Solve s^5-13s^4+40s^3

$$s ^ { 5 } - 13 s ^ { 4 } + 40 s ^ { 3 }$$

Factor

$$\left(s-8\right)\left(s-5\right)s^{3}$$

Solution Steps

Factor out s^{3}.

$$s^{3}\left(s^{2}-13s+40\right)$$

Consider s^{2}-13s+40. Factor the expression by grouping. First, the expression needs to be rewritten as s^{2}+as+bs+40. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 40=40$$

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 40.

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

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -13.

$$a=-8 b=-5$$

Rewrite s^{2}-13s+40 as \left(s^{2}-8s\right)+\left(-5s+40\right).

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

Factor out s in the first and -5 in the second group.

$$s\left(s-8\right)-5\left(s-8\right)$$

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

$$\left(s-8\right)\left(s-5\right)$$

Rewrite the complete factored expression.

$$s^{3}\left(s-8\right)\left(s-5\right)$$

Evaluate

$$\left(s-8\right)\left(s-5\right)s^{3}$$