Solve -b^3-8b^2-16b

$$- b ^ { 3 } - 8 b ^ { 2 } - 16 b$$

Factor

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

Solution Steps

Factor out b.

$$b\left(-b^{2}-8b-16\right)$$

Consider -b^{2}-8b-16. Factor the expression by grouping. First, the expression needs to be rewritten as -b^{2}+pb+qb-16. To find p and q, set up a system to be solved.

$$p+q=-8 pq=-\left(-16\right)=16$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$p=-4 q=-4$$

Rewrite -b^{2}-8b-16 as \left(-b^{2}-4b\right)+\left(-4b-16\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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