Solve 4b^3+12b^2-64b-192

$$4 b ^ { 3 } + 12 b ^ { 2 } - 64 b - 192$$

Factor

$$4\left(b-4\right)\left(b+3\right)\left(b+4\right)$$

Solution Steps

Factor out 4.

$$4\left(b^{3}+3b^{2}-16b-48\right)$$

Consider b^{3}+3b^{2}-16b-48. Do the grouping b^{3}+3b^{2}-16b-48=\left(b^{3}+3b^{2}\right)+\left(-16b-48\right), and factor out b^{2} in the first and -16 in the second group.

$$b^{2}\left(b+3\right)-16\left(b+3\right)$$

Factor out common term b+3 by using distributive property.

$$\left(b+3\right)\left(b^{2}-16\right)$$

Consider b^{2}-16. Rewrite b^{2}-16 as b^{2}-4^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Rewrite the complete factored expression.

$$4\left(b+3\right)\left(b-4\right)\left(b+4\right)$$

Evaluate

$$4\left(b+3\right)\left(b^{2}-16\right)$$