Chapter A.5, Problem 16AYU

To determine

To find: Simplify the given in factored form.

Answer to Problem 16AYU

$\frac{6(x^2+x+1)}{x\left(2x-1\right)}$

Explanation of Solution

Given:

$\frac{12}{x^2+x}\cdot \frac{x^3+1}{\left(4x-2\right)}$

Calculation:

$\frac{12}{x^2+x}\cdot \frac{x^3+1}{\left(4x-2\right)}=\frac{12\left(x+1\right)(x^2+x+1)}{4x^3+2x^2-2x}=\frac{12\left(x+1\right)(x^2+x+1)}{\left(x+1\right)2x\left(2x-1\right)}=\frac{6(x^2+x+1)}{x\left(2x-1\right)}$