What are the solutions to the equation x^4-2x^3+8x^2-32x-128=0 ?

The solutions to the equation x^4-2x^3+8x^2-32x-128=0 are x=-2,x=4,x=4i,x=-4i

Find the zeros of x^4-2x^3+8x^2-32x-128=0

The zeros of x^4-2x^3+8x^2-32x-128=0 are x=-2,x=4,x=4i,x=-4i

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} + 8 x^{2} - 32 x - 128 = 0

x = - 2, x = 4, x = 4 i, x = - 4 i

Solution steps

x^{4} - 2 x^{3} + 8 x^{2} - 32 x - 128 = 0

Factor

x^{4} - 2 x^{3} + 8 x^{2} - 32 x - 128 : (x + 2) (x - 4) (x^{2} + 16)

(x + 2) (x - 4) (x^{2} + 16) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x - 4 = 0 or x^{2} + 16 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x - 4 = 0 : x = 4

Solve

x^{2} + 16 = 0 : x = 4 i, x = - 4 i

The solutions are

x = - 2, x = 4, x = 4 i, x = - 4 i

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

x^{3} = 2^{30}

x^{3} - 16 x^{2} + 55 x + 72 = 0

- x^{4} - 8 x^{3} + 20 x^{2} = 0

12 x^{3} - 3 x^{2} = 0

What are the solutions to the equation x^4-2x^3+8x^2-32x-128=0 ?
The solutions to the equation x^4-2x^3+8x^2-32x-128=0 are x=-2,x=4,x=4i,x=-4i
Find the zeros of x^4-2x^3+8x^2-32x-128=0
The zeros of x^4-2x^3+8x^2-32x-128=0 are x=-2,x=4,x=4i,x=-4i