Question

simplify by removing factors of 1.\\(
\frac{q^{2}-1}{q^{2}-2q + 1}\\)\\(
\frac{q^{2}-1}{q^{2}-2q + 1}=\square\\)

Explanation:

Step1: Factor numerator and denominator

The numerator \( q^2 - 1 \) is a difference of squares, so it factors as \( (q - 1)(q + 1) \). The denominator \( q^2 - 2q + 1 \) is a perfect square trinomial, factoring as \( (q - 1)^2 \).
So we have \(\frac{(q - 1)(q + 1)}{(q - 1)^2}\).

Step2: Remove factor of 1

We can cancel out one \( (q - 1) \) from the numerator and the denominator (since \(\frac{q - 1}{q - 1} = 1\) for \( q
eq 1 \)).
This gives us \(\frac{q + 1}{q - 1}\).

Answer:

\(\frac{q + 1}{q - 1}\)