Solve x^{4}-2x^{3}+2x^2-3x+1=-x^4+3x^3-2x^2+2x-1

$$x ^ { 4 } - 2 x ^ { 3 } + 2 x ^ { 2 } - 3 x + 1 = - x ^ { 4 } + 3 x ^ { 3 } - 2 x ^ { 2 } + 2 x - 1$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=2$$

Solution Steps

Add x^{4} to both sides.

$$x^{4}-2x^{3}+2x^{2}-3x+1+x^{4}=3x^{3}-2x^{2}+2x-1$$

Combine x^{4} and x^{4} to get 2x^{4}.

$$2x^{4}-2x^{3}+2x^{2}-3x+1=3x^{3}-2x^{2}+2x-1$$

Subtract 3x^{3} from both sides.

$$2x^{4}-2x^{3}+2x^{2}-3x+1-3x^{3}=-2x^{2}+2x-1$$

Combine -2x^{3} and -3x^{3} to get -5x^{3}.

$$2x^{4}-5x^{3}+2x^{2}-3x+1=-2x^{2}+2x-1$$

Add 2x^{2} to both sides.

$$2x^{4}-5x^{3}+2x^{2}-3x+1+2x^{2}=2x-1$$

Combine 2x^{2} and 2x^{2} to get 4x^{2}.

$$2x^{4}-5x^{3}+4x^{2}-3x+1=2x-1$$

Subtract 2x from both sides.

$$2x^{4}-5x^{3}+4x^{2}-3x+1-2x=-1$$

Combine -3x and -2x to get -5x.

$$2x^{4}-5x^{3}+4x^{2}-5x+1=-1$$

Add 1 to both sides.

$$2x^{4}-5x^{3}+4x^{2}-5x+1+1=0$$

Add 1 and 1 to get 2.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 2 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±1,±2,±\frac{1}{2}$$

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

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

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 2. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1$$

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.

$$x=\frac{1}{2}$$

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

$$x=2 x=\frac{1}{2}$$

Solve for x (complex solution)

$$x=\frac{1}{2}=0.5$$

$$x=2$$

$$x=i$$

$$x=-i$$