Question

multiply and simplify.\\(
\frac{y^{2} - 36}{y^{2} + 12y + 36} \cdot \frac{3y^{2} - 8y - 3}{y^{2} - 9y + 18}\\)
\\(\square\\), \\(y \
eq -6, 6, 3\\)
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Explanation:

Step1: Factor each quadratic expression

• For \(y^2 - 36\), use difference of squares: \(y^2 - 36=(y - 6)(y + 6)\)
• For \(y^2 + 12y + 36\), use perfect square trinomial: \(y^2 + 12y + 36=(y + 6)^2\)
• For \(3y^2 - 8y - 3\), factor by grouping: \(3y^2 - 8y - 3 = 3y^2 - 9y + y - 3 = 3y(y - 3)+1(y - 3)=(3y + 1)(y - 3)\)
• For \(y^2 - 9y + 18\), factor: \(y^2 - 9y + 18=(y - 3)(y - 6)\)

So the expression becomes: \(\frac{(y - 6)(y + 6)}{(y + 6)^2} \cdot \frac{(3y + 1)(y - 3)}{(y - 3)(y - 6)}\)

Step2: Cancel out common factors

• Cancel \((y - 6)\) from numerator and denominator.
• Cancel \((y + 6)\) from numerator and denominator.
• Cancel \((y - 3)\) from numerator and denominator.

After canceling, we have: \(\frac{3y + 1}{y + 6}\)

Answer:

\(\frac{3y + 1}{y + 6}\)