Solve b^3+b^2-20b

$$b ^ { 3 } + b ^ { 2 } - 20 b$$

Factor

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

Solution Steps

Factor out b.

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

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

$$p+q=1 pq=1\left(-20\right)=-20$$

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 -20.

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

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$p=-4 q=5$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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