How would I divide $\frac{4x^2-2x}{x^2+5x+4}\div \frac{2x}{x^2+2x+1}$?

Question

How would I divide $\frac{4x^2-2x}{x^2+5x+4}\div \frac{2x}{x^2+2x+1}$? Work would be appreciated. What would I do first?

Answer 1

Note first that $$ \frac{4x^2-2x}{x^2+5x+4}\div \frac{2x}{x^2+2x+1} = \frac{4x^2-2x}{x^2+5x+4} \times \frac{x^2+2x+1}{2x}. $$ Now, we need to do some factorization, i.e. $$ 4x^2-2x=2x(2x-1) $$ $$ x^2+5x+4=(x+1)(x+4) $$ $$ x^2+2x+1=(x+1)(x+1) $$ which gives rise to $$ \frac{4x^2-2x}{x^2+5x+4}\div \frac{2x}{x^2+2x+1} = \frac{2x(2x-1)}{(x+1)(x+4)} \times \frac{(x+1)(x+1)}{2x} $$ and after cancellations, you would be good to go.

Answer 2

HINT: Factorise the expression. It's the same as $\frac{4x^2-2x}{2x} × \frac{x^2+2x+1}{x^2+5x+4}$