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Consider the equation $(x^3-3x^2+4x-2)s(x)+(x^2-1)t(x)=x+1$. How can we find polynomials $s(x), t(x)$ which satisfy this equation?

Clearly $x+1$ divides the term involving $t(x)$ but not the first polynomial. If there are solutions then we need $(x+1)$ to divide $s(x)$. I am not sure where to go from here though.

Bernard
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  • You would need the degree of t to be one higher than the degree of s if they existed, because you would need to cancel out the highest degree terms – Alan Apr 23 '21 at 20:23
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    There is no such polynomial because $x-1$ is a factor of the cubic polynomial. – Varadharajan R Apr 23 '21 at 20:27
  • Did you mean the equation ... $=x\color{red}-1$? Then the answer would be $s(x)=\frac15$ and $t(x)=-\frac15(x-3)$ – J. W. Tanner Apr 23 '21 at 20:37
  • By the proofs in the linked dupe it is solvable $\iff$ their gcd $,x-1\mid x+1\iff x+1\mid 2,,$ contradiction (eval at $,x=-1,$ or compare degrees). Note that the proof I give their works in any Euclidean domain since then we have a Bezout equation for the gcd. In particular it works for polynomials over a field (here). – Bill Dubuque Apr 23 '21 at 21:10

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hint

$$x^3-3x^2+4x-2=2(x-1)\Bigl(x^2-2x+2\Bigr)$$ Put $$s(x)=(x+1)R(x)$$ then

$$(x-1)\Bigl((x^2-2x+2)R(x)+t(x)\Bigr)=1$$ which is impossible.

hamam_Abdallah
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