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Factorize the polynomial $x^7-7x^6-x^5+7x^4+x^3-7^2-x+7$

So, I have to factorize this in $\Bbb R[x]$ and $\Bbb C[x]$, but when I'm trying to apply the Ruffini schema, I don't know how to put the coefficients in the cuadratic position.

I must to solve the $7^2$ and sum with the lineal term? or put the $7$ as a cuadratic term?

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I assume the $7^2$ should rather read $7x^2$. – Hagen von Eitzen Jun 5 '13 at 12:02
Is there an $x$ missing, or is it suppodes to be $...-7^2-x+7$? – Dennis Gulko Jun 5 '13 at 12:03
I suspect that the $7^2$ you are asking about is a typo and it is supposed to be $7x^2$. However, if it were actually $7^2$, then it would be a constant term along with the $7$ at the end. – Karl Kronenfeld Jun 5 '13 at 12:04
No, it is $7^2$ – Tomi Sebastián Juárez Jun 5 '13 at 12:05
Maybe the $7^2$ is a typo. See this and this – P.. Jun 5 '13 at 12:12

Note that $1$, $7$ and $-1$ are roots of the polynomial so we find by the euclidean division $$x^7-7x^6-x^5+7x^4+x^3-7x^2-x+7=(x-1)(x-7)(x+1)(x^4+1)$$ moreover we have $$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$$ so we find the decomposition in $\mathbb R[x]$ since the polynomials $(x^2-\sqrt{2}x+1)$ and $(x^2+\sqrt{2}x+1)$ are irreductible in $\mathbb R[x]$

To find the decomposition in $\mathbb C[x]$ we decompose $(x^2-\sqrt{2}x+1)$ and $(x^2+\sqrt{2}x+1)$ by calculating the discriminant to find their roots

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Hint: $\ x^6(x\!-\!7)-x^4(x\!-\!7)+x^2(x\!-\!7)-(x\!-\!7) = (x\!-\!7)(x^6-x^4+x^2-1)$

and $\ \ \ x^6-x^4+x^2-1 = x^4(x^2\!-\!1) + (x^2-1) = (x^2-1)(x^4 + 1)$

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I think the polynomial should be $p(x)=x^7-7x^6-x^5+7x^4+x^3-7x^2-x+7$.

The possible rational roots of $p$ are $\pm7,\pm1$. Three of them are roots of $p$. Now divide $p(x)$ with $(x-1)\ldots$ to find the factorization over $\mathbb Q$. To find the factorization over $\mathbb R$ use that $x^4+1=(x^2+1)^2-2x^2$. To find the factorization over $\mathbb Q$ find the roots of $x^4+1=x^4+e^{\pi \rm{i}}$.

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