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I have been reading about Cordano's solution, tschirnhaus transformation, etc, for the past few days and one thing that I cannot understand, if anything, is the substitution of y=x-3/a or x=y+3/a (if the coefficient in front of cubed x is 1) . I have found pretty good lectures with Professor William Dunham and other links here and in other places, but none go into any details regarding that substitution. I want to know if there is any general form for it, for it feels very specific and since I don't want to just memorize it I would love to have some material (or any explanation here) to read but my searching lead me nowhere, either it has another popular name I am unaware of or there is nothing, which is unlikely. What little I know is that it is used to remove n-1 degree term to ease calculation but I want to read how it was derived or something similar, a read that isn't too advanced. I want to depress a cubic that has a square term in it, but actually know what or how the substitution that I used came about. Thank you!

Dick Armstrong
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    maybe this could help. – The Integrator May 19 '18 at 20:13
  • So, from the image after the translation we get a second term that gets eliminated. Yes, so it is a translation. Now, the depression makes sense, I suppose, and analytically it makes sense but I do not see it or know how to start visualizing it graphically. Would you happen to have any links or material where they explain this with visuals? Thank you for the link it was very helpful. – Dick Armstrong May 19 '18 at 20:37
  • See this and this. – Dave L. Renfro May 19 '18 at 20:42
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    @DickArmstrong i dont know of any visualizations to depressing a polynomial. Maybe you could graph out a few functions and see what you get. – The Integrator May 19 '18 at 20:49
  • @DaveL.Renfro if the two solutions divided by the amount of solutions i.e. 2, for the quadratic, gives -b/2a where a=1 (from the general form ax^2+bx+c=0, a=1 in your first link), then the three solutions divided by the amount of solutions i.e. 3, for the cubic, will give -b/3a or -b/3 if we divide the whole general equation by a to get x^3 from a*x^3. I used the substitution for the quadratic one from your first link (from your post) on the quadratic equation from the second link (from your post), to see the hows and whats and I think I am starting to see the link (get it? link... sorry) – Dick Armstrong May 19 '18 at 22:20
  • @DaveL.Renfro also in your second link you said "note that x-3 is the midpoint of x and x-6" and that connects to -b/2. Glad you said "midpoint" it made me realize the reasoning for the substitution, if I realized the correct thing. Thank you very much! – Dick Armstrong May 19 '18 at 22:31
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    I just noticed that this 25 November 2014 post might actually be better, except MathJax wasn't available there. Incidentally, I think what led me to notice this symmetrizing about zero idea was a math problem I came came across, solved, and posted on 21 August 2008 (continued) – Dave L. Renfro May 20 '18 at 09:46
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    in my Some turbo-charged high school algebra sci.math thread (follow-up here and here). See also Yiyuan Lee’s solution to Solve the equation $(x-3)(x+9)(x+5)(x-7)=385$. – Dave L. Renfro May 20 '18 at 09:46

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