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

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

Factor

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

Solution Steps

Factor out 3.

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

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

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

Consider -a^{2}+4a-4. Factor the expression by grouping. First, the expression needs to be rewritten as -a^{2}+pa+qa-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 -a^{2}+4a-4 as \left(-a^{2}+2a\right)+\left(2a-4\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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