Question

1. 16b² + 60b - 100

Explanation:

Step1: Factor out the greatest - common factor

First, find the GCF of 16, 60, and 100. The GCF of 16, 60, and 100 is 4. So, $16b^{2}+60b - 100=4(4b^{2}+15b - 25)$.

Step2: Factor the quadratic expression inside the parentheses

For the quadratic $4b^{2}+15b - 25$, we need to find two numbers that multiply to $4\times(-25)=-100$ and add up to 15. The numbers are 20 and - 5.
Rewrite the middle term: $4b^{2}+20b-5b - 25$.
Group the terms: $(4b^{2}+20b)-(5b + 25)$.
Factor each group: $4b(b + 5)-5(b + 5)$.
Then, $(4b - 5)(b + 5)$.
So, $16b^{2}+60b - 100=4(4b - 5)(b + 5)$.

Answer:

$4(4b - 5)(b + 5)$