Solve x^4-5x^3+10x^2-10x+4=0

$$x ^ { 4 } - 5 x ^ { 3 } + 10 x ^ { 2 } - 10 x + 4 = 0$$

Solve for x (complex solution)

$$x=1$$

$$x=2$$

$$x=1-i$$

$$x=1+i$$

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

$$±4,±2,±1$$

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}-5x^{3}+10x^{2}-10x+4 by x-1 to get x^{3}-4x^{2}+6x-4. Solve the equation where the result equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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

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

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, -2 for b, and 2 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±\sqrt{-4}}{2}$$

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

$$x=1-i x=1+i$$

List all found solutions.

$$x=1 x=2 x=1-i x=1+i$$

Solve for x

$$x=1$$

$$x=2$$

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

$$±4,±2,±1$$

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}-5x^{3}+10x^{2}-10x+4 by x-1 to get x^{3}-4x^{2}+6x-4. Solve the equation where the result equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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

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

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, -2 for b, and 2 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±\sqrt{-4}}{2}$$

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.

$$x=1 x=2$$