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Mathematics Algebra Wolfram Language

What method is there to retrieve which quantity to replace and complete factorization?

Lee Tao

Posted 2 years ago

What method is there to retrieve which quantity to replace and complete factorization? The following polynomial equation cannot be directly factorized, and the result obtained by directly factoring is consistent with the original equation. -2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0 // Factor -2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0 But there is another identity :y0^2==2p x0 y0^2 == 2 p x0 By using this identity, the 2px0 and y0^2 terms contained in polynomial equations can be replaced with y0^2 and 2px0, respectively, achieving the goal of factorization of polynomial equations. y0^2 -> 2 p x0 2 p x0 -> y0^2 The problem is that before factoring, we are not sure which term or terms in the polynomial equation can be factorized by replacing them with identities. So we can only see if a polynomial equation can be factorized by constantly trying to replace a certain term in the equation So there are the following manual calculation steps : It should be noted that the method and steps of replacing and deforming the original polynomial equation through the isometric relationship of identities are not unique,There are multiple methods! but rather aim to factorize the original polynomial equation after identity transformation. Provide a not very simple path that can ultimately factorize polynomial equations STEP1 : Replace the( -2 m p x0) term in the polynomial equation with (- my0^2), -2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0 /. -2 m p x0 -> -m y0^2 -2 m p s - s^2 t + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 - m y0^2 == 0 -2 m p s - s^2 t + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 - m y0^2 == 0 // Factor -2 m p s - s^2 t + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 - m y0^2 == 0 This substitution cannot factorize polynomial equations STEP2 : Replace the(2 m^2 p t x0) term in the polynomial equation with (m^2 t y0^2) -2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0 /. (2 m^2 p t x0) -> m^2 t y0^2 -2 m p s - s^2 t - 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 + m^2 t y0^2 == 0 -2 m p s - s^2 t - 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 + m^2 t y0^2 == 0 // Factor -2 m p s - s^2 t - 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 + m^2 t y0^2 == 0 This substitution cannot factorize polynomial equations too STEP3 : Replace these two terms in polynomial functions separately.(2 m^2 p t x0)->m^2 t y0^2,(-2m p x0)->-m y0^2 -2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0 /. {(2 m^2 p t x0) -> m^2 t y0^2, (-2 m p x0) -> -m y0^2} -2 m p s - s^2 t + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 - m y0^2 + m^2 t y0^2 == 0 -2 m p s - s^2 t + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 - m y0^2 + m^2 t y0^2 == 0 // Factor -2 m p s - s^2 t + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 - m y0^2 + m^2 t y0^2 == 0 This substitution cannot factorize polynomial equations again too STEP4 : First, replace an item:(2 m^2 p t x0)->m^2 t y0^2 -2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0 /. (2 m^2 p t x0) -> m^2 t y0^2 -2 m p s - s^2 t - 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 + m^2 t y0^2 == 0 Then perform an identical transformation and replacement of a term: -2m p x0==-4 m p x0+2 m p x0==2 m p x0-2m y0^2 THEN -2 m p s - s^2 t - 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 + m^2 t y0^2 == 0 /. (-2 m p x0) -> 2 m p x0 - 2 m y0^2 -2 m p s - s^2 t + 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 - 2 m y0^2 + m^2 t y0^2 == 0 -2 m p s - s^2 t + 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 - 2 m y0^2 + m^2 t y0^2 == 0 // Factor -((s - x0 + m y0) (2 m p + s t - t x0 + 2 y0 - m t y0)) == 0 After several attempts, the polynomial can finally be factorized after completing its final identity transformation All the processes combined are: -2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0 /. {(2 m^2 p t x0) -> m^2 t y0^2, (-2 m p x0) -> 2 m p x0 - 2 m y0^2} // Factor -((s - x0 + m y0) (2 m p + s t - t x0 + 2 y0 - m t y0)) == 0 My requirement is that after knowing a polynomial equation and an identity, the polynomial equation must be able to factorize after undergoing a series of identity transformations. So, is there a method for mathematica to find the path of identical deformation and ultimately factorize successfully? -2 m p s-s^2 t-2 m p x0+2 m^2 p t x0+2 s t x0-t x0^2-2 m^2 p y0-2 s y0+2 x0 y0==0ANDy0^2==2p x0

What method is there to retrieve which quantity to replace and complete factorization?

The following polynomial equation cannot be directly factorized, and the result obtained by directly factoring is consistent with the original equation.

-2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 
   2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0 // Factor



-2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 
  2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0

But there is another identity :y0^2==2p x0

y0^2 == 2 p x0

By using this identity, the 2px0 and y0^2 terms contained in polynomial equations can be replaced with y0^2 and 2px0, respectively, achieving the goal of factorization of polynomial equations.

y0^2 -> 2 p x0

2 p x0 -> y0^2

The problem is that before factoring, we are not sure which term or terms in the polynomial equation can be factorized by replacing them with identities. So we can only see if a polynomial equation can be factorized by constantly trying to replace a certain term in the equation

So there are the following manual calculation steps :

It should be noted that the method and steps of replacing and deforming the original polynomial equation through the isometric relationship of identities are not unique,There are multiple methods! but rather aim to factorize the original polynomial equation after identity transformation.

Provide a not very simple path that can ultimately factorize polynomial equations

STEP1 : Replace the( -2 m p x0) term in the polynomial equation with (- my0^2),

-2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 
   2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0 /. -2 m p x0 -> -m y0^2


-2 m p s - s^2 t + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 
  2 s y0 + 2 x0 y0 - m y0^2 == 0


-2 m p s - s^2 t + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 
   2 s y0 + 2 x0 y0 - m y0^2 == 0 // Factor


-2 m p s - s^2 t + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 
  2 s y0 + 2 x0 y0 - m y0^2 == 0

This substitution cannot factorize polynomial equations

STEP2 : Replace the(2 m^2 p t x0) term in the polynomial equation with (m^2 t y0^2)

-2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 
   2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0 /. (2 m^2 p t x0) -> m^2 t y0^2


-2 m p s - s^2 t - 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 
  2 s y0 + 2 x0 y0 + m^2 t y0^2 == 0



-2 m p s - s^2 t - 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 
   2 s y0 + 2 x0 y0 + m^2 t y0^2 == 0 // Factor

-2 m p s - s^2 t - 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 
  2 s y0 + 2 x0 y0 + m^2 t y0^2 == 0

This substitution cannot factorize polynomial equations too

STEP3 : Replace these two terms in polynomial functions separately.(2 m^2 p t x0)->m^2 t y0^2,(-2m p x0)->-m y0^2

-2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 
   2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 
  0 /. {(2 m^2 p t x0) -> m^2 t y0^2, (-2 m p x0) -> -m y0^2}


-2 m p s - s^2 t + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 -
   m y0^2 + m^2 t y0^2 == 0


-2 m p s - s^2 t + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 -
    m y0^2 + m^2 t y0^2 == 0 // Factor


-2 m p s - s^2 t + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 -
   m y0^2 + m^2 t y0^2 == 0

This substitution cannot factorize polynomial equations again too

STEP4 : First, replace an item:(2 m^2 p t x0)->m^2 t y0^2

-2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 
   2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 0 /. (2 m^2 p t x0) -> m^2 t y0^2




-2 m p s - s^2 t - 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 
  2 s y0 + 2 x0 y0 + m^2 t y0^2 == 0

Then perform an identical transformation and replacement of a term:

-2m p x0==-4 m p x0+2 m p x0==2 m p x0-2m y0^2

THEN

-2 m p s - s^2 t - 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 
   2 s y0 + 2 x0 y0 + m^2 t y0^2 == 0 /. (-2 m p x0) -> 
  2 m p x0 - 2 m y0^2


-2 m p s - s^2 t + 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 
  2 s y0 + 2 x0 y0 - 2 m y0^2 + m^2 t y0^2 == 0




-2 m p s - s^2 t + 2 m p x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 
   2 s y0 + 2 x0 y0 - 2 m y0^2 + m^2 t y0^2 == 0 // Factor



-((s - x0 + m y0) (2 m p + s t - t x0 + 2 y0 - m t y0)) == 0

After several attempts, the polynomial can finally be factorized after completing its final identity transformation

All the processes combined are:

-2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - 
t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0 == 
0 /. {(2 m^2 p t x0) -> m^2 t y0^2, (-2 m p x0) -> 
2 m p x0 - 2 m y0^2} // Factor



-((s - x0 + m y0) (2 m p + s t - t x0 + 2 y0 - m t y0)) == 0

My requirement is that after knowing a polynomial equation and an identity, the polynomial equation must be able to factorize after undergoing a series of identity transformations. So, is there a method for mathematica to find the path of identical deformation and ultimately factorize successfully?

-2 m p s-s^2 t-2 m p x0+2 m^2 p t x0+2 s t x0-t x0^2-2 m^2 p y0-2 s y0+2 x0 y0==0ANDy0^2==2p x0

POSTED BY: Lee Tao

6 Replies

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Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 2 years ago

From the relation `y0^2 == 2 p x0` we get that `p == y0^2/(2x0)`. Replacing this into the polynomial we can factorize, but it is not a polynomial any more. We see that if we replace `y0^2 -> 2 p x0` the x0 at the denominator disappears: pol = -2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0; rel = y0^2 == 2 p x0; fac0 = Factor[pol /. Solve[rel, p][[1]]] fac1 = Factor[fac0 /. Rule @@ rel] I have no idea how to generalize this.

POSTED BY: Gianluca Gorni

Lee Tao

Posted 2 years ago

I came across a new example today. For polynomials： -2 a^2 b^2 k + a^2 b^2 t - a^2 m^2 t - 2 b^2 m x0 - 2 a^2 k m t x0 - b^2 t x0^2 - a^2 k^2 t x0^2 + 2 a^2 k m y0 + 2 b^2 x0 y0 + 2 a^2 k^2 x0 y0 There is an identity relation： x0^2/a^2+y0^2/b^2==1 The polynomial deformation can certainly be decomposed by the identity relation. But the specific path is not clear. The result from the above code is not very satisfactory. correct answer is： (x0 k + m - y0) (a^2 (2 y0 - x0 t) k - a^2 t m - 2 b^2 x0 - a^2 y0 t) How to modify the previous code when faced with this problem？

POSTED BY: Lee Tao

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 2 years ago

Here is a way pol = -2 a^2 b^2 k + a^2 b^2 t - a^2 m^2 t - 2 b^2 m x0 - 2 a^2 k m t x0 - b^2 t x0^2 - a^2 k^2 t x0^2 + 2 a^2 k m y0 + 2 b^2 x0 y0 + 2 a^2 k^2 x0 y0; rel = x0^2/a^2 + y0^2/b^2 == 1; sost1 = Solve[rel /. (1/a^2 -> 1/temp), temp][[1, 1]] /. temp -> a^2 raz1 = Factor[pol /. sost1] // ExpandDenominator raz2 = (x0*raz1 /. Reverse[sost1])/x0 sost2 = Rule @@ Expand[MultiplySides[Equal @@ sost1, 2 (b^2 - y0^2)/a^2, GenerateConditions -> False]] pol2 = Factor[raz2 /. sost2] Simplify[pol2 - pol, rel]

Here is a way

pol = -2 a^2 b^2 k + a^2 b^2 t - a^2 m^2 t -
   2 b^2 m x0 - 2 a^2 k m t x0 - b^2 t x0^2 -
   a^2 k^2 t x0^2 + 2 a^2 k m y0 + 2 b^2 x0 y0 +
   2 a^2 k^2 x0 y0;
rel = x0^2/a^2 + y0^2/b^2 == 1;
sost1 = Solve[rel /. (1/a^2 -> 1/temp), temp][[1, 1]] /. temp -> a^2
raz1 = Factor[pol /. sost1] // ExpandDenominator
raz2 = (x0*raz1 /. Reverse[sost1])/x0
sost2 = Rule @@ Expand[MultiplySides[Equal @@ sost1,
    2 (b^2 - y0^2)/a^2,
    GenerateConditions -> False]]
pol2 = Factor[raz2 /. sost2]
Simplify[pol2 - pol, rel]

POSTED BY: Gianluca Gorni

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 2 years ago

I don't know the answer. I just played around with your polynomial: pol = -2 m p s - s^2 t - 2 m p x0 + 2 m^2 p t x0 + 2 s t x0 - t x0^2 - 2 m^2 p y0 - 2 s y0 + 2 x0 y0; rel = y0^2 == 2 p x0; fac0 = Factor[pol /. Solve[rel, p][[1]]] fac1 = fac0 /. Rule @@ rel fac2 = fac1[[3]] (fac1/fac1[[3]] // Expand) Simplify[pol - fac2, rel]

I don't know the answer. I just played around with your polynomial:

pol = -2 m p s - s^2 t - 2 m p x0 +
   2 m^2 p t x0 + 2 s t x0 - t x0^2 -
   2 m^2 p y0 - 2 s y0 + 2 x0 y0;
rel = y0^2 == 2 p x0;
fac0 = Factor[pol /. Solve[rel, p][[1]]]
fac1 = fac0 /. Rule @@ rel
fac2 = fac1[[3]] (fac1/fac1[[3]] // Expand)
Simplify[pol - fac2, rel]

POSTED BY: Gianluca Gorni

Lee Tao

Posted 2 years ago

What is the idea behind your code to achieve factorization of the original polynomial through identity transformation? Can you explain it in detail? Thank you!

POSTED BY: Lee Tao

Lee Tao

Posted 2 years ago

To go a step further. For any polynomial and an identity. If the polynomial can achieve the purpose of factorization through the transformation of identity. Can there be a universal algorithm program to solve this kind of problem?

POSTED BY: Lee Tao

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