Solve x^4-3x^3+8x^2-18x+12

$$x ^ { 4 } - 3 x ^ { 3 } + 8 x ^ { 2 } - 18 x + 12$$

Factor

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

Solution Steps

To factor the expression, solve the equation where it equals to 0.

$$x^{4}-3x^{3}+8x^{2}-18x+12=0$$

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. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-3x^{3}+8x^{2}-18x+12 by x-1 to get x^{3}-2x^{2}+6x-12. To factor the result, solve the equation where it equals to 0.

$$x^{3}-2x^{2}+6x-12=0$$

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. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-2x^{2}+6x-12 by x-2 to get x^{2}+6. To factor the result, solve the equation where it equals to 0.

$$x^{2}+6=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 0 for b, and 6 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-24}}{2}$$

Polynomial x^{2}+6 is not factored since it does not have any rational roots.

$$x^{2}+6$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

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