Question

write the polynomial expression in simplest form: $\frac{x^2 - 9}{x^2 - 4} \cdot \frac{4x - 8}{12x + 36}$. (1 point) \\(\frac{(x - 3)(x - 2)}{3(x + 2)(x + 2)}\\) \\(\frac{x - 3}{3(x + 2)}\\) \\(\frac{1}{3}\\) \\(\frac{2(x - 3)}{6(x + 2)}\\)

Explanation:

Step1: Factor each polynomial

Factor \(x^2 - 9\) using difference of squares: \(x^2 - 9=(x - 3)(x + 3)\)
Factor \(x^2 - 4\) using difference of squares: \(x^2 - 4=(x - 2)(x + 2)\)
Factor \(4x - 8\) by taking out GCF: \(4x - 8 = 4(x - 2)\)
Factor \(12x + 36\) by taking out GCF: \(12x + 36=12(x + 3)=3\times4(x + 3)\)

So the expression becomes:
\(\frac{(x - 3)(x + 3)}{(x - 2)(x + 2)}\cdot\frac{4(x - 2)}{12(x + 3)}\)

Step2: Cancel common factors

Cancel \((x + 3)\) from numerator and denominator:
\(\frac{(x - 3)\cancel{(x + 3)}}{(x - 2)(x + 2)}\cdot\frac{4(x - 2)}{12\cancel{(x + 3)}}\)

Cancel \((x - 2)\) from numerator and denominator:
\(\frac{(x - 3)\cancel{(x + 3)}}{\cancel{(x - 2)}(x + 2)}\cdot\frac{4\cancel{(x - 2)}}{12\cancel{(x + 3)}}\)

Simplify the constants: \(\frac{4}{12}=\frac{1}{3}\)

Now we have: \(\frac{(x - 3)}{(x + 2)}\cdot\frac{1}{3}=\frac{x - 3}{3(x + 2)}\)

Answer:

\(\frac{x - 3}{3(x + 2)}\) (corresponding to the option \(\boldsymbol{\frac{x - 3}{3(x + 2)}}\))