Solve a^3-3a^2-4a+12=

$$a ^ { 3 } - 3 a ^ { 2 } - 4 a + 12 =$$

Factor

$$\left(a-3\right)\left(a-2\right)\left(a+2\right)$$

Solution Steps

Do the grouping a^{3}-3a^{2}-4a+12=\left(a^{3}-3a^{2}\right)+\left(-4a+12\right), and factor out a^{2} in the first and -4 in the second group.

$$a^{2}\left(a-3\right)-4\left(a-3\right)$$

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

$$\left(a-3\right)\left(a^{2}-4\right)$$

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

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

Rewrite the complete factored expression.

$$\left(a-3\right)\left(a-2\right)\left(a+2\right)$$

Evaluate

$$\left(a-3\right)\left(a^{2}-4\right)$$