Solve 6a^3-18a^2-2040a

$$6 a ^ { 3 } - 18 a ^ { 2 } - 2040 a$$

Factor

$$6a\left(a-20\right)\left(a+17\right)$$

Solution Steps

Factor out 6.

$$6\left(a^{3}-3a^{2}-340a\right)$$

Consider a^{3}-3a^{2}-340a. Factor out a.

$$a\left(a^{2}-3a-340\right)$$

Consider a^{2}-3a-340. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-340. To find p and q, set up a system to be solved.

$$p+q=-3 pq=1\left(-340\right)=-340$$

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. List all such integer pairs that give product -340.

$$1,-340 2,-170 4,-85 5,-68 10,-34 17,-20$$

Calculate the sum for each pair.

$$1-340=-339 2-170=-168 4-85=-81 5-68=-63 10-34=-24 17-20=-3$$

The solution is the pair that gives sum -3.

$$p=-20 q=17$$

Rewrite a^{2}-3a-340 as \left(a^{2}-20a\right)+\left(17a-340\right).

$$\left(a^{2}-20a\right)+\left(17a-340\right)$$

Factor out a in the first and 17 in the second group.

$$a\left(a-20\right)+17\left(a-20\right)$$

Factor out common term a-20 by using distributive property.

$$\left(a-20\right)\left(a+17\right)$$

Rewrite the complete factored expression.

$$6a\left(a-20\right)\left(a+17\right)$$

Evaluate

$$6a\left(a-20\right)\left(a+17\right)$$