Question

perform the multiplication.
\\(\frac{x^2 - 16}{6} cdot \frac{x^2 - x - 12}{x^2 - 8x + 16}\\)
\\(\frac{x^2 - 16}{6} cdot \frac{x^2 - x - 12}{x^2 - 8x + 16} = square\\)

Explanation:

Step1: Factor each expression

First, factor the numerators and denominators:

• \(x^2 - 16\) is a difference of squares: \(x^2 - 16=(x + 4)(x - 4)\)
• \(x^2 - x - 12\) factors to \((x - 4)(x + 3)\) (by finding two numbers that multiply to -12 and add to -1: -4 and 3)
• \(x^2 - 8x + 16\) is a perfect square trinomial: \(x^2 - 8x + 16=(x - 4)^2\)

So the expression becomes:
\[
\frac{(x + 4)(x - 4)}{6} \cdot \frac{(x - 4)(x + 3)}{(x - 4)^2}
\]

Step2: Cancel common factors

Now, cancel out the common factors in the numerators and denominators:

• We can cancel one \((x - 4)\) from the numerator of the first fraction and the denominator of the second fraction.
• We can also cancel one \((x - 4)\) from the numerator of the second fraction and the denominator of the second fraction (since \((x - 4)^2=(x - 4)(x - 4)\)).

After canceling, we have:
\[
\frac{(x + 4)}{6} \cdot \frac{(x + 3)}{(x - 4)}
\]

Step3: Multiply the remaining factors

Multiply the numerators together and the denominators together:
\[
\frac{(x + 4)(x + 3)}{6(x - 4)}
\]

Expand the numerator: \((x + 4)(x + 3)=x^2 + 3x + 4x + 12=x^2 + 7x + 12\)

So the simplified expression is:
\[
\frac{x^2 + 7x + 12}{6(x - 4)}
\]
Or we can also write it as \(\frac{(x + 3)(x + 4)}{6(x - 4)}\) (either form is correct, but expanding the numerator is also acceptable)

Answer:

\(\frac{(x + 3)(x + 4)}{6(x - 4)}\) (or \(\frac{x^2 + 7x + 12}{6x - 24}\) after expanding the denominator, but the factored form is more simplified)