Solve 80a^3+52a^2-60a

$$80 a ^ { 3 } + 52 a ^ { 2 } - 60 a$$

Factor

$$4a\left(5a-3\right)\left(4a+5\right)$$

Solution Steps

Factor out 4.

$$4\left(20a^{3}+13a^{2}-15a\right)$$

Consider 20a^{3}+13a^{2}-15a. Factor out a.

$$a\left(20a^{2}+13a-15\right)$$

Consider 20a^{2}+13a-15. Factor the expression by grouping. First, the expression needs to be rewritten as 20a^{2}+pa+qa-15. To find p and q, set up a system to be solved.

$$p+q=13 pq=20\left(-15\right)=-300$$

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 -300.

$$-1,300 -2,150 -3,100 -4,75 -5,60 -6,50 -10,30 -12,25 -15,20$$

Calculate the sum for each pair.

$$-1+300=299 -2+150=148 -3+100=97 -4+75=71 -5+60=55 -6+50=44 -10+30=20 -12+25=13 -15+20=5$$

The solution is the pair that gives sum 13.

$$p=-12 q=25$$

Rewrite 20a^{2}+13a-15 as \left(20a^{2}-12a\right)+\left(25a-15\right).

$$\left(20a^{2}-12a\right)+\left(25a-15\right)$$

Factor out 4a in the first and 5 in the second group.

$$4a\left(5a-3\right)+5\left(5a-3\right)$$

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

$$\left(5a-3\right)\left(4a+5\right)$$

Rewrite the complete factored expression.

$$4a\left(5a-3\right)\left(4a+5\right)$$

Evaluate

$$4a\left(5a-3\right)\left(4a+5\right)$$