How can I fully factor 4x - 5x^3 + 6?

If you want to factorize the polynomial, so have to find the roots:

4x - 5x³ + 6 = 0

- 5x³ + 4x + 6 = 0 → let: x = (u + v)

- 5.(u + v)³ + 4.(u + v) + 6 = 0

- 5.[(u + v)².(u + v)] + 4.(u + v) + 6 = 0

- 5.[(u² + 2uv + v²).(u + v)] + 4.(u + v) + 6 = 0

- 5.[u³ + u²v + 2u²v + 2uv² + uv² + v³] + 4.(u + v) + 6 = 0

- 5.[u³ + v³ + 3u²v + 3uv²] + 4.(u + v) + 6 = 0

- 5.[(u³ + v³) + (3u²v + 3uv²)] + 4.(u + v) + 6 = 0

- 5.[(u³ + v³) + 3uv.(u + v)] + 4.(u + v) + 6 = 0

- 5.(u³ + v³) - 15uv.(u + v) + 4.(u + v) + 6 = 0 → you factorize (u + v)

- 5.(u³ + v³) - (u + v).(15uv - 4) + 6 = 0 → suppose that: (15uv - 4) = 0 ← equation (1)

- 5.(u³ + v³) - (u + v).(0) + 6 = 0

- 5.(u³ + v³) + 6 = 0 ← equation (2)



You can get a system of 2 equations:

(1) : (15uv - 4) = 0

(1) : 15uv = 4

(1) : uv = 4/15

(1) : u³v³ = (4/15)³


(2) : - 5.(u³ + v³) + 6 = 0

(2) : u³ + v³ = 6/5


Let: U = u³

Let: V = v³


You can get a new system of 2 equations:

(1) : UV = (4/15)³ ← this is the product P

(2) : U + V = 6/5 ← this is the sum S


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

x² - Sx + P = 0

x² - (6/5).x + (4/15)³ = 0


Δ = (6/5)² - [4 * (4/15)³]

Δ = (36/25) - (256/3375)

Δ = 4604/3375

Δ = (1151 * 2²)/(15 * 15²)

Δ = (1151/15) * (2/15)²


x₁ = [(6/5) + (2/15)√(1151/15)]/2 = (3/5) + (1/15)√(1151/15) ← this is U

x₂ = [(6/5) - (2/15)√(1151/15)]/2 = (3/5) - (1/15)√(1151/15) ← this is V


Recall: u³ = U → u = U^(1/3)

u = [(3/5) + (1/15)√(1151/15)]^(1/3)

u ≈ 1.05790965895423



Recall: v³ = V → v = V^(1/3)

v = [(3/5) - (1/15)√(1151/15)]^(1/3)

v ≈ 0.252069412931036


Recall: x = (u + v)

x ≈ 1.05790965895423 + 0.252069412931036

x ≈ 1,30997907188527 ← this is the unique root xo

xo ≈ 1,31 → so you can factorize (x - xo)



= - 5x³ + 4x + 6

= (x - xo).(- 5x² + ax + b) → you expand

= - 5x³ + ax² + bx + 5x².xo - ax.xo - b.xo → you group

= - 5x³ + x².(a + 5xo) + x.(b - a.xo) - b.xo → you compare to: - 5x³ + 4x + 6


- b.xo = 6 → b = - 6/xo

b - a.xo = 4 → a.xo = b - 4 → a = (b - 4)/xo

a + 5xo = 0 → a = - 5xo


Resume:

b = - 6/xo → where: xo ≈ 1,31

b = - 6/1.31

b ≈ - 4.58


a = (b - 4)/xo → where: xo ≈ 1,31

a = (b - 4)/1.31 → where: b ≈ - 4.58

a = (- 4.58 - 4)/1.31

a ≈ - 6.55


a + 5xo = 0

a = - 5xo → where: xo ≈ 1,31

a ≈ - 6.55 of course because the previous result is a ≈ - 6.55




= - 5x³ + 4x + 6

= (x - xo).(- 5x² + ax + b) → where: xo ≈ 1,31

= (x - 1.31).(- 5x² + ax + b) → where: a ≈ - 6.55

= (x - 1.31).(- 5x² - 6.55x + b) → where: b ≈ - 4.58

= (x - 1.31).(- 5x² - 6.55x - 4.58)

= - (x - 1.31).(5x² + 6.55x + 4.58) ← this is the factorization

It is a cubic with no rational roots and so does not factor.

(note: 4th and higher degree polynomials may still factor when they have no rational roots)