What are the solutions to the equation roots x^4+6x^3+15x^2+20x+12 ?

The solutions to the equation roots x^4+6x^3+15x^2+20x+12 are x=-2,x=-1+sqrt(2)i,x=-1-sqrt(2)i

Find the zeros of roots x^4+6x^3+15x^2+20x+12

The zeros of roots x^4+6x^3+15x^2+20x+12 are x=-2,x=-1+sqrt(2)i,x=-1-sqrt(2)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

roots

x^{4} + 6 x^{3} + 15 x^{2} + 20 x + 12

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

Solution steps

x^{4} + 6 x^{3} + 15 x^{2} + 20 x + 12

The roots are the intercepts with the x-axis

(y = 0)

x^{4} + 6 x^{3} + 15 x^{2} + 20 x + 12 = 0

Factor

x^{4} + 6 x^{3} + 15 x^{2} + 20 x + 12 : (x + 2)^{2} (x^{2} + 2 x + 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 2 = 0 : x = - 2

Solve

x^{2} + 2 x + 3 = 0 : x = - 1 + 2 i, x = - 1 - 2 i

The solutions are

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

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

- 2 x^{3} + 6 x^{2} + 20 x = 0

x^{3} + 3 x^{2} - 6 x + 8 = 0

(x + 3)^{4} = 0

x^{3} = 9 x - 1

What are the solutions to the equation roots x^4+6x^3+15x^2+20x+12 ?
The solutions to the equation roots x^4+6x^3+15x^2+20x+12 are x=-2,x=-1+sqrt(2)i,x=-1-sqrt(2)i
Find the zeros of roots x^4+6x^3+15x^2+20x+12
The zeros of roots x^4+6x^3+15x^2+20x+12 are x=-2,x=-1+sqrt(2)i,x=-1-sqrt(2)i