Solve c^3-3c^2-40c

$$c ^ { 3 } - 3 c ^ { 2 } - 40 c$$

Factor

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

Solution Steps

Factor out c.

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

Consider c^{2}-3c-40. Factor the expression by grouping. First, the expression needs to be rewritten as c^{2}+ac+bc-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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-8 b=5$$

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

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

Factor out c in the first and 5 in the second group.

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

Factor out common term c-8 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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