Question

q6 simplify the expression. \\(\frac{4b^2 - 16}{b + 2}\\) \\(\frac{2(b^2 - 4)}{b}\\) \\(4(b - 2)\\) \\(4(b + 2)\\) \\(2(b - 8)\\)

Explanation:

Step1: Factor the numerator

First, factor out the greatest common factor from the numerator \(4b^2 - 16\). The GCF of \(4b^2\) and \(16\) is \(4\), so we can write \(4b^2 - 16 = 4(b^2 - 4)\). Then, notice that \(b^2 - 4\) is a difference of squares, which factors as \((b - 2)(b + 2)\). So the numerator becomes \(4(b - 2)(b + 2)\).
\[
\frac{4b^2 - 16}{b + 2}=\frac{4(b^2 - 4)}{b + 2}=\frac{4(b - 2)(b + 2)}{b + 2}
\]

Step2: Cancel common factors

Now, we can cancel out the common factor of \(b + 2\) from the numerator and the denominator (assuming \(b
eq - 2\) to avoid division by zero).
\[
\frac{4(b - 2)(b + 2)}{b + 2}=4(b - 2)
\]

Answer:

\(4(b - 2)\) (the option with this expression)