Solve x^4+x^3+x^2+x=left(x+1right)^2

$${ x }^{ 4 } + { x }^{ 3 } + { x }^{ 2 } +x = { \left(x+1 \right) }^{ 2 }$$

Solve for x (complex solution)

$$x=-1$$

$$x=1$$

$$x=\frac{-\sqrt{3}i-1}{2}\approx -0.5-0.866025404i$$

$$x=\frac{-1+\sqrt{3}i}{2}\approx -0.5+0.866025404i$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Subtract x^{2} from both sides.

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

Combine x^{2} and -x^{2} to get 0.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Subtract 1 from both sides.

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

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

$$±1$$

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

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

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

$$±1$$

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

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

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

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

Do the calculations.

$$x=\frac{-1±\sqrt{-3}}{2}$$

Solve the equation x^{2}+x+1=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for x

$$x=-1$$

$$x=1$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Subtract x^{2} from both sides.

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

Combine x^{2} and -x^{2} to get 0.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Subtract 1 from both sides.

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

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

$$±1$$

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

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

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

$$±1$$

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

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

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

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

Do the calculations.

$$x=\frac{-1±\sqrt{-3}}{2}$$

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.

$$x=1 x=-1$$