2\left(x^2-3^2\right)}\cdot \frac{\left(x-3\right)\left(x^2+3\right)}{3x\left(x^2-4\right)}\\ =\frac{\left(x+3\right)\left(x^2-4\right)}{2\left(x-3\right)\left(x+3\right)}\cdot \frac{\left(x-3\right)\left(x^2+3\right)}{3x\left(x^2-4\right)}\\ =\frac{1}{2}\cdot \frac{\left(x^2+3\right)}{3x}\end{array}$

Recall that fractions are multiplied as $\frac{a}{b}\cdot \frac{c}{d}=\frac{ab}{cd}$.

Apply it,

  $\begin{array}{c}\frac{x^3+3x^2-4x-12}{2x^2-18}\cdot \frac{x^3-3x^2+3x-9}{3x^3-12x}=\frac{\left(x+3\right)\left(x^2-4\right)}{2\left(x^2-3^2\right)}\cdot \frac{\left(x-3\right)\left(x^2+3\right)}{3x\left(x^2-4\right)}\\ =\frac{\left(x+3\right)\left(x^2-4\right)}{2\left(x-3\right)\left(x+3\right)}\cdot \frac{\left(x-3\right)\left(x^2+3\right)}{3x\left(x^2-4\right)}\\ =\frac{1}{2}\cdot \frac{\left(x^2+3\right)}{3x}\\ =\frac{x^2+3}{6x}\end{array}$

Thus, the simplified form of the expression $\frac{x^3+3x^2-4x-12}{2x^2-18}\cdot \frac{x^3-3x^2+3x-9}{3x^3-12x}$ is $\frac{x^2+3}{6x}$.