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

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

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

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

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

x = - 1, x = \frac{3 + 5}{2}, x = \frac{3 - 5}{2}

Decimal Notation

x = - 1, x = 2.61803 \dots, x = 0.38196 \dots

Solution steps

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

Factor

x^{4} - x^{3} - 4 x^{2} - x + 1 : (x + 1)^{2} (x^{2} - 3 x + 1)

(x + 1)^{2} (x^{2} - 3 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x^{2} - 3 x + 1 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x^{2} - 3 x + 1 = 0 : x = \frac{3 + 5}{2}, x = \frac{3 - 5}{2}

The solutions are

x = - 1, x = \frac{3 + 5}{2}, x = \frac{3 - 5}{2}

4 x^{3} - 12 x = 0

60 x^{3} - 30 x = 0

2 x^{3} + x^{2} - 1 = 0

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

4 x^{4} + 6 x^{2} - 2 = 0

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