What are the solutions to the equation x(x^2+2x+3)=x^3-1 ?

The solutions to the equation x(x^2+2x+3)=x^3-1 are x=-1/2 ,x=-1

Find the zeros of x(x^2+2x+3)=x^3-1

The zeros of x(x^2+2x+3)=x^3-1 are x=-1/2 ,x=-1

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x (x^{2} + 2 x + 3) = x^{3} - 1

x = - \frac{1}{2}, x = - 1

Solution steps

x (x^{2} + 2 x + 3) = x^{3} - 1

Expand

x (x^{2} + 2 x + 3) : x^{3} + 2 x^{2} + 3 x

x^{3} + 2 x^{2} + 3 x = x^{3} - 1

Move

1

to the left side

x^{3} + 2 x^{2} + 3 x + 1 = x^{3}

Move

x^{3}

to the left side

2 x^{2} + 3 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

3^{2} - 4 \cdot 2 \cdot 1 = 1

x_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 1}{2 \cdot 2}, x_{2} = \frac{- 3 - 1}{2 \cdot 2}

x = \frac{- 3 + 1}{2 \cdot 2} : - \frac{1}{2}

x = \frac{- 3 - 1}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

x = - \frac{1}{2}, x = - 1

2 x^{2} = - 9 x - 10

15 m (m + 5) = 38 m - 20

x^{2} + 2 4^{2} = 2 6^{2}

x^{2} - 4 x - 3 = 0

6 x^{2} + 450 = 0

What are the solutions to the equation x(x^2+2x+3)=x^3-1 ?
The solutions to the equation x(x^2+2x+3)=x^3-1 are x=-1/2 ,x=-1
Find the zeros of x(x^2+2x+3)=x^3-1
The zeros of x(x^2+2x+3)=x^3-1 are x=-1/2 ,x=-1