Solve x^5-2x^4-4x^3+17x^2-36x+36=0

$$x ^ { 5 } - 2 x ^ { 4 } - 4 x ^ { 3 } + 17 x ^ { 2 } - 36 x + 36 = 0$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{11}i}{2}\approx 0.5+1.658312395i$$

$$x=\frac{-\sqrt{11}i+1}{2}\approx 0.5-1.658312395i$$

$$x=-3$$

$$x=2$$

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 36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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^{5}-2x^{4}-4x^{3}+17x^{2}-36x+36 by x-2 to get x^{4}-4x^{2}+9x-18. Solve the equation where the result equals to 0.

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

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^{4}-4x^{2}+9x-18 by x-2 to get x^{3}+2x^{2}+9. Solve the equation where the result equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+2x^{2}+9 by x+3 to get x^{2}-x+3. Solve the equation where the result equals to 0.

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

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, -1 for b, and 3 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-11}}{2}$$

Solve the equation x^{2}-x+3=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{11}i+1}{2} x=\frac{1+\sqrt{11}i}{2}$$

List all found solutions.

$$x=2 x=-3 x=\frac{-\sqrt{11}i+1}{2} x=\frac{1+\sqrt{11}i}{2}$$

Solve for x

$$x=2$$

$$x=-3$$

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 36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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^{5}-2x^{4}-4x^{3}+17x^{2}-36x+36 by x-2 to get x^{4}-4x^{2}+9x-18. Solve the equation where the result equals to 0.

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

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^{4}-4x^{2}+9x-18 by x-2 to get x^{3}+2x^{2}+9. Solve the equation where the result equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+2x^{2}+9 by x+3 to get x^{2}-x+3. Solve the equation where the result equals to 0.

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

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, -1 for b, and 3 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-11}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=2 x=-3$$