Solve 12a^3+22a^2-70a

$$12 a ^ { 3 } + 22 a ^ { 2 } - 70 a$$

Factor

$$2a\left(3a-5\right)\left(2a+7\right)$$

Solution Steps

Factor out 2.

$$2\left(6a^{3}+11a^{2}-35a\right)$$

Consider 6a^{3}+11a^{2}-35a. Factor out a.

$$a\left(6a^{2}+11a-35\right)$$

Consider 6a^{2}+11a-35. Factor the expression by grouping. First, the expression needs to be rewritten as 6a^{2}+pa+qa-35. To find p and q, set up a system to be solved.

$$p+q=11 pq=6\left(-35\right)=-210$$

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. List all such integer pairs that give product -210.

$$-1,210 -2,105 -3,70 -5,42 -6,35 -7,30 -10,21 -14,15$$

Calculate the sum for each pair.

$$-1+210=209 -2+105=103 -3+70=67 -5+42=37 -6+35=29 -7+30=23 -10+21=11 -14+15=1$$

The solution is the pair that gives sum 11.

$$p=-10 q=21$$

Rewrite 6a^{2}+11a-35 as \left(6a^{2}-10a\right)+\left(21a-35\right).

$$\left(6a^{2}-10a\right)+\left(21a-35\right)$$

Factor out 2a in the first and 7 in the second group.

$$2a\left(3a-5\right)+7\left(3a-5\right)$$

Factor out common term 3a-5 by using distributive property.

$$\left(3a-5\right)\left(2a+7\right)$$

Rewrite the complete factored expression.

$$2a\left(3a-5\right)\left(2a+7\right)$$

Evaluate

$$2a\left(3a-5\right)\left(2a+7\right)$$