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

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

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

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

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

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

Solution steps

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

The solutions are

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

2 x^{3} - 11 x^{2} + 17 x - 6 = 0

x (2 x - 3) - (2 x + 6) (3 - 2 x)^{2} = 0

y, x = (y + 5)^{3}

y^{4} = 15 y^{2} - 44

10 x^{3} + 70 x^{2} + 120 x = 0

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