Solve 6ab^3-33ab^2+36ab

$$6 a b ^ { 3 } - 33 a b ^ { 2 } + 36 a b$$

Factor

$$3ab\left(b-4\right)\left(2b-3\right)$$

Solution Steps

Factor out 3.

$$3\left(2ab^{3}-11ab^{2}+12ab\right)$$

Consider 2ab^{3}-11ab^{2}+12ab. Factor out ab.

$$ab\left(2b^{2}-11b+12\right)$$

Consider 2b^{2}-11b+12. Factor the expression by grouping. First, the expression needs to be rewritten as 2b^{2}+pb+qb+12. To find p and q, set up a system to be solved.

$$p+q=-11 pq=2\times 12=24$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$p=-8 q=-3$$

Rewrite 2b^{2}-11b+12 as \left(2b^{2}-8b\right)+\left(-3b+12\right).

$$\left(2b^{2}-8b\right)+\left(-3b+12\right)$$

Factor out 2b in the first and -3 in the second group.

$$2b\left(b-4\right)-3\left(b-4\right)$$

Factor out common term b-4 by using distributive property.

$$\left(b-4\right)\left(2b-3\right)$$

Rewrite the complete factored expression.

$$3ab\left(b-4\right)\left(2b-3\right)$$

Evaluate

$$3ab\left(b-4\right)\left(2b-3\right)$$