Solve x^4-7x^2+4x+20=0

$$x ^ { 4 } - 7 x ^ { 2 } + 4 x + 20 = 0$$

Solve for x (complex solution)

$$x=2+i$$

$$x=2-i$$

$$x=-2$$

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

$$±20,±10,±5,±4,±2,±1$$

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^{4}-7x^{2}+4x+20 by x+2 to get x^{3}-2x^{2}-3x+10. Solve the equation where the result equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 10 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±10,±5,±2,±1$$

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

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

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, -4 for b, and 5 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-4}}{2}$$

Solve the equation x^{2}-4x+5=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for x

$$x=-2$$

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

$$±20,±10,±5,±4,±2,±1$$

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^{4}-7x^{2}+4x+20 by x+2 to get x^{3}-2x^{2}-3x+10. Solve the equation where the result equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 10 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±10,±5,±2,±1$$

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

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

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, -4 for b, and 5 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-4}}{2}$$

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.

$$x=-2$$