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

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

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

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

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

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

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

x + 2 = 0 : x = - 2

Solve

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

The solutions are

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

p^{3} + 3 p = 4

x^{6} = 81 x^{4}

8 x^{3} - 6 x = 0

2 x^{3} - 4 x - 7 x = - x^{2} + x + 3

48.46 x^{3} + 12.6 x - 0.005 = 0

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