Solve 8x^3-68x^2+120x

$$8 x ^ { 3 } - 68 x ^ { 2 } + 120 x$$

Factor

$$4x\left(x-6\right)\left(2x-5\right)$$

Solution Steps

Factor out 4.

$$4\left(2x^{3}-17x^{2}+30x\right)$$

Consider 2x^{3}-17x^{2}+30x. Factor out x.

$$x\left(2x^{2}-17x+30\right)$$

Consider 2x^{2}-17x+30. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=-17 ab=2\times 30=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -17.

$$a=-12 b=-5$$

Rewrite 2x^{2}-17x+30 as \left(2x^{2}-12x\right)+\left(-5x+30\right).

$$\left(2x^{2}-12x\right)+\left(-5x+30\right)$$

Factor out 2x in the first and -5 in the second group.

$$2x\left(x-6\right)-5\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(2x-5\right)$$

Rewrite the complete factored expression.

$$4x\left(x-6\right)\left(2x-5\right)$$

Evaluate

$$4x\left(x-6\right)\left(2x-5\right)$$