Solve -b^3+5b^2+24b

$$- b ^ { 3 } + 5 b ^ { 2 } + 24 b$$

Factor

$$-b\left(b-8\right)\left(b+3\right)$$

Solution Steps

Factor out b.

$$b\left(-b^{2}+5b+24\right)$$

Consider -b^{2}+5b+24. Factor the expression by grouping. First, the expression needs to be rewritten as -b^{2}+pb+qb+24. To find p and q, set up a system to be solved.

$$p+q=5 pq=-24=-24$$

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 -24.

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

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$p=8 q=-3$$

Rewrite -b^{2}+5b+24 as \left(-b^{2}+8b\right)+\left(-3b+24\right).

$$\left(-b^{2}+8b\right)+\left(-3b+24\right)$$

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

$$-b\left(b-8\right)-3\left(b-8\right)$$

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

$$\left(b-8\right)\left(-b-3\right)$$

Rewrite the complete factored expression.

$$b\left(b-8\right)\left(-b-3\right)$$

Evaluate

$$-b\left(b-8\right)\left(b+3\right)$$