Solve -3ab^3+12ab^2-12ab

$$- 3 a b ^ { 3 } + 12 a b ^ { 2 } - 12 a b$$

Factor

$$-3ab\left(2-b\right)^{2}$$

Solution Steps

Factor out 3.

$$3\left(-ab^{3}+4ab^{2}-4ab\right)$$

Consider -ab^{3}+4ab^{2}-4ab. Factor out ab.

$$ab\left(-b^{2}+4b-4\right)$$

Consider -b^{2}+4b-4. Factor the expression by grouping. First, the expression needs to be rewritten as -b^{2}+pb+qb-4. To find p and q, set up a system to be solved.

$$p+q=4 pq=-\left(-4\right)=4$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$p=2 q=2$$

Rewrite -b^{2}+4b-4 as \left(-b^{2}+2b\right)+\left(2b-4\right).

$$\left(-b^{2}+2b\right)+\left(2b-4\right)$$

Factor out -b in the first and 2 in the second group.

$$-b\left(b-2\right)+2\left(b-2\right)$$

Factor out common term b-2 by using distributive property.

$$\left(b-2\right)\left(-b+2\right)$$

Rewrite the complete factored expression.

$$3ab\left(b-2\right)\left(-b+2\right)$$

Evaluate

$$-3ab\left(2-b\right)^{2}$$