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Suppose $\lambda _1,\lambda _2,\lambda _3$ is the roots of $f(x)=x^3+x^2+x-1$, $g(x)=x^2+x+1$, solve for a polynomial $p(x)$ with rational coefficients such that $g\left(\lambda _1\right),g\left(\lambda _2\right),g\left(\lambda _3\right)$ is the root of it.

I have no idea, need some hints.

Seems I can get the result: $\lambda _1\cdot \lambda _2\cdot \lambda _3$ is the factor of $-1$?

And polynomials of polynomials, I've only see that in the chapter of symmtric polynomials of my text.

forlorn
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  • Look at $f$ and $g$ to find another way to express $g(\lambda_k)$. Since $f(x) = (x-\lambda_1)(x-\lambda_2)(x-\lambda_3)$, that should help you find $p(x) = (x-g(\lambda_1))(x-g(\lambda_2))(x-g(\lambda_3))$. – Daniel Fischer Aug 07 '13 at 13:29

2 Answers2

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For every root $\lambda$ of $f$, $\lambda\ne0$ and $g(\lambda)=1/\lambda$. Let $p(x)=1+x+x^2-x^3$. Then $f(x)=x^3p(1/x)$ hence for every such $\lambda$, $p(g(\lambda))=p(1/\lambda)=f(\lambda)/\lambda^3=0$, QED.

Did
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  • Yes, the key is that $f(x)=g(x)x-1$. – Thomas Andrews Aug 07 '13 at 13:45
  • Oh, I've understood, how to directly construct $p(x)$, any guidelines? – forlorn Aug 07 '13 at 14:30
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    As written in the post, $p(x)=x^3f(1/x)$. – Did Aug 07 '13 at 14:42
  • @mathdummy You are looking for a polynomial whose roots are the multiplicative inverses of the roots of $f(x)$, and $p(x)=x^3f(1/x)$ obviously satisfies that - it is a polynomial and $p(1/\lambda)=1/\lambda^3f(\lambda)=0$, at least when $\lambda\neq 0$ - so in general, you need $f(0)\neq 0$. – Thomas Andrews Aug 07 '13 at 15:15
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This problem has an "easy" solution, given by @Did, using that $f(x)=g(x)x-1$.

But a more general approach can be found by taking a single root, of $f(x)$, $\lambda,$ and computing the $g(\lambda)^2$, $g(\lambda)^3$:

$$\begin{align}g(\lambda)^0 &= 0\cdot \lambda^2 + &0\cdot \lambda + &1\\ g(\lambda)^1 &= 1\cdot \lambda^2 + &1\cdot\lambda + &1\\ g(\lambda)^2 &= 1\cdot \lambda^2 + &2\cdot \lambda + &2\\ g(\lambda)^3 &= 2\cdot\lambda^2 + &3\cdot\lambda + &4 \end{align}$$

(We compute these simply by multiplying and then reducing using that $\lambda^3=-\lambda^2-\lambda + 1$.)

So we want to write the vector $(2,3,4)$ as a linear combination of the vectors $(0,0,1)$, $(1,1,1)$, and $(1,2,2)$.

We can quickly see that $(2,3,4) = (1,2,2) + (1,1,1) + (0,0,1)$, so we get the equation:

$$g(\lambda)^3 = g(\lambda)^2 +g(\lambda) + 1$$

and thus our polynomial:

$$x^3-x-x-1$$

has $g(\lambda)$ as a root. Since $\lambda$ was chosen arbitrarily, this means that $g(\lambda_i)$ is a root for all $i$.

Thomas Andrews
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  • Sorry, I do not know your answer by now :( What's the meaning of the power of $g(\lambda)$ – forlorn Aug 07 '13 at 14:35
  • $g(\lambda)^k$ is a "power" of $g(\lambda)$. Standard meaning of $X^k$. So $g(\lambda)^2 = g(\lambda)g(\lambda),,g(\lambda)^3 =g(\lambda)g(\lambda)^2,\dots$ @mathdummy – Thomas Andrews Aug 07 '13 at 14:39
  • I don't know how you're calculating them, seems just changing their coefficients by some rules. And if $g(\lambda)$ is an polynomial with 2 degree, then $g(\lambda)^2$ is an polynomial with 4 degree? – forlorn Aug 07 '13 at 14:44
  • Right $g(\lambda)^2 = \lambda^4 + 2\lambda^3 + 3\lambda^2 + 2\lambda+1$. But we know that $\lambda$ is a root of $f(x)$, so $\lambda^3=-\lambda^2-\lambda+1$and $\lambda^4 = -\lambda^3-\lambda^2+\lambda = 2\lambda-1$. So we can reduce $g(\lambda)^2$ to $\lambda^2+2\lambda+2$ relatively easily. @mathdummy – Thomas Andrews Aug 07 '13 at 14:47
  • You can also just think of the computation as dividing $g(x)^k$ by $f(x)$ and computing a remainder:$$(1+x+x^2)^k = f(x)q_k(x) + r_k(x)$$ where the degree of $r_k(x)$ is less than $3$. Then $g(\lambda)^k = f(\lambda)q_k(\lambda) + r_k(\lambda) = r_k(\lambda)$ since $f(\lambda)=0$. – Thomas Andrews Aug 07 '13 at 14:58