Solve 3x^3+x^2-48x-16

$$3 x ^ { 3 } + x ^ { 2 } - 48 x - 16$$

Factor

$$\left(x-4\right)\left(x+4\right)\left(3x+1\right)$$

Solution Steps

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

$$x^{2}\left(3x+1\right)-16\left(3x+1\right)$$

Factor out common term 3x+1 by using distributive property.

$$\left(3x+1\right)\left(x^{2}-16\right)$$

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

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

Rewrite the complete factored expression.

$$\left(x-4\right)\left(3x+1\right)\left(x+4\right)$$

Evaluate

$$\left(3x+1\right)\left(x^{2}-16\right)$$