Solve 9m^3-36m^2+20m

$$9 m ^ { 3 } - 36 m ^ { 2 } + 20 m$$

Factor

$$m\left(3m-10\right)\left(3m-2\right)$$

Solution Steps

Factor out m.

$$m\left(9m^{2}-36m+20\right)$$

Consider 9m^{2}-36m+20. Factor the expression by grouping. First, the expression needs to be rewritten as 9m^{2}+am+bm+20. To find a and b, set up a system to be solved.

$$a+b=-36 ab=9\times 20=180$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 180.

$$-1,-180 -2,-90 -3,-60 -4,-45 -5,-36 -6,-30 -9,-20 -10,-18 -12,-15$$

Calculate the sum for each pair.

$$-1-180=-181 -2-90=-92 -3-60=-63 -4-45=-49 -5-36=-41 -6-30=-36 -9-20=-29 -10-18=-28 -12-15=-27$$

The solution is the pair that gives sum -36.

$$a=-30 b=-6$$

Rewrite 9m^{2}-36m+20 as \left(9m^{2}-30m\right)+\left(-6m+20\right).

$$\left(9m^{2}-30m\right)+\left(-6m+20\right)$$

Factor out 3m in the first and -2 in the second group.

$$3m\left(3m-10\right)-2\left(3m-10\right)$$

Factor out common term 3m-10 by using distributive property.

$$\left(3m-10\right)\left(3m-2\right)$$

Rewrite the complete factored expression.

$$m\left(3m-10\right)\left(3m-2\right)$$

Evaluate

$$m\left(3m-10\right)\left(3m-2\right)$$