Factor -3bk 2 + 9bk - 6b.

a: -3b(k + 1)(k - 2)
b: -3b(k - 1)(k + 2)
c: -3b(k - 1)(k - 2)

1. Find the greatest common factor (GCF)
What is the largest number that divides evenly into 3bk^2, -9bk, and 6b?
It is 3.
What is the highest degree of b that divides evenly into 3bk^2, -9bk, and 6b?
It is b.
What is the highest degree of k that divides evenly into 3bk^2, -9bk, and 6b?
It is 1, since k is not in every term.
Multiplying the results above,
The GCF is 3b

2. Factor out the GCF (Write the GCF first. Then, in parentheses, divide each term by the GCF).
3b(3bk^2/3b + -9bk/3b + 6b/3b)

3. Simplify each term in the parentheses
-3b(k^2-3k+2)

4. Factor k^2-3k+2
Ask: Which two numbers add up to -3 and multiply to 2?
-2 and -1
Rewrite the expression using the above
(k-2)(k-1)

-3b(k-2)(k-1)

Your answer is B, have a nice day :D

Distributive property