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

The solutions to the equation (x+2)(x^2-2x+4)=0 are x=-2,x=1+sqrt(3)i,x=1-sqrt(3)i

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

The zeros of (x+2)(x^2-2x+4)=0 are x=-2,x=1+sqrt(3)i,x=1-sqrt(3)i

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

x = - 2, x = 1 + 3 i, x = 1 - 3 i

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 2 = 0 : x = - 2

Solve

x^{2} - 2 x + 4 = 0 : x = 1 + 3 i, x = 1 - 3 i

The solutions are

x = - 2, x = 1 + 3 i, x = 1 - 3 i

x^{4} + 4 x^{2} - 45 = 0

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

(x^{2} - 3 x + 1)^{6} = 1

x^{4} + x^{2} - 42 = 0

x, 4 x^{3} - 4 y = 0

What are the solutions to the equation (x+2)(x^2-2x+4)=0 ?
The solutions to the equation (x+2)(x^2-2x+4)=0 are x=-2,x=1+sqrt(3)i,x=1-sqrt(3)i
Find the zeros of (x+2)(x^2-2x+4)=0
The zeros of (x+2)(x^2-2x+4)=0 are x=-2,x=1+sqrt(3)i,x=1-sqrt(3)i