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

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

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

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

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

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

Solution steps

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

Move

2 (x + 1)

to the left side

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

Solve

x + 2 = 0 : x = - 2

The solutions are

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

y, 2 x^{3} y^{1} = y (2 x^{2} - y^{2})

x^{10} + 6 x^{5} - 16 = 0

x^{3} - 7 x - 6

(\frac{1}{3} x)^{2} = (\frac{1}{6} x)^{3}

z^{3} - z + 6 = 0

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