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Let $$p(x) = x^4 + 7x^3 + 14x^2 + 7x+1 $$$$q(x) = x^4 +10x^3 + 23x^2 + 10x+1$$

Find polynomials $f(x),g(x)$ with rational coefficients such that $$f(x)p(x) + g(x)q(x) = 2x^2 + 6x +2$$

I totally have no idea to solve this problem... Please help me to do it. Thank you!!

flawr
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Nhay
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  • It reminds me of the sort of thing you would use the extended Euclidean algorithm for, to write a linear combination of two ordinary integers as some multiple of their GCD. Could you try something like that here? – pjs36 Mar 25 '16 at 22:43
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    First things first, what is $\gcd(p,q)?$ – Will Jagy Mar 25 '16 at 22:55

2 Answers2

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If you factor both polynomials in $\mathbb Q[x]$ you'll get

$p(x) = (x^2+3x+1) (x^2+4x+1)$

$q(x) = (x^2+3x+1) (x^2+7x+1)$

These factors are irreducible (Why?).

Therefore $gcd(p,q) = x^2+3x+1$. Now with the Euclidean algorithm (or some guessing) you should be able to get two polynomials $r,s$ such that $$r(x^2+4 x+1)+s(x^2+7 x+1) = gcd(x^2+4 x+1,x^2+7 x+1) = 1$$

If you multiply this by $2(x^2+3x+1)$ you get exactly what you want:

$$(2r) p + (2s)q = 2x^2+6x+2.$$

Hint 1:

$r,s$ are of degree 1

Hint 2:

$x+7$,$x+4$

flawr
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You can use Grobner basis to solve this,

$f1=x^4+7x^3+14x^2+7x+1;$,

$f2=x^4+10x^3+23x^2+10x+1;$

$f3=(f2-f1)/3=x^2+3x^2+x$

$f4=-3(f2-(x+7)f3)=x^2+3x+1$

Since, $2x^2+6x+2 = 2*f4$ Now just replace it as $f4->f3->f2$ till everything is in terms of $f1$ and $f2$,

Answer: $2/3(-(x+4)f2+(x+7)f1)$

Pankaj Sejwal
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