What is the product? StartFraction 2 y Over y minus 3 EndFraction divided by StartFraction 4 y minus 12 Over 2 y + 6 EndFraction

(y^2 + 3y) / (y^2 - 6y + 9)

Factored form: [y*(y+3)] / [(y-3)^2]

Step-by-step explanation:

We need to find the following product:

(2y / y-3) / ((4y-12) / (2y+6))

We have a division of fractions. We can transform this in a product, where we have the first fraction multiplied by the inverse of the second fraction. So, we have that:

(2y / y-3) * ((2y+6) / (4y-12))

= 2y*(2y+6) / ((y-3)*(4y-12))

= [2y * 2 * (y+3)] / [4 * (y-3) * (y-3)]

= [y*(y+3)] / [(y-3) * (y-3)]

= (y^2 + 3y) / (y^2 - 6y + 9)

when studying polynomials, you often hear the terms zeros, roots, factors and x x the multiplicity of a zero is important because it tells us how the graph of the let's look at the graph of a function that has the same zeros

step-by-step explanation:

2 pecks

step-by-step explanation: