Chapter 6.2, Problem 40WE

To determine

To calculate: The simplified form of the expression $\frac{2y}{3y^2+15y}\cdot \left(3y^3-75y\right)$.

Answer to Problem 40WE

The simplified form of the expression $\frac{2y}{3y^2+15y}\cdot \left(3y^3-75y\right)$ is $2y\left(y-5\right)$.

Explanation of Solution

Given information:

The expression $\frac{2y}{3y^2+15y}\cdot \left(3y^3-75y\right)$.

Formula used:

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

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

Calculation:

Consider the expression $\frac{2y}{3y^2+15y}\cdot \left(3y^3-75y\right)$.

Take out the common factor that is $3y$ from the second term and from the denominator of first term,

  $\begin{array}{c}\frac{2y}{3y^2+15y}\cdot \left(3y^3-75y\right)=\frac{2y}{3y^2+15y}\cdot 3y\left(y^2-25\right)\\ =\frac{2y}{3y\left(y+5\right)}\cdot 3y\left(y^2-25\right)\end{array}$

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

Apply it in the second term,

  $\begin{array}{c}\frac{2y}{3y^2+15y}\cdot \left(3y^3-75y\right)=\frac{2y}{3y^2+15y}\cdot 3y\left(y^2-25\right)\\ =\frac{2y}{3y\left(y+5\right)}\cdot 3y\left(y^2-5^2\right)\\ =\frac{2y}{3y\left(y+5\right)}\cdot 3y\left(y-5\right)\left(y+5\right)\end{array}$

Strike off the similar terms,

  $\begin{array}{c}\frac{2y}{3y^2+15y}\cdot \left(3y^3-75y\right)=\frac{2y}{3y^2+15y}\cdot 3y\left(y^2-25\right)\\ =\frac{2y}{3y\left(y+5\right)}\cdot 3y\left(y^2-5^2\right)\\ =\frac{2y}{3y\left(y+5\right)}\cdot 3y\left(y-5\right)\left(y+5\right)\\ =\frac{2}{1}\cdot y\left(y-5\right)\end{array}$

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

Apply it,

  $\begin{array}{c}\frac{2y}{3y^2+15y}\cdot \left(3y^3-75y\right)=\frac{2y}{3y\left(y+5\right)}\cdot 3y\left(y-5\right)\left(y+5\right)\\ =\frac{2}{1}\cdot y\left(y-5\right)\\ =2y\left(y-5\right)\end{array}$

Thus, the simplified form of the expression $\frac{2y}{3y^2+15y}\cdot \left(3y^3-75y\right)$ is $2y\left(y-5\right)$.