What are the solutions to the equation (x^2+1)(4+x)=0 ?

The solutions to the equation (x^2+1)(4+x)=0 are x=i,x=-i,x=-4

Find the zeros of (x^2+1)(4+x)=0

The zeros of (x^2+1)(4+x)=0 are x=i,x=-i,x=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^{2} + 1) (4 + x) = 0

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

Solution steps

(x^{2} + 1) (4 + x) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x^{2} + 1 = 0 or 4 + x = 0

Solve

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

Solve

4 + x = 0 : x = - 4

The solutions are

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

x^{3} - 2 x^{2} - 9 = 0

x^{3} + 125

x^{3} = - 8 i

2 x^{3} - 10 x^{2} - 72 x = 0

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

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