Solve x^5+3x^4=x+3

$${ x }^{ 5 } +3 { x }^{ 4 } =x+3$$

Solve for x

$$x=-3$$

$$x=1$$

$$x=-1$$

Solution Steps

Subtract x from both sides.

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

Subtract 3 from both sides.

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

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

$$±3,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{5}+3x^{4}-x-3 by x-1 to get x^{4}+4x^{3}+4x^{2}+4x+3. Solve the equation where the result equals to 0.

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

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

$$±3,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+4x^{3}+4x^{2}+4x+3 by x+1 to get x^{3}+3x^{2}+x+3. Solve the equation where the result equals to 0.

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

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

$$±3,±1$$

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}+x+3 by x+3 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.

$$x=1 x=-1 x=-3$$

Solve for x (complex solution)

$$x=-1$$

$$x=1$$

$$x=-i$$

$$x=i$$

$$x=-3$$