Solve 27a^3+18a^2-9a

$$27 a ^ { 3 } + 18 a ^ { 2 } - 9 a$$

Factor

$$9a\left(3a-1\right)\left(a+1\right)$$

Solution Steps

Factor out 9.

$$9\left(3a^{3}+2a^{2}-a\right)$$

Consider 3a^{3}+2a^{2}-a. Factor out a.

$$a\left(3a^{2}+2a-1\right)$$

Consider 3a^{2}+2a-1. Factor the expression by grouping. First, the expression needs to be rewritten as 3a^{2}+pa+qa-1. To find p and q, set up a system to be solved.

$$p+q=2 pq=3\left(-1\right)=-3$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$p=-1 q=3$$

Rewrite 3a^{2}+2a-1 as \left(3a^{2}-a\right)+\left(3a-1\right).

$$\left(3a^{2}-a\right)+\left(3a-1\right)$$

Factor out a in 3a^{2}-a.

$$a\left(3a-1\right)+3a-1$$

Factor out common term 3a-1 by using distributive property.

$$\left(3a-1\right)\left(a+1\right)$$

Rewrite the complete factored expression.

$$9a\left(3a-1\right)\left(a+1\right)$$

Evaluate

$$9a\left(3a-1\right)\left(a+1\right)$$