Solve f(x)=x^4-x^3+x^2-3x-6

$$f ( x ) = x ^ { 4 } - x ^ { 3 } + x ^ { 2 } - 3 x - 6$$

Factor

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

Solution Steps

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

$$x^{4}-x^{3}+x^{2}-3x-6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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}-x^{3}+x^{2}-3x-6 by x+1 to get x^{3}-2x^{2}+3x-6. To factor the result, solve the equation where it equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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}+3x-6 by x-2 to get x^{2}+3. To factor the result, solve the equation where it equals to 0.

$$x^{2}+3=0$$

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 3 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-12}}{2}$$

Polynomial x^{2}+3 is not factored since it does not have any rational roots.

$$x^{2}+3$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

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