Solve s^3-16s^2-4s+64

$$s ^ { 3 } - 16 s ^ { 2 } - 4 s + 64$$

Factor

$$\left(s-16\right)\left(s-2\right)\left(s+2\right)$$

Solution Steps

Do the grouping s^{3}-16s^{2}-4s+64=\left(s^{3}-16s^{2}\right)+\left(-4s+64\right), and factor out s^{2} in the first and -4 in the second group.

$$s^{2}\left(s-16\right)-4\left(s-16\right)$$

Factor out common term s-16 by using distributive property.

$$\left(s-16\right)\left(s^{2}-4\right)$$

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

$$\left(s-2\right)\left(s+2\right)$$

Rewrite the complete factored expression.

$$\left(s-16\right)\left(s-2\right)\left(s+2\right)$$

Evaluate

$$\left(s-16\right)\left(s^{2}-4\right)$$