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

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

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

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

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

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

Decimal Notation

x = 1, x = - 2, x = - 0.38196 \dots, x = - 2.61803 \dots

Solution steps

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

Switch sides

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

Factor

x^{4} + 4 x^{3} + 2 x^{2} - 5 x - 2 : (x - 1) (x + 2) (x^{2} + 3 x + 1)

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

Solve

x - 1 = 0 : x = 1

Solve

x + 2 = 0 : x = - 2

Solve

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

The solutions are

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

65 = b^{5}

3 x^{3} - 30 x^{2} + 63 x = 0

x^{4} - 25 x + 144 = 0

(2 - x) \cdot (2 - x) \cdot (4 - x) = 0

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

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