Solve (x^4+x^3-11x^2-9x+18

$$( x ^ { 4 } + x ^ { 3 } - 11 x ^ { 2 } - 9 x + 18$$

Factor

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

Solution Steps

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

$$x^{4}+x^{3}-11x^{2}-9x+18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 18 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±18,±9,±6,±3,±2,±1$$

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}-11x^{2}-9x+18 by x-1 to get x^{3}+2x^{2}-9x-18. To factor the result, solve the equation where it equals to 0.

$$x^{3}+2x^{2}-9x-18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -18 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±18,±9,±6,±3,±2,±1$$

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}-9x-18 by x+2 to get x^{2}-9. To factor the result, solve the equation where it equals to 0.

$$x^{2}-9=0$$

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 -9 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$x=\frac{0±6}{2}$$

Solve the equation x^{2}-9=0 when ± is plus and when ± is minus.

$$x=-3 x=3$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

$$\left(x-1\right)\left(x+2\right)\left(x^{2}-9\right)$$