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If I have that $p_{k,j}: F \to F$ is given by:

$p_{1,0}=(x-2)^3$

$p_{2,0}=(x-1)$

and I have a polynomial function $f_0(x): F \to F$ given by $f_0(x)=1$

How would I find polynomials $h_1$ and $h_2$ so that $f_0(x)=p_{1,0}(x)h_1(x)+p_{2,0}h_2(x)$ for every $x \in F$?

Alex Ravsky
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user141745
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    http://en.wikipedia.org/wiki/Polynomial_greatest_common_divisor#B.C3.A9zout.27s_identity_and_extended_GCD_algorithm – Alex Ravsky Dec 10 '14 at 22:14
  • Thanks for the reference, I'm a little confused how this works though. I looked through some examples and couldn't find a reason for them picking what to multiply the the polynomials by for this to be true. – user141745 Dec 11 '14 at 02:51
  • In the equation in your last line, is $p_{1,0}$ being multiplied by $x$ or is $p_{1,0}$ written as a function of x? It looks confusing because $p_{2,0}$ doesn't have '$(x)$' after it. – Radial Arm Saw Jun 16 '20 at 01:19

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