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

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

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

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

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

x = 0, x = - 1, x = 1

Solution steps

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 1 = 0 or x - 1 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

The solutions are

x = 0, x = - 1, x = 1

16 x^{3} + 80 x^{2} - x - 5 = 0

75 = - 5 + 10 b^{3}

15 x^{4} + 7 x^{2} = 2

14 x^{2} - 24 x - 2 x^{3} = 0

x^{4} - 49 x^{2} = 0

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