Solve x^4-5x^3+7x^2-7x-20=0

$${ x }^{ 4 } -5 { x }^{ 3 } +7 { x }^{ 2 } -7x-20 = 0$$

Solve for x (complex solution)

$$x=-1$$

$$x=4$$

$$x=1-2i$$

$$x=1+2i$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-5x^{3}+7x^{2}-7x-20 by x+1 to get x^{3}-6x^{2}+13x-20. Solve the equation where the result equals to 0.

$$x^{3}-6x^{2}+13x-20=0$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-6x^{2}+13x-20 by x-4 to get x^{2}-2x+5. Solve the equation where the result equals to 0.

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

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, -2 for b, and 5 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±\sqrt{-16}}{2}$$

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

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

List all found solutions.

$$x=-1 x=4 x=1-2i x=1+2i$$

Solve for x

$$x=-1$$

$$x=4$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-5x^{3}+7x^{2}-7x-20 by x+1 to get x^{3}-6x^{2}+13x-20. Solve the equation where the result equals to 0.

$$x^{3}-6x^{2}+13x-20=0$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-6x^{2}+13x-20 by x-4 to get x^{2}-2x+5. Solve the equation where the result equals to 0.

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

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, -2 for b, and 5 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±\sqrt{-16}}{2}$$

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.

$$x=-1 x=4$$