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I am working on an exercize and I know how to more bruteforcely solve it through pure algebra in its simplest form, but it's such a massive mess to demonstrate so I would like to see if there is another way to solve this

Let $x^3 + ax^2+bx+c$ be a given polynomial (with $a,b,c\in Z$ being acctual numbers but for here I am curiuos in the general case) and we know that it has the roots $x_1,x_2,x_3$ which are given, now determine the polynomial that has the following roots $x_1^2$, $x_2^2$ and $x_3^2$

There are several more but I am mostly curious at how one would go about this in a more elegant mannar than just applying the cubical solution formula and then use $p(x)=(x-x_1^2)(x-x_2^2)(x-x_3^2)$ simplifying and such, while possible it is such a hazzle. Is there a more elegant manner?

For the sake of an example, let's take the polynomial $x^3-7x^2+6x-3$ and it has the roots $x_1$ etc, how would I find the coefficients for the new polynomial with the roots of the stated polynomials being squared?

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  • $\begingroup$ An idea would be to factorize $p(x^2)$ as a product of $\;x^3 + ax^2+bx+c\;$ and the 'conjugate' $\;x^3-ax^2+bx-c\;$ (when the sign of the $x_i$ are changed). Replacing $x$ by $x^2$ in the result should return the wished polynomial if I didn't make any error at least... $\endgroup$ – Raymond Manzoni Feb 26 '15 at 14:58
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Consider $z:=x^2$ and factorize $\;p(z)=p(x^2)=(x^2-x_1^2)(x^2-x_2^2)(x^2-x_3^2)\;$ as a product of $\;(x^3 + ax^2+bx+c)\;$ and the 'conjugate' $\;(x^3-ax^2+bx-c)=(x+x_1)(x+x_2)(x+x_3)\;$ (obtained when the signs of the three $x_i$ are reversed as you may notice by replacing $x$ with $-x$).

The substitution of $x^2$ by $z$ in the result will return the wished polynomial (since the roots of $\;p(z)=0\,$ are indeed $z=x_i^2\,$).

I obtained thus $\;p(z)=z^3+(2b-a^2)\,z^2+(b^2-2ac)\,z-c^2$.

I considered the simple example $\;(x-1)(x-3)(x-5)=x^3-9\,x^2+23\,x-15\;$ and used the previous formula to get following result (as confirmed by Wolfram Alpha) : $$p(x)=x^3-35\,x^2+259\,x-225$$

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    $\begingroup$ Agrees with my answer $\ x(x+b)^2-(ax+c)^2\ \ $ $\endgroup$ – Bill Dubuque Feb 26 '15 at 15:32
  • $\begingroup$ @Bill Dubuque: Yes :-) Thanks for the confirmation (both points of view are interesting!). $\endgroup$ – Raymond Manzoni Feb 26 '15 at 15:33
  • $\begingroup$ This is so brilliant! $\endgroup$ – Zelos Malum Feb 26 '15 at 17:36
  • $\begingroup$ Would this method be possible for permutations of the roots and other powers? Like $x_1^3$, $x_1^4$ etc? or $x_1 x_2^2$? Or does an entirely idfferent approch need to be used? $\endgroup$ – Zelos Malum Feb 26 '15 at 19:54
  • $\begingroup$ Thanks @ZelosMalum, see too the method proposed by Bill Dubuque. Well, I don't know concerning the general case but for $(x_i)^n$ for example I would try to replace the $(x,-x)$ that I considered by $(x\,e^{2\pi i k/n})_{k=0,1,\cdots,n-1}$ that is use a complete complex factorization (the method proposed could thus be powerful!). Wishing you fine generalizations, $\endgroup$ – Raymond Manzoni Feb 26 '15 at 22:12
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$\!\begin{align} {\rm If}\ \ \color{#c00}{z = x^2}\ \ {\rm then}\ \ 0 = &\ \ \color{#c00}{x^2}x\!+a\color{#c00}{x^2}\!+bx+c\\ = &\ \ \ \color{#c00}z\ x+a\ \color{#c00}z+bx+c\\ \Rightarrow\!&\ \ (z\!+\!b)\,x = -(az\!+\!c)\\ \Rightarrow\!&\ \ (z\!+\!b)^2 z =\ \ (az\!+\!c)^2\ \ \text{by squaring prior}\\ \end{align}$

The prior equation yields a cubic in $z$ having roots $\, z = x^2$

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There is a very simple way to do so.
In the cubic polynomial, just replace is $x$ by $\sqrt{x}$
The final answer will be $$\sqrt{x}^3+ax+b\sqrt x+c$$
The reason why that works is this -
In the original polynomial, the roots are $x=x_1,x_2,x_3$
If we want the roots to be $x'=x_1^2,x_2^2,x_3^2$, we define $x$ in terms of the variable $x'$ like this (by taking the square root) $$\sqrt{x'}=x_1,x_2,x_3=x$$ Now we plug $\sqrt{x'}$ in the original polynomial, replace $x'$ by $x$ and we are done!

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  • $\begingroup$ Enlighten me further because in your supposed solution I cannot seem to be getting what hte coefficients should be $\endgroup$ – Zelos Malum Feb 26 '15 at 15:11
  • $\begingroup$ The "polynomial" $\sqrt{x}^3+ax+b\sqrt x+c$ isn't a polynomial. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 26 '15 at 15:15
  • $\begingroup$ @ZelosMalum The roots of the polynomial will be $x=x_1,x_2,x_3$. If we define a function in terms of a new variable $x'$, which has roots $x'=x_1^2,x_2^2,x_3^2$, we can get the $x$ in terms of $x'$. When we get $x$ in terms of $x'$, we can plug it back in the original cubic and get our answer. $\endgroup$ – AvZ Feb 26 '15 at 15:17
  • $\begingroup$ @Martín-BlasPérezPinilla Thanks for pointing that out. Fixed. $\endgroup$ – AvZ Feb 26 '15 at 15:18

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