Solve b^3-3b^2-40b

$$b ^ { 3 } - 3 b ^ { 2 } - 40 b$$

Factor

$$b\left(b-8\right)\left(b+5\right)$$

Solution Steps

Factor out b.

$$b\left(b^{2}-3b-40\right)$$

Consider b^{2}-3b-40. Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb-40. To find p and q, set up a system to be solved.

$$p+q=-3 pq=1\left(-40\right)=-40$$

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 -40.

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

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -3.

$$p=-8 q=5$$

Rewrite b^{2}-3b-40 as \left(b^{2}-8b\right)+\left(5b-40\right).

$$\left(b^{2}-8b\right)+\left(5b-40\right)$$

Factor out b in the first and 5 in the second group.

$$b\left(b-8\right)+5\left(b-8\right)$$

Factor out common term b-8 by using distributive property.

$$\left(b-8\right)\left(b+5\right)$$

Rewrite the complete factored expression.

$$b\left(b-8\right)\left(b+5\right)$$

Evaluate

$$b\left(b-8\right)\left(b+5\right)$$