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It's well known that given a polynomial $p(x)$ of the form$$\sum_{k=0}^\infty a_kx^n=a_nx^n+a_{n-1}x^{n-1}+\ldots a_1x+a_0\tag1$$ We can substitute $x=\frac{y-a_{n-1}}{n}$ to eleminate $a_{n-1}x^{n-1}$ term to get its reduced or depressed form. But, I'm now wondering if it's possible to eliminate other terms, such as $a_{n-2}x^{n-2}$ or even $a_{n-4}x^{n-4}$, with the same type of substitution method.

Question: Is it possible to substitute $x$ with a value into a polynomial$$p(x)=a_nx^n+a_{n-2}x^{n-2}+a_{n-3}x^{n-3}+\ldots+a_1x+a_0\tag2$$ To eliminate other terms such as $a_{n-3}x^{n-3}$?

I tried substituting $x$ with $\alpha+\beta$ into $x^3+ax+b=0$, but that got me nowhere. Perhaps you have to substitute $x$ with something else such as $x=\alpha+\beta^2$?

Any sort of help is appreciated!

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Yes. Kinda. Example: Let $p(x)=x^3+ax+b$. Then $p(x+\sqrt{-a/3}) = \sqrt{-3a}\cdot x^2+(2 \sqrt{-a} a)/(3 \sqrt{3})+b+x^3$ is a polynomial without linear term. Similar should be possible to make other terms vanish. But note

  1. In general it only works with complex numbers. I.e. the neccessary shift to make the $x^{n-k}$ term vanish is the root of a polynomial of degree $k$. This may not exist in real numbers. And for higher $k$ it will be a really ugly expression.
  2. If you make one term vanish, all others will generally re-appear. So the whole scheme is not that useful when trying to solve a polynomial.
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  • $\begingroup$ Man, I was so excited about this... Anyways, is there a way to find the substitution that will get rid of the term that you so desire? Similar to substituting $x=\frac {y-a_{n-1}}{n}$. Something easy to remember. $\endgroup$
    – Frank
    Nov 10 '16 at 22:23
  • $\begingroup$ And is it possible to keep things in the rational? Like what if $a>3$. We would then introduce complex numbers. My goal, is to minimize the amount of complex figures I use in this polynomial... $\endgroup$
    – Frank
    Nov 10 '16 at 22:34
  • $\begingroup$ "easy to remember"? I don't think so. But using some computeralgebra, it is not hard to find. Expand $p(x+\beta)$ into powers of $x$ and solve for $a_k=0$. Thats how I found this particular case: wolframalpha.com/input/?i=(x%2Bsqrt(-a%2F3))%5E3%2Ba*(x%2Bsqrt(-a%2F3))%2Bb $\endgroup$
    – Simon
    Nov 10 '16 at 22:38

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