Question

identify the excluded values of this product. then rewrite the product in simplest form.\\(\frac{6y^2 + 18y - 60}{3y^3 - 12y} cdot \frac{y^2 - 16}{y^2 + 2y - 8}\\)type the correct answer in the box. do not type the excluded values.

Explanation:

Step1: Factor all polynomials

First numerator: $6y^2 + 18y - 60 = 6(y^2 + 3y - 10) = 6(y+5)(y-2)$
First denominator: $3y^2 - 12y = 3y(y - 4)$
Second numerator: $y^2 - 16 = (y+4)(y-4)$
Second denominator: $y^2 + 2y - 8 = (y+4)(y-2)$

Substitute into the product:
$$\frac{6(y+5)(y-2)}{3y(y-4)} \cdot \frac{(y+4)(y-4)}{(y+4)(y-2)}$$

Step2: Cancel common factors

Cancel $3$ (from $6/3$), $(y-2)$, $(y-4)$, and $(y+4)$:
$$\frac{2(y+5)}{y} \cdot \frac{1}{1} = \frac{2(y+5)}{y}$$

Step3: Simplify the expression

Distribute the 2 in the numerator:
$$\frac{2y + 10}{y}$$

Answer:

$\frac{2y+10}{y}$ (or $\frac{2(y+5)}{y}$ is also acceptable)