Solve 2m^3+16m^2-18m

$$2 m ^ { 3 } + 16 m ^ { 2 } - 18 m$$

Factor

$$2m\left(m-1\right)\left(m+9\right)$$

Solution Steps

Factor out 2.

$$2\left(m^{3}+8m^{2}-9m\right)$$

Consider m^{3}+8m^{2}-9m. Factor out m.

$$m\left(m^{2}+8m-9\right)$$

Consider m^{2}+8m-9. Factor the expression by grouping. First, the expression needs to be rewritten as m^{2}+am+bm-9. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\left(-9\right)=-9$$

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 -9.

$$-1,9 -3,3$$

Calculate the sum for each pair.

$$-1+9=8 -3+3=0$$

The solution is the pair that gives sum 8.

$$a=-1 b=9$$

Rewrite m^{2}+8m-9 as \left(m^{2}-m\right)+\left(9m-9\right).

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

Factor out m in the first and 9 in the second group.

$$m\left(m-1\right)+9\left(m-1\right)$$

Factor out common term m-1 by using distributive property.

$$\left(m-1\right)\left(m+9\right)$$

Rewrite the complete factored expression.

$$2m\left(m-1\right)\left(m+9\right)$$

Evaluate

$$2m\left(m-1\right)\left(m+9\right)$$