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Solve for the zeros of the cubic polynomial

 

𝑥3 + 23𝑥2 + 3𝑥 − 5

 Oct 23, 2018
 #1
avatar+98196 
+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

 

 

cool cool cool

 Oct 23, 2018
 #2
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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

 Oct 23, 2018

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