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

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

Factor

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

Solution Steps

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

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

Factor out common term p-3 by using distributive property.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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