Solve 6x^4-5x^3-2x^2+x=

$$6 x ^ { 4 } - 5 x ^ { 3 } - 2 x ^ { 2 } + x =$$

Factor

$$x\left(x-1\right)\left(3x-1\right)\left(2x+1\right)$$

Solution Steps

Factor out x.

$$x\left(6x^{3}-5x^{2}-2x+1\right)$$

Consider 6x^{3}-5x^{2}-2x+1. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient 6. One such root is -\frac{1}{2}. Factor the polynomial by dividing it by 2x+1.

$$\left(2x+1\right)\left(3x^{2}-4x+1\right)$$

Consider 3x^{2}-4x+1. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-4 ab=3\times 1=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite 3x^{2}-4x+1 as \left(3x^{2}-3x\right)+\left(-x+1\right).

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

Factor out 3x in the first and -1 in the second group.

$$3x\left(x-1\right)-\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(3x-1\right)$$

Rewrite the complete factored expression.

$$x\left(2x+1\right)\left(x-1\right)\left(3x-1\right)$$

Evaluate

$$x\left(x-1\right)\left(3x-1\right)\left(2x+1\right)$$