Solve 16r^3+44r^2-80r

$$16 r ^ { 3 } + 44 r ^ { 2 } - 80 r$$

Factor

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

Solution Steps

Factor out 4.

$$4\left(4r^{3}+11r^{2}-20r\right)$$

Consider 4r^{3}+11r^{2}-20r. Factor out r.

$$r\left(4r^{2}+11r-20\right)$$

Consider 4r^{2}+11r-20. Factor the expression by grouping. First, the expression needs to be rewritten as 4r^{2}+ar+br-20. To find a and b, set up a system to be solved.

$$a+b=11 ab=4\left(-20\right)=-80$$

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 -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 11.

$$a=-5 b=16$$

Rewrite 4r^{2}+11r-20 as \left(4r^{2}-5r\right)+\left(16r-20\right).

$$\left(4r^{2}-5r\right)+\left(16r-20\right)$$

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

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

Factor out common term 4r-5 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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