Question

simplify.
\\(\frac{q^2 - 16x^2}{q^2 + 8qx + 16x^2}\\)
\\(\frac{q^2 - 16x^2}{q^2 + 8qx + 16x^2} = \square\\) (simplify your answer.)

Explanation:

Step1: Factor numerator and denominator

The numerator \( q^2 - 16x^2 \) is a difference of squares, so it factors as \( (q - 4x)(q + 4x) \) (since \( a^2 - b^2=(a - b)(a + b) \), here \( a = q \), \( b = 4x \) as \( (4x)^2=16x^2 \)).

The denominator \( q^2 + 8qx + 16x^2 \) is a perfect square trinomial, which factors as \( (q + 4x)^2 \) (since \( (a + b)^2=a^2 + 2ab + b^2 \), here \( a = q \), \( b = 4x \) as \( 2\times q\times4x = 8qx \) and \( (4x)^2 = 16x^2 \)).

So the expression becomes \( \frac{(q - 4x)(q + 4x)}{(q + 4x)^2} \).

Step2: Cancel common factors

We can cancel out the common factor \( (q + 4x) \) from the numerator and the denominator (assuming \( q + 4x
eq0 \) to avoid division by zero).

After canceling, we get \( \frac{q - 4x}{q + 4x} \).

Answer:

\( \frac{q - 4x}{q + 4x} \)