Solve 12a^4-8a^3-4a^2

$$12 a ^ { 4 } - 8 a ^ { 3 } - 4 a ^ { 2 }$$

Factor

$$4\left(a-1\right)\left(3a+1\right)a^{2}$$

Solution Steps

Factor out 4.

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

Consider 3a^{4}-2a^{3}-a^{2}. Factor out a^{2}.

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

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 negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$p=-3 q=1$$

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

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

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

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

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

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

Rewrite the complete factored expression.

$$4a^{2}\left(a-1\right)\left(3a+1\right)$$

Evaluate

$$4\left(a-1\right)\left(3a+1\right)a^{2}$$