I have a problem that I cannot solve, the problem is as follows:
Find a polynomial with the lowest power $p(x)$ such that its remainder by dividing it with $(x-1)^2$ is $2x$, and the remainder by dividing it by $(x-2)^3$ is $3x$. Thanks in advance.
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$\begingroup$ Show your attempts, please. $\endgroup$– Lion HeartCommented Apr 12, 2022 at 14:39
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$\begingroup$ @Mathology Welcome to Math Stack Exchange. What are your thoughts on this problem? Generally questions showing no effort by the person asking, don't get a lot of answers. $\endgroup$– paw88789Commented Apr 12, 2022 at 14:40
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$\begingroup$ Oh okay. Well I have tried setting the equations and switching x=1 and x=2, but I get stuck there. I get that p(1)=2 and p(2)=6 but I don't know how to finish the problem. $\endgroup$– MathologyCommented Apr 12, 2022 at 14:46
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$\begingroup$ @Mathology Please write the equations so we can see where you are stuck. $\endgroup$– MiguelCommented Apr 12, 2022 at 14:53
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1$\begingroup$ @Mathology: your approach doesn't account power in $(x-1)^2$ and $(x-2)^3$. According to polynomial division remainder definition $p(x)=a(x) (x-2)^3 +3x$. From problem statement follows that $a(x)$ must have lower possible power. Start with $a(x)=a_0$ and check if it is consistent with remainder $2x$ from dividing by $(x-1)^2$. $\endgroup$– Ivan KaznacheyeuCommented Apr 12, 2022 at 15:03
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