Solve 9x^4-6x^3+10x^2-6x+1=0

$$9 x ^ { 4 } - 6 x ^ { 3 } + 10 x ^ { 2 } - 6 x + 1 = 0$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

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 1 and q divides the leading coefficient 9. List all candidates \frac{p}{q}.

$$±\frac{1}{9},±\frac{1}{3},±1$$

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.

$$x=\frac{1}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 9x^{4}-6x^{3}+10x^{2}-6x+1 by 3\left(x-\frac{1}{3}\right)=3x-1 to get 3x^{3}-x^{2}+3x-1. Solve the equation where the result equals to 0.

$$3x^{3}-x^{2}+3x-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{1}{3},±1$$

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.

$$x=\frac{1}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}-x^{2}+3x-1 by 3\left(x-\frac{1}{3}\right)=3x-1 to get x^{2}+1. Solve the equation where the result equals to 0.

$$x^{2}+1=0$$

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 1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{0±\sqrt{-4}}{2}$$

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.

$$x=\frac{1}{3}$$

Solve for x (complex solution)

$$x=i$$

$$x=-i$$

$$x=\frac{1}{3}\approx 0.333333333$$