Chapter A.3, Problem 50E

Solution

Simplify.

  $\frac{1}{y}$ .

Given:

The given fraction is $\frac{y^3+2y^2+4y}{y^3+2y^2}\cdot \frac{y^2-4}{y^3-8}$ .

Concept Used:

• Remove any grouping symbol such as brackets and parentheses by multiplying factors.
• Use the exponent rule to remove grouping if the terms are containing exponents.
• Combine the like terms by addition or subtraction.
• Combine the constants.

Calculation:

The given fraction is $\frac{y^3+2y^2+4y}{y^3+2y^2}\cdot \frac{y^2-4}{y^3-8}$ .

  $\begin{array}{c}\frac{y^3+2y^2+4y}{y^3+2y^2}\cdot \frac{y^2-4}{y^3-8}=\frac{y(y^2+2y+4)}{y^2(y+2)}\cdot \frac{y^2-2^2}{y^3-2^3}\\ =\frac{(y^2+2y+4)}{y(y+2)}\cdot \frac{(y+2)(y-2)}{(y-2)(y^2+y\cdot 2+2^2)}\\ =\frac{(y^2+2y+4)}{y(y+2)}\cdot \frac{(y+2)(y-2)}{(y-2)(y^2+2y+4)}\\ =\frac{(y^2+2y+4)(y+2)(y-2)}{y(y^2+2y+4)(y+2)(y-2)}\\ =\frac{1}{y}\end{array}$

Thus by simplifying $\frac{y^3+2y^2+4y}{y^3+2y^2}\cdot \frac{y^2-4}{y^3-8}$ to a single fraction it comes $\frac{1}{y}$ .