## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{6(x^2-x+1)}{x(2x-1)}$
By factoring both the denominator and numerator of each rational function and cancelling the common factors, we have $$\frac{12}{x^2+x}\cdot \frac{x^3+1}{4x-2} = \frac{12}{x(x+1)}\cdot \frac{(x+1)(x^2-x+1)}{2(2x-1)}\\ = \frac{6(x^2-x+1)}{x(2x-1)} .$$