#1**+1 **

x^3 + 23x^2 + 3x - 5 = 0

This isn't capable of being factored

There is a "formula" for solving a cubic, but it's messy

Here's a graph of the approximate solutions : https://www.desmos.com/calculator/bqq37uwev1

The approximate solutions are x ≈ -22.859 , -.543, .403

CPhill Oct 23, 2018

#2**0 **

**Mathematica 11 Home Edition gives this "monstrous" solution!!!**

**Solve for x: x^3 + 23 x^2 + 3 x - 5 = 0**

**Look for a simple substitution that eliminates the quadratic term of x^3 + 23 x^2 + 3 x - 5. Eliminate the quadratic term by substituting y = x + 23/3: -5 + 3 (y - 23/3) + 23 (y - 23/3)^2 + (y - 23/3)^3 = 0**

**Write the cubic polynomial on the left-hand side in standard form. Expand out terms of the left hand side: y^3 - (520 y)/3 + 23578/27 = 0**

**Perform the substitution y = z + λ/z. Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later: 23578/27 - 520/3 (z + λ/z) + (z + λ/z)^3 = 0**

**Transform the rational equation into a polynomial equation. Multiply both sides by z^3 and collect in terms of z: z^6 + z^4 (3 λ - 520/3) + (23578 z^3)/27 + z^2 (3 λ^2 - (520 λ)/3) + λ^3 = 0**

**Find an appropriate value for λ in order to make the coefficients of z^2 and z^4 both zero. Substitute λ = 520/9 and then u = z^3, yielding a quadratic equation in the variable u: u^2 + (23578 u)/27 + 140608000/729 = 0**

**Solve for u. Find the positive solution to the quadratic equation: u = 1/27 i (11789 i + 3 sqrt(180831))**

**Perform back substitution on u = 1/27 i (11789 i + 3 sqrt(180831)). Substitute back for u = z^3: z^3 = 1/27 i (11789 i + 3 sqrt(180831))**

**Take the cube root of both sides. Taking cube roots gives 1/3 (i (11789 i + 3 sqrt(180831)))^(1/3) times the third roots of unity: z = 1/3 (i (11789 i + 3 sqrt(180831)))^(1/3) or z = -1/3 (-1)^(1/3) (i (11789 i + 3 sqrt(180831)))^(1/3) or z = 1/3 (-1)^(2/3) (i (11789 i + 3 sqrt(180831)))^(1/3)**

**Perform back substitution with y = z + 520/(9 z).**

**Substitute each value of z into y = z + 520/(9 z): y = 520/(3 (i (3 sqrt(180831) + 11789 i))^(1/3)) + 1/3 (i (11789 i + 3 sqrt(180831)))^(1/3) or y = (520 (-1)^(2/3))/(3 (i (3 sqrt(180831) + 11789 i))^(1/3)) - 1/3 (-1)^(1/3) (i (11789 i + 3 sqrt(180831)))^(1/3) or y = 1/3 (-1)^(2/3) (i (11789 i + 3 sqrt(180831)))^(1/3) - (520 (-1)^(1/3))/(3 (i (3 sqrt(180831) + 11789 i))^(1/3))**

**Simplify each solution. Bring each solution to a common denominator and simplify: y = ((3 i sqrt(180831) - 11789)^(2/3) + 520)/(3 (3 i sqrt(180831) - 11789)^(1/3)) or y = (520 (-1)^(2/3) - (-1)^(1/3) (3 i sqrt(180831) - 11789)^(2/3))/(3 (3 i sqrt(180831) - 11789)^(1/3)) or y = 1/3 ((-1)^(1/3) (3 i sqrt(180831) - 11789)^(2/3) - 520) (i/(11789 i + 3 sqrt(180831)))^(1/3)**

**Perform back substitution on the three roots. Substitute back for x = y - 23/3:**

**x = ((3 i sqrt(180831) - 11789)^(2/3) + 520)/(3 (3 i sqrt(180831) - 11789)^(1/3)) - 23/3 x = (520 (-1)^(2/3) - (-1)^(1/3) (3 i sqrt(180831) - 11789)^(2/3))/(3 (3 i sqrt(180831) - 11789)^(1/3)) - 23/3 x = 1/3 (i/(11789 i + 3 sqrt(180831)))^(1/3) ((-1)^(1/3) (3 i sqrt(180831) - 11789)^(2/3) - 520) - 23/3**

Guest Oct 23, 2018