Solve c(m)=m^3+5m^2-4m-20

$$c ( m ) = m ^ { 3 } + 5 m ^ { 2 } - 4 m - 20$$

Factor

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

Solution Steps

Do the grouping m^{3}+5m^{2}-4m-20=\left(m^{3}+5m^{2}\right)+\left(-4m-20\right), and factor out m^{2} in the first and -4 in the second group.

$$m^{2}\left(m+5\right)-4\left(m+5\right)$$

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

$$\left(m+5\right)\left(m^{2}-4\right)$$

Consider m^{2}-4. Rewrite m^{2}-4 as m^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Rewrite the complete factored expression.

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

Evaluate

$$\left(m+5\right)\left(m^{2}-4\right)$$