Solve 20r^3+92r^2-40r

$$20 r ^ { 3 } + 92 r ^ { 2 } - 40 r$$

Factor

$$4r\left(5r-2\right)\left(r+5\right)$$

Solution Steps

Factor out 4.

$$4\left(5r^{3}+23r^{2}-10r\right)$$

Consider 5r^{3}+23r^{2}-10r. Factor out r.

$$r\left(5r^{2}+23r-10\right)$$

Consider 5r^{2}+23r-10. Factor the expression by grouping. First, the expression needs to be rewritten as 5r^{2}+ar+br-10. To find a and b, set up a system to be solved.

$$a+b=23 ab=5\left(-10\right)=-50$$

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 -50.

$$-1,50 -2,25 -5,10$$

Calculate the sum for each pair.

$$-1+50=49 -2+25=23 -5+10=5$$

The solution is the pair that gives sum 23.

$$a=-2 b=25$$

Rewrite 5r^{2}+23r-10 as \left(5r^{2}-2r\right)+\left(25r-10\right).

$$\left(5r^{2}-2r\right)+\left(25r-10\right)$$

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

$$r\left(5r-2\right)+5\left(5r-2\right)$$

Factor out common term 5r-2 by using distributive property.

$$\left(5r-2\right)\left(r+5\right)$$

Rewrite the complete factored expression.

$$4r\left(5r-2\right)\left(r+5\right)$$

Evaluate

$$4r\left(5r-2\right)\left(r+5\right)$$