How do solve for 4x^3+2x-8=0?

4x³ + 2x - 8 = 0 → let: x = (u + v)

4.(u + v)³ + 2.(u + v) - 8 = 0

4.[(u + v)².(u + v)] + 2.(u + v) - 8 = 0

4.[(u² + 2uv + v²).(u + v)] + 2.(u + v) - 8 = 0

4.[u³ + u²v + 2u²v + 2uv² + uv² + v³] + 2.(u + v) - 8 = 0

4.[u³ + v³ + 3u²v + 3uv²] + 2.(u + v) - 8 = 0

4.[(u³ + v³) + (3u²v + 3uv²)] + 2.(u + v) - 8 = 0

4.[(u³ + v³) + 3uv.(u + v)] + 2.(u + v) - 8 = 0

4.(u³ + v³) + 12uv.(u + v) + 2.(u + v) - 8 = 0 → you factorize: (u + v)

4.(u³ + v³) + (u + v).(12uv + 2) - 8 = 0 → suppose that: (12uv + 2) = 0 ←equation (1)

4.(u³ + v³) + (u + v).(0) - 8 = 0

4.(u³ + v³) - 8 = 0 ←equation (2)


You can get a system of 2 equations:

(1) : (12uv + 2) = 0

(1) : 12uv = - 2

(1) : uv = - 1/6

(1) : u³v³ = (- 1/6)³


(2) : 4.(u³ + v³) - 8 = 0

(2) : 4.(u³ + v³) = 8

(2) : u³ + v³ = 2


Let: U = u³

Let: V = v³


You can get a new system of 2 equations:

(1) : UV = (- 1/6)³ ← this is the product P

(2) : U + V = 2 ← this is the sum S


You know that the values S & P are the solutions of the following equation:

x² - Sx + P = 0

x² - 2x + (- 1/6)³ = 0


Δ = (- 2)² - [4 * (- 1/6)³]

Δ = 4 + (4/6³)

Δ = 868/6³

Δ = 217/54 = (1/3)².(217/6)


x₁ = [2 - (1/3).√(217/6)]/2 = 1 - (1/6).√(217/6) ← this is U

x₂ = [2 + (1/3).√(217/6)]/2 = 1 + (1/6).√(217/6) ← this is V


Recall: u³ = U → u = U^(1/3)

Recall: v³ = V → v = V^(1/3)


Recall: x = u + v

x = [1 - (1/6).√(217/6)]^(1/3) + [1 + (1/6).√(217/6)]^(1/3)

x ≈ (- 0.132232483912045) + (1.26040638227384)

x ≈ 1.1281738983618

Hello,

Solve through Cardano's method:
   4𝑥³ + 2𝑥 – 8 = 0
   216𝑥³ + 108𝑥 – 432 = 0    ←←← Multiply by 54 to remove future fractions
   (6𝑥)³ + 18(6𝑥) – 432 = 0

By setting 6𝑥=𝑢+𝑣 and expanding the result:
   (𝑢 + 𝑣)³ + 18(𝑢 + 𝑣) – 432 = 0
   𝑢³ + 𝑣³ + 3(𝑢 + 𝑣)𝑢𝑣 + 18(𝑢 + 𝑣) – 432 = 0
   𝑢³ + 𝑣³ + 3(𝑢 + 𝑣)(𝑢𝑣 + 6) – 432 = 0
   𝑢³ + 𝑣³ = 432 – 3(𝑢 + 𝑣)(𝑢𝑣 + 6)

Adding the condition 𝑢𝑣=-6, we get:
   𝑢³ + 𝑣³ = 432 – 3(𝑢 + 𝑣)(-6 + 6) = 432 – 0 = 432
and
   𝑢³𝑣³ = (𝑢𝑣)³ = (-6)³ = -216
resulting in the system
   { 𝑢³ + 𝑣³ = 432
   { 𝑢³𝑣³ = -216

This makes 𝑢³ and 𝑣³ the roots of the quadratic equation
   𝑋² – 432𝑋 – 216 = 0

Applying quadratic formula:
   ∆ = (-432)² – 4×1×(-216) = 187488 = (12√1302)²
we get
   { 𝑢³ = (432 – 12√1302)/2 = 216 – 6√1302
   { 𝑣³ = (432 + 12√1302)/2 = 216 + 6√1302

Thus
   { 𝑢 = ∛(216 – 6√1302)
   { 𝑣 = ∛(216 + 6√1302)

And since 6𝑥=𝑢+𝑣 :
   6𝑥 = 𝑢 + 𝑣
   6𝑥 = ∛(216 – 6√1302) + ∛(216 + 6√1302)
   𝑥 = [  ∛(216 – 6√1302) + ∛(216 + 6√1302)  ] / 6    ◄◄◄ANSWER
which is the EXACT solution.

Its APPROXIMATE value is:
   𝑥 ≈ 1.128173898

https://www.wolframalpha.com/input/?i=(%E2%88%9B(216+-+6%E2%88%9A1302)+%2B+%E2%88%9B(216+%2B+6%E2%88%9A1302))%2F6

Regards,
Dragon.Jade :-)

4x³ + 2x - 8 = 0

The first thing you can do is simplify this by dividing both sides by 2:

2x³ + x - 4 = 0

Next step is to generate a list of all possible rational roots, which is a factor of the constant term (4) over a factor of the high-degree coefficient (2).

that list is:

±1, ±2, ±4, ±1/2

So of the 8 possible outcomes, test them to see if any are zeroes. If we find one, then we can turn it into a factor, determine the other quadratic factor, then from that, find the other two zeroes:

So, after doing some work off-screen, found that none of them are a rational root, so that plan gets thrown out the window.

The next approach you can take is to use Newton's Method, which has you start with a guess (such as x = 1), then subtract the value of the function over the derivative of the function from the original value. Do this process a few times and you will get a value close.

To show this first step, we get:

f(x) = 2x³ + x - 4
f'(x) = 6x² + 1

So starting with x₀ = 1, solve for x₁ of this equation:

x₁ = x₀ - f(x₀) / f'(x₀)

So we get:

x₁ = x₀ - (2x₀³ + x₀ - 4) / (6x₀² + 1)
x₁ = 1 - (2 * 1³ + 1 - 4) / (6 * 1² + 1)
x₁ = 1 - (2 * 1 - 3) / (6 * 1 + 1)
x₁ = 1 - (2 - 3) / (6 + 1)
x₁ = 1 - (-1) / 7
x₁ = 1 + 0.142857
x₁ = 1.142857

That gets you closer to the real solution. Do the same process a few more times to get closer each loop.

You will find that a decimal approximation of the solution is:

x ≈ 1.1282