Solve 4x^3-52x^2+160x

$$4 { x }^{ 3 } -52 { x }^{ 2 } +160x$$

Factor

$$4x\left(x-8\right)\left(x-5\right)$$

Solution Steps

Factor out 4.

$$4\left(x^{3}-13x^{2}+40x\right)$$

Consider x^{3}-13x^{2}+40x. Factor out x.

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

Consider x^{2}-13x+40. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+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 x^{2}-13x+40 as \left(x^{2}-8x\right)+\left(-5x+40\right).

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

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

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

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

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

Rewrite the complete factored expression.

$$4x\left(x-8\right)\left(x-5\right)$$

Evaluate

$$4x\left(x-8\right)\left(x-5\right)$$