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

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

Factor

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

Solution Steps

Factor out 4.

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

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

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

Consider 20r^{2}+13r-15. Factor the expression by grouping. First, the expression needs to be rewritten as 20r^{2}+ar+br-15. To find a and b, set up a system to be solved.

$$a+b=13 ab=20\left(-15\right)=-300$$

Since ab is negative, a and b have the opposite signs. Since a+b 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.

$$a=-12 b=25$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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