What are the solutions to the equation 3s^3-5s+2=0 ?

The solutions to the equation 3s^3-5s+2=0 are s=1,s=(-3+sqrt(33))/6 ,s=(-3-sqrt(33))/6

Find the zeros of 3s^3-5s+2=0

The zeros of 3s^3-5s+2=0 are s=1,s=(-3+sqrt(33))/6 ,s=(-3-sqrt(33))/6

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3 s^{3} - 5 s + 2 = 0

s = 1, s = \frac{- 3 + 33}{6}, s = \frac{- 3 - 33}{6}

Decimal Notation

s = 1, s = 0.45742 \dots, s = - 1.45742 \dots

Solution steps

3 s^{3} - 5 s + 2 = 0

Factor

3 s^{3} - 5 s + 2 : (s - 1) (3 s^{2} + 3 s - 2)

(s - 1) (3 s^{2} + 3 s - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

s - 1 = 0 or 3 s^{2} + 3 s - 2 = 0

Solve

s - 1 = 0 : s = 1

Solve

3 s^{2} + 3 s - 2 = 0 : s = \frac{- 3 + 33}{6}, s = \frac{- 3 - 33}{6}

The solutions are

s = 1, s = \frac{- 3 + 33}{6}, s = \frac{- 3 - 33}{6}

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

(x^{2} + 4 x)^{2} - 17 (x^{2} + 4 x) = 60

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

x, x^{3} - 6 x^{2} + 6 x + 5 = 0

(z^{3} + i - 1) (1 - 2 z + 2 z^{2}) = 0

What are the solutions to the equation 3s^3-5s+2=0 ?
The solutions to the equation 3s^3-5s+2=0 are s=1,s=(-3+sqrt(33))/6 ,s=(-3-sqrt(33))/6
Find the zeros of 3s^3-5s+2=0
The zeros of 3s^3-5s+2=0 are s=1,s=(-3+sqrt(33))/6 ,s=(-3-sqrt(33))/6