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

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

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

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

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

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

Solution steps

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

Switch sides

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

x - 2 = 0 : x = 2

Solve

x + 1 = 0 : x = - 1

The solutions are

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

7 x^{4} - 21 x = 0

x, f = (x^{2} - 2)^{2}

x, x^{8} + 15 x^{4} = 16

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

0 = 4 x^{3} - 45 x^{2} - 4 x + 40

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