Solve 12b^3+33b^2-60b

$$12 b ^ { 3 } + 33 b ^ { 2 } - 60 b$$

Factor

$$3b\left(4b-5\right)\left(b+4\right)$$

Solution Steps

Factor out 3.

$$3\left(4b^{3}+11b^{2}-20b\right)$$

Consider 4b^{3}+11b^{2}-20b. Factor out b.

$$b\left(4b^{2}+11b-20\right)$$

Consider 4b^{2}+11b-20. Factor the expression by grouping. First, the expression needs to be rewritten as 4b^{2}+pb+qb-20. To find p and q, set up a system to be solved.

$$p+q=11 pq=4\left(-20\right)=-80$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 11.

$$p=-5 q=16$$

Rewrite 4b^{2}+11b-20 as \left(4b^{2}-5b\right)+\left(16b-20\right).

$$\left(4b^{2}-5b\right)+\left(16b-20\right)$$

Factor out b in the first and 4 in the second group.

$$b\left(4b-5\right)+4\left(4b-5\right)$$

Factor out common term 4b-5 by using distributive property.

$$\left(4b-5\right)\left(b+4\right)$$

Rewrite the complete factored expression.

$$3b\left(4b-5\right)\left(b+4\right)$$

Evaluate

$$3b\left(4b-5\right)\left(b+4\right)$$