What are the solutions to the equation x^4-4x^3-5x^2+16x+4=0 ?

The solutions to the equation x^4-4x^3-5x^2+16x+4=0 are x=-2,x=2,x=2+sqrt(5),x=2-sqrt(5)

Find the zeros of x^4-4x^3-5x^2+16x+4=0

The zeros of x^4-4x^3-5x^2+16x+4=0 are x=-2,x=2,x=2+sqrt(5),x=2-sqrt(5)

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x^{4} - 4 x^{3} - 5 x^{2} + 16 x + 4 = 0

x = - 2, x = 2, x = 2 + 5, x = 2 - 5

Decimal Notation

x = - 2, x = 2, x = 4.23606 \dots, x = - 0.23606 \dots

Solution steps

x^{4} - 4 x^{3} - 5 x^{2} + 16 x + 4 = 0

Factor

x^{4} - 4 x^{3} - 5 x^{2} + 16 x + 4 : (x + 2) (x - 2) (x^{2} - 4 x - 1)

(x + 2) (x - 2) (x^{2} - 4 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

Solve

x^{2} - 4 x - 1 = 0 : x = 2 + 5, x = 2 - 5

The solutions are

x = - 2, x = 2, x = 2 + 5, x = 2 - 5

x, (x^{2} + y^{2})^{2} = 4 x^{2} y

x, x^{3} = 7 x

4 x^{3} + 6 x^{2} - 4 x = 0

x^{4} + 2 x^{2} - 35 = 0

6 x^{7} = 8 x - 13

What are the solutions to the equation x^4-4x^3-5x^2+16x+4=0 ?
The solutions to the equation x^4-4x^3-5x^2+16x+4=0 are x=-2,x=2,x=2+sqrt(5),x=2-sqrt(5)
Find the zeros of x^4-4x^3-5x^2+16x+4=0
The zeros of x^4-4x^3-5x^2+16x+4=0 are x=-2,x=2,x=2+sqrt(5),x=2-sqrt(5)