Solve 42a^3+45a^2-27a

$$42 a ^ { 3 } + 45 a ^ { 2 } - 27 a$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(14a^{3}+15a^{2}-9a\right)$$

Consider 14a^{3}+15a^{2}-9a. Factor out a.

$$a\left(14a^{2}+15a-9\right)$$

Consider 14a^{2}+15a-9. Factor the expression by grouping. First, the expression needs to be rewritten as 14a^{2}+pa+qa-9. To find p and q, set up a system to be solved.

$$p+q=15 pq=14\left(-9\right)=-126$$

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 -126.

$$-1,126 -2,63 -3,42 -6,21 -7,18 -9,14$$

Calculate the sum for each pair.

$$-1+126=125 -2+63=61 -3+42=39 -6+21=15 -7+18=11 -9+14=5$$

The solution is the pair that gives sum 15.

$$p=-6 q=21$$

Rewrite 14a^{2}+15a-9 as \left(14a^{2}-6a\right)+\left(21a-9\right).

$$\left(14a^{2}-6a\right)+\left(21a-9\right)$$

Factor out 2a in the first and 3 in the second group.

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

Factor out common term 7a-3 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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