Solve c^3-4c^2-4c+16

$$c ^ { 3 } - 4 c ^ { 2 } - 4 c + 16$$

Factor

$$\left(c-4\right)\left(c-2\right)\left(c+2\right)$$

Solution Steps

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

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

Factor out common term c-4 by using distributive property.

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

Consider c^{2}-4. Rewrite c^{2}-4 as c^{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).

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

Rewrite the complete factored expression.

$$\left(c-4\right)\left(c-2\right)\left(c+2\right)$$

Evaluate

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