What are the solutions to the equation 3x^3-4x+12=0 ?

The solutions to the equation 3x^3-4x+12=0 are x\approx-1.86497…

Find the zeros of 3x^3-4x+12=0

The zeros of 3x^3-4x+12=0 are x\approx-1.86497…

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3 x^{3} - 4 x + 12 = 0

x \approx - 1.86497 \dots

Solution steps

3 x^{3} - 4 x + 12 = 0

Find one solution for

3 x^{3} - 4 x + 12 = 0

using Newton-Raphson:

x \approx - 1.86497 \dots

Apply long division:

\frac{3 x ^{3} - 4 x + 12}{x + 1.86497 \dots} = 3 x^{2} - 5.59492 \dots x + 6.43440 \dots

3 x^{2} - 5.59492 \dots x + 6.43440 \dots \approx 0

Find one solution for

3 x^{2} - 5.59492 \dots x + 6.43440 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

x \approx - 1.86497 \dots

7 x^{4} - 42 x^{2} - 35 x = 0

x^{4} - x^{2} = 1

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

2 t^{4} + 432 t = 2 t

48 x^{2} = 3 x^{4}

What are the solutions to the equation 3x^3-4x+12=0 ?
The solutions to the equation 3x^3-4x+12=0 are x\approx-1.86497…
Find the zeros of 3x^3-4x+12=0
The zeros of 3x^3-4x+12=0 are x\approx-1.86497…