Solve a^3-4a^2-9a+36

$$a ^ { 3 } - 4 a ^ { 2 } - 9 a + 36$$

Factor

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

Solution Steps

Do the grouping a^{3}-4a^{2}-9a+36=\left(a^{3}-4a^{2}\right)+\left(-9a+36\right), and factor out a^{2} in the first and -9 in the second group.

$$a^{2}\left(a-4\right)-9\left(a-4\right)$$

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

$$\left(a-4\right)\left(a^{2}-9\right)$$

Consider a^{2}-9. Rewrite a^{2}-9 as a^{2}-3^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Rewrite the complete factored expression.

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

Evaluate

$$\left(a-4\right)\left(a^{2}-9\right)$$