What are the solutions to the equation 2x^3+2x^2-8x+12=0 ?

The solutions to the equation 2x^3+2x^2-8x+12=0 are x=-3,x=1+i,x=1-i

Find the zeros of 2x^3+2x^2-8x+12=0

The zeros of 2x^3+2x^2-8x+12=0 are x=-3,x=1+i,x=1-i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{3} + 2 x^{2} - 8 x + 12 = 0

x = - 3, x = 1 + i, x = 1 - i

Solution steps

2 x^{3} + 2 x^{2} - 8 x + 12 = 0

Factor

2 x^{3} + 2 x^{2} - 8 x + 12 : 2 (x + 3) (x^{2} - 2 x + 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 3 = 0 or x^{2} - 2 x + 2 = 0

Solve

x + 3 = 0 : x = - 3

Solve

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

The solutions are

x = - 3, x = 1 + i, x = 1 - i

x^{4} + 16 = 0

n^{3} - 3 n^{2} + 2 n - 336 = 0

x, x^{3} - y^{3} + z^{3} = 4

m^{4} + 2 m^{3} + 3 m^{2} + 2 m + 1 = 0

x^{3} - 2 x + 1 = 0

What are the solutions to the equation 2x^3+2x^2-8x+12=0 ?
The solutions to the equation 2x^3+2x^2-8x+12=0 are x=-3,x=1+i,x=1-i
Find the zeros of 2x^3+2x^2-8x+12=0
The zeros of 2x^3+2x^2-8x+12=0 are x=-3,x=1+i,x=1-i