While solving a cubic equation, how do I divide a cubic polynomial by a linear one (Python 3)?

Question

I'm making a solver of cubic equations in Python that includes division of polynomials.

from sympy import symbols
# Coefficients
a = int(input("1st coef: "))
b = int(input("2nd coef: "))
c = int(input("3rd coef: "))
d = int(input("Const: "))
# Polynomial of x
def P(x):
    return a*x**3 + b*x**2 + c*x + d
x = symbols('x')
# Find 1 root by Cardano
R = (3*a*c - b**2) / (9*a**2)
Q = (3*b**3 - 9*a*b*c + 27*a**2*d) / (54*a**3)
Delta = R**3 + Q**2
y = (-Q + sqrt(Delta))**(1/3) + (-Q - sqrt(Delta))**(1/3)
x_1 = y - b/(3*a)
# Division
P(x) / (x - x_1) = p(x)
print(p(x)) # Just a placeholder

The program returns an error: "cannot assign to operator" and highlights the P(x) after the # Division comment (worded poorly, yes, but I'm from Russia so idc). What I tried doing was to assign a variable to a polynomial and then dividing:

z = P(x)
w = x - x_1
p = z / w
print(p)

But alas: it just returns a plain old quotient (a = 1, b = 4, c = -9, d = -36):

(x**3 + 4*x**2 - 9*x - 36)/(x - 2.94254537742264)

Does anyone out here knows what to do in this situation (not to mention the non-exact value of x_1: the roots of x^3+4x^2-9x-36=0 are 3, -4, and -3, no floaty-irrational-messy-ugly things in sight)?

tl;dr: Polynomial division confusion and non-exact roots

Answer 1

I am not sure what exactly your question is but here is an attempt at an answer

The line

P(x) / (x - x_1) = p(x)

is problematic for multiple reasons. First of all it's important to know that the = operator in python (and a lot of other modern programming languages) is an assignment operator. You seem to come from more of a math background, so consider it to be something like the := operator. The direction of this is always fixed, i.e. with a = b you are always assigning the value of b to the variable a. In your case you are basically assigning an expression the value of p which does not make much sense:

1. Python can't assign anything to an expression (At least not as far as I know)
2. p(x) is not yet defined

The second problem is that you are mixing python functions with math functions.

A python function looks something like this:

def some_function(some_parameter)

  print("Some important Thing!: ", some_parameter)

  some_return_value = 42
  return some_return_value

It (can) take some variable(s) as input, do a bunch of things with them, and then (can) return something else. They are generally called with the bracket operator (). I.e. some_function(42) translates to execute some_function and substitute the first parameter with the value 42. An expression in sympy however is as far as python is concerned just an object/variable.

So basically you could have just written P = a*x**3 + b*x**2 + c*x + d. What your P(x) function is doing is basically taking the expression a*x**3 + b*x**2 + c*x + d, substituting x for whatever you have put in the brackets, and then giving it back in as a sympy expression. (It's important to understand, that the x in your P python function has nothing to do with the x you define later! Because of that, one usually tries to avoid such "false friends" in coding)

Also, a math function in sympy is really just an expression formed from sympy symbols. As far as sympy is concerned, the return value of the P function is a (mathematical) function of the symbols a,b,c,d and the symbol you put into the brackets. This is why, whenever you want to integrate or differentiate, you will need to specify by which symbol to do that.

So the line should have looked something like this.

p = P(x) /  (x - x_1)

Or you leave replace the P(x) function with P = a*x**3 + b*x**2 + c*x + d and end up with

p = P / (x - x_1)

Thirdly if you would like to have the expression simplified you should take a look here (https://docs.sympy.org/latest/tutorial/simplification.html). There are multiple ways here of simplifying expressions, depending on what sort of expression you want as a result. To make for faster code sympy will only simplify your expression if you specifically ask for it.

You might however be disappointed with the results, as the line

y = (-Q + sqrt(Delta))**(1/3) + (-Q - sqrt(Delta))**(1/3)

will do an implicit conversion to floating point numbers, and you are going to end up with rounding problems. To blame is the (1/3) part which will evaluate to 0.33333333 before ever seeing sympy. One possible fix for this would be

y = (-Q + sqrt(Delta))**(sympy.Rational(1,3)) + (-Q - sqrt(Delta))**(sympy.Rational(1,3))

(You might need to add import sympy at the top)

Generally, it might be worth learning a bit more about python. It's a language that mostly tries to get out of your way with annoying technical details. This unfortunately however also means that things can get very confusing when using libraries like sympy, that heavily rely on stuff like classes and operator overloading. Learning a bit more python might give you a better idea about what's going on under the hood, and might make the distinction between python stuff and sympy specific stuff easier. Basically, you want to make sure to read and understand this (https://docs.sympy.org/latest/gotchas.html).

Let me know if you have any questions, or need some resources :)