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

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

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

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

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

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

Solution steps

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

Factor

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

x (x - 1) (x + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 1 = 0 or x + 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 2 = 0 : x = - 2

The solutions are

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

x^{3} = x + 12

x^{3} - 6 x^{2} - 36 x + 40 = 0

x^{3} = 6 + 3 x

16 = x^{2} + 24 x - 6 x^{3}

x^{3} + x^{2} = \frac{( 11 x + 10 )}{3}

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