Question

Divide:$\frac{y^2-6y+8}{y^2-4y}÷\left(3y^2-12y\right)$.

Short answer

The required answer is$\frac{y-2}{3y^2\left(y-4\right)}$.

Step-by-step solution

Step 1. Given information.

The given expression is:

$\frac{y^2-6y+8}{y^2-4y}÷\left(3y^2-12y\right)$

Step 2. Simplify the expression.

$\frac{y^2-6y+8}{y^2-4y}÷\left(3y^2-12y\right)$

Rewrite the division as the product of the first rational expression and the reciprocal of the second.

$\frac{y^2-6y+8}{y^2-4y}·\frac{1}{3y^2-12y}$

Factor the numerators and denominators and then multiply.

$$\begin{gathered} \frac{\left(y-2\right)\left(y-4\right)}{y\left(y-4\right)}·\frac{1}{3y\left(y-4\right)} \\ \frac{\left(y-2\right)\left(y-4\right)}{3y·y\left(y-4\right)\left(y-4\right)} \end{gathered}$$

Simplify by dividing out common factors.

$\frac{y-2}{3y^2\left(y-4\right)}$