Question

what is the product? state any restrictions.
\\(\frac{y^2}{y - 3} \cdot \frac{y^2 - y - 6}{y^2 + y}\\)
a \\(\frac{y + 2}{y + 1}, y \
eq 3, -1\\)
b \\(\frac{y + 2}{y + 1}, y \
eq 3, 0, -1\\)
c \\(\frac{y^2 + 2y}{y + 1}, y \
eq 3, -1\\)
d \\(\frac{y^2 + 2y}{y + 1}, y \
eq 3, 0, -1\\)

Explanation:

Step1: Factor the expressions

Factor \(y^2 - y - 6\) as \((y - 3)(y + 2)\) and \(y^2 + y\) as \(y(y + 1)\). So the expression becomes \(\frac{y^2}{y - 3} \cdot \frac{(y - 3)(y + 2)}{y(y + 1)}\).

Step2: Cancel common factors

Cancel \(y - 3\) from numerator and denominator, and cancel one \(y\) from \(y^2\) and \(y\). We get \(\frac{y(y + 2)}{y + 1}\), which simplifies to \(\frac{y^2 + 2y}{y + 1}\).

Step3: Determine restrictions

For the original fractions, denominators \(y - 3\), \(y^2 + y = y(y + 1)\) cannot be zero. So \(y
eq 3\), \(y
eq 0\), \(y
eq - 1\).

Answer:

D. \(\frac{y^2 + 2y}{y + 1}, y
eq 3, 0, -1\)