1

A similar question appears here, though I wish to ask for polynomials of higher degree.

Let's say we have the polynomial $f(x)=x^4-3x^3-3x^2+4x-6$, and we wish to find a polynomial with roots 3 greater than those of $f(x)$. I can see that putting $g(x)=f(x-3)$ works, but then we have the expression $$(x-3)^4-3(x-3)^3-3(x-3)^2-4(x-3)+6$$ which will take quite a while to evaluate. (It comes out to $\boldsymbol{x^4-15x^3+78x^2-175x+441}$).

Is there a shortcut to find the polynomial above through some 'trick' (e.g. some form of synthetic division?)

Furthermore, if we wish to find a polynomial with roots $k$ greater than a given polynomial, will this new polynomial have a given 'form' of $k$? E.g. what if I wanted to solve the above question, but with my roots being $4,5,$ or $6$ greater than those of the given polynomial?

palimpsest
  • 51
  • 1
    Is there a shortcut Not really. What you have is the shortcut. – dxiv Mar 04 '18 at 00:47
  • You say "which will take quite a while to evaluate"? That isn't a very precise criterion. If you are interested in the computational complexity, then computing the coefficients of $f(x - 3)$ using Horner's rule is quite efficient. – Rob Arthan Mar 04 '18 at 01:50

1 Answers1

1

Well, your roots will be shifted $3$ units to the right. From there, you can calculate the new roots, using the sum and product of roots formulae to calculate the coefficients of each term. This can be considered a shortcut of sorts. For a polynomial $f(x)$ where $$f(x)=a_nx^n+...+a_1x+a_0$$ $$\alpha\beta\gamma...=(-1)^n\frac{a_0}{a_n}$$ $$\alpha+\beta+\gamma+...=-\frac{a_{n-1}}{a_n}$$ The greek letters represent the roots.

There are other formulae for finding $a_{n-2}$ and so forth. I will give $a_{n-2}$ in cyclic sum notation, but the rest are easily accessible and directly quotable. $$\sum \alpha\beta=\frac{a_{n-2}}{a_n}$$

Harry Alli
  • 2,091
  • apologies, but what do the greek letters on the LHS of the 2nd line denote? – palimpsest Mar 04 '18 at 00:48
  • 1
    Well, his method is more efficient since he doesn't have to calculate the (complex) roots. – Paolo Leonetti Mar 04 '18 at 00:48
  • @Mathematrix see the edit above please – Harry Alli Mar 04 '18 at 00:52
  • How does your method work for the given example without first finding all the roots? – Rob Arthan Mar 04 '18 at 01:52
  • 1
    @RobArthan I guess what the answer is getting at is using Vieta's relations for the symmetric polynomials, which does not require actually finding the original roots e.g. $;\sum (\alpha+3)$ $=\sum \alpha + 4 \cdot 3$ $= 3+12=15,$, then $;\sum (\alpha+3)(\beta+3)$ $= \sum \alpha \beta + 6 \cdot \sum \alpha + 6 \cdot 9 = \ldots;$ It is a legitimate approach, though not necessarily much of a shortcut vs. the direct expansion. – dxiv Mar 04 '18 at 02:07
  • @dxiv I was just about to type something similar, thanks! But yes, I guess this is more of an alternative which would be somewhat of a shortcut in some cases and vice versa in others, depending on whether the numbers are easy to work with – Harry Alli Mar 04 '18 at 02:10
  • Thanks for the clarification. I think it's really down to the OP to explain what criteria are relevant in his/her application. – Rob Arthan Mar 04 '18 at 02:20