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

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

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

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

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

x = 0, x = - 1, x = 1, x = 2 i, x = - 2 i

Solution steps

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

Factor

- x (x^{2} - 1) (x^{2} + 4) : - x (x + 1) (x - 1) (x^{2} + 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

Solve

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

The solutions are

x = 0, x = - 1, x = 1, x = 2 i, x = - 2 i

x^{5} - 3 x + 1 = 0

- 2 (9 x^{2} - 49) (x^{2} - 5) = 0

x^{4} - 7 x^{3} + 12 x^{2} + 4 x - 16 = 0

x^{4} - 38 x^{2} + 72 = 0

- 9 x^{4} - 46 x^{2} + 5 = 0

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