Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
2  
Maybe the $7^2$ is a typo. See this and this –  P.. Jun 5 '13 at 12:12

3 Answers 3

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

share|improve this answer

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

share|improve this answer

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 $x^4+1=x^4+e^{\pi \rm{i}}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.