Chapter 6.2, Problem 59WE

To determine

To calculate: The values of x such that the expression $\frac{2x^3-2x}{3x^3}\cdot \frac{9}{x^2+3x+2}$ is equal to 0.

Answer to Problem 59WE

The value of x such that the expression $\frac{2x^3-2x}{3x^3}\cdot \frac{9}{x^2+3x+2}$ is equal to 0is $1$.

Explanation of Solution

Given information:

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

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 $\frac{2x^3-2x}{3x^3}\cdot \frac{9}{x^2+3x+2}$.

Take the trinomial $x^2+3x+2$.

Recall that 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.

Rewrite the trinomial and take out the common terms,

  $\begin{array}{c}{x}^2+3x+2=x^2+2x+x+2\\ =x\left(x+2\right)+1\left(x+2\right)\\ =\left(x+2\right)\left(x+1\right)\end{array}$

Take the term $2x$ common from numerator of first term.

Rewrite the original expression as,

  $\frac{2x^3-2x}{3x^3}\cdot \frac{9}{x^2+3x+2}=\frac{2x\left(x^2-1\right)}{3x^3}\cdot \frac{9}{\left(x+2\right)\left(x+1\right)}$

Take the binomial $x^2-1$.

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

Apply it,

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

Strike off the similar terms,

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

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

Apply it,

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

Now, equate the result obtained to 0.

  $\begin{array}{c}\frac{6\left(x-1\right)}{x^2\left(x+3\right)}=0\\ 6\left(x-1\right)=0\\ x-1=0\\ x=1\end{array}$

Thus, the value of x such that the expression $\frac{2x^3-2x}{3x^3}\cdot \frac{9}{x^2+3x+2}$ is equal to 0 is $1$.