What are the solutions to the equation x^4+2x^3-5x^2-4x+6=0 ?

The solutions to the equation x^4+2x^3-5x^2-4x+6=0 are x=1,x=-3,x=-sqrt(2),x=sqrt(2)

Find the zeros of x^4+2x^3-5x^2-4x+6=0

The zeros of x^4+2x^3-5x^2-4x+6=0 are x=1,x=-3,x=-sqrt(2),x=sqrt(2)

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x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6 = 0

x = 1, x = - 3, x = - 2, x = 2

Decimal Notation

x = 1, x = - 3, x = - 1.41421 \dots, x = 1.41421 \dots

Solution steps

x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6 = 0

Factor

x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6 : (x - 1) (x + 3) (x + 2) (x - 2)

(x - 1) (x + 3) (x + 2) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x + 3 = 0 or x + 2 = 0 or x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 3 = 0 : x = - 3

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

x = 1, x = - 3, x = - 2, x = 2

8 x^{3} - 1 = 0

w^{3} = i

z^{3} = \frac{i}{8}

x^{2} (x - 6) = 0

z^{4} = 1 + i

What are the solutions to the equation x^4+2x^3-5x^2-4x+6=0 ?
The solutions to the equation x^4+2x^3-5x^2-4x+6=0 are x=1,x=-3,x=-sqrt(2),x=sqrt(2)
Find the zeros of x^4+2x^3-5x^2-4x+6=0
The zeros of x^4+2x^3-5x^2-4x+6=0 are x=1,x=-3,x=-sqrt(2),x=sqrt(2)