What are the solutions to the equation (x+3)(x^2+1)^2=0 ?

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

Find the zeros of (x+3)(x^2+1)^2=0

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

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(x + 3) (x^{2} + 1)^{2} = 0

x = - 3, x = i, x = - i

Solution steps

(x + 3) (x^{2} + 1)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 3 = 0 : x = - 3

Solve

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

The solutions are

x = - 3, x = i, x = - i

2 t^{4} + 432 t = 2 t

48 x^{2} = 3 x^{4}

α^{5} - 5 α + 4 = 0

x, y = x^{6}

x, (2 x + 3)^{4} = 3 y^{4}

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