Chapter A.3, Problem 52E

Solution

Simplify.

  $\frac{y(y+4)}{(y-1)}$ .

Given:

The given fraction is $\frac{y^2+8y+16}{3y^2-y-2}\cdot \frac{3y^2+2y}{y+4}$ .

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 $\frac{y^2+8y+16}{3y^2-y-2}\cdot \frac{3y^2+2y}{y+4}$ .

  $\begin{array}{c}\frac{y^2+8y+16}{3y^2-y-2}\cdot \frac{3y^2+2y}{y+4}=\frac{y^2+8y+16}{3y^2-y-2}\cdot \frac{y(3y+2)}{y+4}\\ solve{y}^2+8y+16=y^2+(4+4)y+16\\ =y^2+4y+4y+16\\ =y(y+4)+4(y+4)\\ =(y+4)(y+4)\\ solve3y^2-y-2=3y^2-(3-2)y-2\\ =3y^2-3y+2y-2\\ =3y(y-1)+2(y-1)\\ =(3y+2)(y-1)\\ putthevaluesabove\\ \frac{y^2+8y+16}{3y^2-y-2}\cdot \frac{y(3y+2)}{y+4}=\frac{(y+4)(y+4)}{(3y+2)(y-1)}\cdot \frac{y(3y+2)}{y+4}\\ =\frac{y(y+4)(y+4)(3y+2)}{(y+4)(3y+2)(y-1)}\\ =\frac{y(y+4)}{(y-1)}\end{array}$

Thus by simplifying $\frac{y^2+8y+16}{3y^2-y-2}\cdot \frac{3y^2+2y}{y+4}$ to a single fraction it comes $\frac{y(y+4)}{(y-1)}$ .