What are the solutions to the equation x^4+4x^3+5x^2+4x+1=0 ?

The solutions to the equation x^4+4x^3+5x^2+4x+1=0 are x=-(-sqrt(5)+3)/2 ,x=-(sqrt(5)+3)/2

Find the zeros of x^4+4x^3+5x^2+4x+1=0

The zeros of x^4+4x^3+5x^2+4x+1=0 are x=-(-sqrt(5)+3)/2 ,x=-(sqrt(5)+3)/2

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x^{4} + 4 x^{3} + 5 x^{2} + 4 x + 1 = 0

x = - \frac{- 5 + 3}{2}, x = - \frac{5 + 3}{2}

Decimal Notation

x = - 0.38196 \dots, x = - 2.61803 \dots

Solution steps

x^{4} + 4 x^{3} + 5 x^{2} + 4 x + 1 = 0

Find one solution for

x^{4} + 4 x^{3} + 5 x^{2} + 4 x + 1 = 0

using Newton-Raphson:

x \approx - 0.38196 \dots

Apply long division:

\frac{x ^{4} + 4 x ^{3} + 5 x ^{2} + 4 x + 1}{x + 0.38196 \dots} = x^{3} + 3.61803 \dots x^{2} + 3.61803 \dots x + 2.61803 \dots

x^{3} + 3.61803 \dots x^{2} + 3.61803 \dots x + 2.61803 \dots \approx 0

Find one solution for

x^{3} + 3.61803 \dots x^{2} + 3.61803 \dots x + 2.61803 \dots = 0

using Newton-Raphson:

x \approx - 2.61803 \dots

Apply long division:

\frac{x ^{3} + 3.61803 \dots x ^{2} + 3.61803 \dots x + 2.61803 \dots}{x + 2.61803 \dots} = x^{2} + x + 1

x^{2} + x + 1 \approx 0

Find one solution for

x^{2} + x + 1 = 0

using Newton-Raphson:

No Solution for

x \in R

Approximate to algebraic form

x \approx - 0.38196 \dots : x = - \frac{- 5 + 3}{2}

x \approx - 2.61803 \dots : x = - \frac{5 + 3}{2}

x = - \frac{- 5 + 3}{2}, x = - \frac{5 + 3}{2}

The solutions are

x = - \frac{- 5 + 3}{2}, x = - \frac{5 + 3}{2}

12 x^{3} - 9 x^{2} = 0

- x^{3} + 2 x^{2} + 6 x + 8 = 0

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

5 x^{3} + 27 x^{2} - 11 x + 5

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

What are the solutions to the equation x^4+4x^3+5x^2+4x+1=0 ?
The solutions to the equation x^4+4x^3+5x^2+4x+1=0 are x=-(-sqrt(5)+3)/2 ,x=-(sqrt(5)+3)/2
Find the zeros of x^4+4x^3+5x^2+4x+1=0
The zeros of x^4+4x^3+5x^2+4x+1=0 are x=-(-sqrt(5)+3)/2 ,x=-(sqrt(5)+3)/2