What are the solutions to the equation x^4+2x^3-5x^2-8x+4=0 ?

The solutions to the equation x^4+2x^3-5x^2-8x+4=0 are x=-2,x=2,x=-1+sqrt(2),x=-1-sqrt(2)

Find the zeros of x^4+2x^3-5x^2-8x+4=0

The zeros of x^4+2x^3-5x^2-8x+4=0 are x=-2,x=2,x=-1+sqrt(2),x=-1-sqrt(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{4} + 2 x^{3} - 5 x^{2} - 8 x + 4 = 0

x = - 2, x = 2, x = - 1 + 2, x = - 1 - 2

Decimal Notation

x = - 2, x = 2, x = 0.41421 \dots, x = - 2.41421 \dots

Solution steps

x^{4} + 2 x^{3} - 5 x^{2} - 8 x + 4 = 0

Factor

x^{4} + 2 x^{3} - 5 x^{2} - 8 x + 4 : (x + 2) (x - 2) (x^{2} + 2 x - 1)

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

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

Solve

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

The solutions are

x = - 2, x = 2, x = - 1 + 2, x = - 1 - 2

x^{4} + 40 = 14 x^{2}

z^{3} = (1 - 3 i)

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

x^{4} - 4 x^{3} + 8 x^{2} - 16 x + 16

- 1 = m^{3}

What are the solutions to the equation x^4+2x^3-5x^2-8x+4=0 ?
The solutions to the equation x^4+2x^3-5x^2-8x+4=0 are x=-2,x=2,x=-1+sqrt(2),x=-1-sqrt(2)
Find the zeros of x^4+2x^3-5x^2-8x+4=0
The zeros of x^4+2x^3-5x^2-8x+4=0 are x=-2,x=2,x=-1+sqrt(2),x=-1-sqrt(2)