Chapter A.3, Problem 41E

Solution

Write the expression in reduced form.

  $\frac{\left(4z^2+2z+1\right)}{(z+3)}$ .

Given:

The given expression is $\frac{8z^3-1}{2z^2+5z-3}$ .

Concept Used:

• Factor the numerator and the denominator.
• List restricted values.
• Cancel common factors.
• Reduce to lowest terms and note any restricted values not implied by the expression.

Calculation:

The given expression is $\frac{8z^3-1}{2z^2+5z-3}$ .

  $\begin{array}{c}\frac{8z^3-1}{2z^2+5z-3}=\frac{{(2z)}^3-{(1)}^3}{2z^2+5z-3}\\ =\frac{(2z-1)\left({(2z)}^2+2z\cdot 1+1^2\right)}{2z^2+5z-3}\\ =\frac{(2z-1)\left(4z^2+2z+1\right)}{2z^2+5z-3}\\ solve2z^2+5z-3=2z^2+(6-1)z-3\\ =2z^2+6z-z-3\\ =2z(z+3)-1(z+3)\\ =(2z-1)(z+3)\\ putthevalueabove\\ \frac{8z^3-1}{2z^2+5z-3}=\frac{(2z-1)\left(4z^2+2z+1\right)}{2z^2+5z-3}\\ =\frac{(2z-1)\left(4z^2+2z+1\right)}{(2z-1)(z+3)}\\ =\frac{\left(4z^2+2z+1\right)}{(z+3)}\end{array}$

Thus the reduced form of $\frac{8z^3-1}{2z^2+5z-3}$ is $\frac{\left(4z^2+2z+1\right)}{(z+3)}$ .