Solve r^4+2r^3+4r^2-2r-5=0

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

Solve for r (complex solution)

$$r=-1$$

$$r=1$$

$$r=-1-2i$$

$$r=-1+2i$$

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 -5 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±5,±1$$

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.

$$r=1$$

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

$$r^{3}+3r^{2}+7r+5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 5 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±5,±1$$

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.

$$r=-1$$

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

$$r^{2}+2r+5=0$$

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.

$$r=\frac{-2±\sqrt{2^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$r=\frac{-2±\sqrt{-16}}{2}$$

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

$$r=-1-2i r=-1+2i$$

List all found solutions.

$$r=1 r=-1 r=-1-2i r=-1+2i$$

Solve for r

$$r=-1$$

$$r=1$$

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 -5 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±5,±1$$

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.

$$r=1$$

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

$$r^{3}+3r^{2}+7r+5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 5 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±5,±1$$

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.

$$r=-1$$

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

$$r^{2}+2r+5=0$$

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.

$$r=\frac{-2±\sqrt{2^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$r=\frac{-2±\sqrt{-16}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$r\in \emptyset$$

List all found solutions.

$$r=1 r=-1$$