Solve 12a^3+2a^2-192a-32

$$12 a ^ { 3 } + 2 a ^ { 2 } - 192 a - 32$$

Factor

$$2\left(a-4\right)\left(a+4\right)\left(6a+1\right)$$

Solution Steps

Factor out 2.

$$2\left(6a^{3}+a^{2}-96a-16\right)$$

Consider 6a^{3}+a^{2}-96a-16. Do the grouping 6a^{3}+a^{2}-96a-16=\left(6a^{3}+a^{2}\right)+\left(-96a-16\right), and factor out a^{2} in the first and -16 in the second group.

$$a^{2}\left(6a+1\right)-16\left(6a+1\right)$$

Factor out common term 6a+1 by using distributive property.

$$\left(6a+1\right)\left(a^{2}-16\right)$$

Consider a^{2}-16. Rewrite a^{2}-16 as a^{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).

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

Rewrite the complete factored expression.

$$2\left(6a+1\right)\left(a-4\right)\left(a+4\right)$$

Evaluate

$$2\left(6a+1\right)\left(a^{2}-16\right)$$