Solve b^3-5b^2-36b

$$b ^ { 3 } - 5 b ^ { 2 } - 36 b$$

Factor

$$b\left(b-9\right)\left(b+4\right)$$

Solution Steps

Factor out b.

$$b\left(b^{2}-5b-36\right)$$

Consider b^{2}-5b-36. Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb-36. To find p and q, set up a system to be solved.

$$p+q=-5 pq=1\left(-36\right)=-36$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$p=-9 q=4$$

Rewrite b^{2}-5b-36 as \left(b^{2}-9b\right)+\left(4b-36\right).

$$\left(b^{2}-9b\right)+\left(4b-36\right)$$

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

$$b\left(b-9\right)+4\left(b-9\right)$$

Factor out common term b-9 by using distributive property.

$$\left(b-9\right)\left(b+4\right)$$

Rewrite the complete factored expression.

$$b\left(b-9\right)\left(b+4\right)$$

Evaluate

$$b\left(b-9\right)\left(b+4\right)$$