Solve 1/6x^3-1/3x^2-11/6x+2

$$\frac{ 1 }{ 6 } { x }^{ 3 } - \frac{ 1 }{ 3 } { x }^{ 2 } - \frac{ 11 }{ 6 } x+2$$

Factor

$$\frac{\left(x-4\right)\left(x-1\right)\left(x+3\right)}{6}$$

Solution Steps

Factor out \frac{1}{6}.

$$\frac{x^{3}-2x^{2}-11x+12}{6}$$

Consider x^{3}-2x^{2}-11x+12. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 12 and q divides the leading coefficient 1. One such root is 4. Factor the polynomial by dividing it by x-4.

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

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

$$a+b=2 ab=1\left(-3\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

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

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

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

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

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

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

Rewrite the complete factored expression.

$$\frac{\left(x-4\right)\left(x-1\right)\left(x+3\right)}{6}$$

Evaluate

$$\frac{\left(x-4\right)\left(x-1\right)\left(x+3\right)}{6}$$