Solve 2x^3-3x^2-32x+48=

$$2 x ^ { 3 } - 3 x ^ { 2 } - 32 x + 48 =$$

Factor

$$\left(x-4\right)\left(2x-3\right)\left(x+4\right)$$

Solution Steps

Do the grouping 2x^{3}-3x^{2}-32x+48=\left(2x^{3}-3x^{2}\right)+\left(-32x+48\right), and factor out x^{2} in the first and -16 in the second group.

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

Factor out common term 2x-3 by using distributive property.

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

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.

$$\left(x-4\right)\left(2x-3\right)\left(x+4\right)$$

Evaluate

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