Solve 40x^3-82x^2+40x

$$40 x ^ { 3 } - 82 x ^ { 2 } + 40 x$$

Factor

$$2x\left(4x-5\right)\left(5x-4\right)$$

Solution Steps

Factor out 2.

$$2\left(20x^{3}-41x^{2}+20x\right)$$

Consider 20x^{3}-41x^{2}+20x. Factor out x.

$$x\left(20x^{2}-41x+20\right)$$

Consider 20x^{2}-41x+20. Factor the expression by grouping. First, the expression needs to be rewritten as 20x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=-41 ab=20\times 20=400$$

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 400.

$$-1,-400 -2,-200 -4,-100 -5,-80 -8,-50 -10,-40 -16,-25 -20,-20$$

Calculate the sum for each pair.

$$-1-400=-401 -2-200=-202 -4-100=-104 -5-80=-85 -8-50=-58 -10-40=-50 -16-25=-41 -20-20=-40$$

The solution is the pair that gives sum -41.

$$a=-25 b=-16$$

Rewrite 20x^{2}-41x+20 as \left(20x^{2}-25x\right)+\left(-16x+20\right).

$$\left(20x^{2}-25x\right)+\left(-16x+20\right)$$

Factor out 5x in the first and -4 in the second group.

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

Factor out common term 4x-5 by using distributive property.

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

Rewrite the complete factored expression.

$$2x\left(4x-5\right)\left(5x-4\right)$$

Evaluate

$$2x\left(4x-5\right)\left(5x-4\right)$$