Solve 6x^5-39x^4+18x^3

$$6 x ^ { 5 } - 39 x ^ { 4 } + 18 x ^ { 3 }$$

Factor

$$3\left(x-6\right)\left(2x-1\right)x^{3}$$

Solution Steps

Factor out 3.

$$3\left(2x^{5}-13x^{4}+6x^{3}\right)$$

Consider 2x^{5}-13x^{4}+6x^{3}. Factor out x^{3}.

$$x^{3}\left(2x^{2}-13x+6\right)$$

Consider 2x^{2}-13x+6. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-13 ab=2\times 6=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -13.

$$a=-12 b=-1$$

Rewrite 2x^{2}-13x+6 as \left(2x^{2}-12x\right)+\left(-x+6\right).

$$\left(2x^{2}-12x\right)+\left(-x+6\right)$$

Factor out 2x in the first and -1 in the second group.

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

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

$$\left(x-6\right)\left(2x-1\right)$$

Rewrite the complete factored expression.

$$3x^{3}\left(x-6\right)\left(2x-1\right)$$

Evaluate

$$3\left(x-6\right)\left(2x-1\right)x^{3}$$