Solve 5m^3+15m^2-200m

$$5 m ^ { 3 } + 15 m ^ { 2 } - 200 m$$

Factor

$$5m\left(m-5\right)\left(m+8\right)$$

Solution Steps

Factor out 5.

$$5\left(m^{3}+3m^{2}-40m\right)$$

Consider m^{3}+3m^{2}-40m. Factor out m.

$$m\left(m^{2}+3m-40\right)$$

Consider m^{2}+3m-40. Factor the expression by grouping. First, the expression needs to be rewritten as m^{2}+am+bm-40. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-40\right)=-40$$

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 -40.

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

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 3.

$$a=-5 b=8$$

Rewrite m^{2}+3m-40 as \left(m^{2}-5m\right)+\left(8m-40\right).

$$\left(m^{2}-5m\right)+\left(8m-40\right)$$

Factor out m in the first and 8 in the second group.

$$m\left(m-5\right)+8\left(m-5\right)$$

Factor out common term m-5 by using distributive property.

$$\left(m-5\right)\left(m+8\right)$$

Rewrite the complete factored expression.

$$5m\left(m-5\right)\left(m+8\right)$$

Evaluate

$$5m\left(m-5\right)\left(m+8\right)$$