Solve V(x)=4x^3-130x^2+1000x

$$V ( x ) = 4 x ^ { 3 } - 130 x ^ { 2 } + 1000 x$$

Factor

$$2x\left(x-20\right)\left(2x-25\right)$$

Solution Steps

Factor out 2.

$$2\left(2x^{3}-65x^{2}+500x\right)$$

Consider 2x^{3}-65x^{2}+500x. Factor out x.

$$x\left(2x^{2}-65x+500\right)$$

Consider 2x^{2}-65x+500. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+500. To find a and b, set up a system to be solved.

$$a+b=-65 ab=2\times 500=1000$$

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 1000.

$$-1,-1000 -2,-500 -4,-250 -5,-200 -8,-125 -10,-100 -20,-50 -25,-40$$

Calculate the sum for each pair.

$$-1-1000=-1001 -2-500=-502 -4-250=-254 -5-200=-205 -8-125=-133 -10-100=-110 -20-50=-70 -25-40=-65$$

The solution is the pair that gives sum -65.

$$a=-40 b=-25$$

Rewrite 2x^{2}-65x+500 as \left(2x^{2}-40x\right)+\left(-25x+500\right).

$$\left(2x^{2}-40x\right)+\left(-25x+500\right)$$

Factor out 2x in the first and -25 in the second group.

$$2x\left(x-20\right)-25\left(x-20\right)$$

Factor out common term x-20 by using distributive property.

$$\left(x-20\right)\left(2x-25\right)$$

Rewrite the complete factored expression.

$$2x\left(x-20\right)\left(2x-25\right)$$

Evaluate

$$2x\left(x-20\right)\left(2x-25\right)$$