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

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

Factor

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

Solution Steps

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

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

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

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

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.

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

Evaluate

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