Solve x^6-4x^5+4x^4-81x^2+324x-324=0

$$x ^ { 6 } - 4 x ^ { 5 } + 4 x ^ { 4 } - 81 x ^ { 2 } + 324 x - 324 = 0$$

Solve for x

$$x=-3$$

$$x=3$$

$$x=2$$

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 -324 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±324,±162,±108,±81,±54,±36,±27,±18,±12,±9,±6,±4,±3,±2,±1$$

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^{6}-4x^{5}+4x^{4}-81x^{2}+324x-324 by x-2 to get x^{5}-2x^{4}-81x+162. Solve the equation where the result equals to 0.

$$x^{5}-2x^{4}-81x+162=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 162 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±162,±81,±54,±27,±18,±9,±6,±3,±2,±1$$

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^{5}-2x^{4}-81x+162 by x-2 to get x^{4}-81. Solve the equation where the result equals to 0.

$$x^{4}-81=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -81 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±81,±27,±9,±3,±1$$

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^{4}-81 by x-3 to get x^{3}+3x^{2}+9x+27. Solve the equation where the result equals to 0.

$$x^{3}+3x^{2}+9x+27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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}+3x^{2}+9x+27 by x+3 to get x^{2}+9. Solve the equation where the result 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.

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

Do the calculations.

$$x=\frac{0±\sqrt{-36}}{2}$$

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.

$$x=2 x=3 x=-3$$

Solve for x (complex solution)

$$x=-3$$

$$x=3$$

$$x=-3i$$

$$x=3i$$

$$x=2$$