Solve 12x^3-16x^2+3x+1

$$12 { x }^{ 3 } -16 { x }^{ 2 } +3x+1$$

Factor

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

Solution Steps

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 12. One such root is \frac{1}{2}. Factor the polynomial by dividing it by 2x-1.

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

Consider 6x^{2}-5x-1. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-5 ab=6\left(-1\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite 6x^{2}-5x-1 as \left(6x^{2}-6x\right)+\left(x-1\right).

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

Factor out 6x in 6x^{2}-6x.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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