What are the solutions to the equation 10=-2x^3-2x+14 ?

The solutions to the equation 10=-2x^3-2x+14 are x=1,x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2

Find the zeros of 10=-2x^3-2x+14

The zeros of 10=-2x^3-2x+14 are x=1,x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2

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10 = - 2 x^{3} - 2 x + 14

x = 1, x = - \frac{1}{2} + i \frac{7}{2}, x = - \frac{1}{2} - i \frac{7}{2}

Solution steps

10 = - 2 x^{3} - 2 x + 14

Switch sides

- 2 x^{3} - 2 x + 14 = 10

Move

10

to the left side

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

Factor

- 2 x^{3} - 2 x + 4 : - 2 (x - 1) (x^{2} + x + 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x^{2} + x + 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

x^{2} + x + 2 = 0 : x = - \frac{1}{2} + i \frac{7}{2}, x = - \frac{1}{2} - i \frac{7}{2}

The solutions are

x = 1, x = - \frac{1}{2} + i \frac{7}{2}, x = - \frac{1}{2} - i \frac{7}{2}

9 x^{4} - 5 x^{2} - 4 = 0

3 x^{3} + 15 x^{2} - 12 x = 0

2 x^{3} + 5 x^{2} - 6 x - 9

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

a, a^{4} = \frac{4}{3} a^{2} b + \frac{4}{9} b^{2}

What are the solutions to the equation 10=-2x^3-2x+14 ?
The solutions to the equation 10=-2x^3-2x+14 are x=1,x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2
Find the zeros of 10=-2x^3-2x+14
The zeros of 10=-2x^3-2x+14 are x=1,x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2