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Find the remainder when $p(x) = (x+2)^{101} + (x+3)^{200}$ is divided by $ x^2 +5x + 6 $.

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  • $\begingroup$ No idea. It's a quiz question. $\endgroup$ – Gunners98 Aug 6 '15 at 18:00
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    $\begingroup$ Hints: $(x+2)^2+(x+2)=x^2+5x+6$ so $(x+2)^2\equiv -(x+2)$. Similarly $(x+3)^2-(x+3)=x^2+5x+6.$ $\endgroup$ – Jyrki Lahtonen Aug 6 '15 at 18:01
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    $\begingroup$ Well...there are two values for $x$ which seem especially interesting for both $p(x)$ and the quadratic. I'd start there. $\endgroup$ – lulu Aug 6 '15 at 18:02
  • $\begingroup$ Hint: factor $x^2+5x+6$ and divide the two terms of $p(x)$ separately. $\endgroup$ – Alex G. Aug 6 '15 at 18:03
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    $\begingroup$ Without knowing the context of your quiz, it's impossible to provide appropriate help. It's legitimate to answer your question with "Use the Chinese remainder theorem in the ring $\Bbb R[x]/\langle (x+2)(x+3) \rangle$", but we have no idea whether that's the right level for you. $\endgroup$ – Greg Martin Aug 6 '15 at 18:05
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Note $x^2+5x+6=(x+2)(x+3)$. Write the Euclidean division by $x^2+5x+6$: $$p(x)=q(x)(x^2+5x+6)=q(x)(x+2)(x+3)+r(x),\quad \deg r(x)\le 1.$$ This implies: $$\begin{cases}p(-2)=1=r(-2)\\p(-3)=-1=r(-3)\end{cases}$$ Thus we have to find a polynomial $r$ of degree $\le 1$ such that $r(-2)=1,\enspace r(-3)=-1$. This is solved by the interpolation polynomial:

$$r(x)=1\cdot\frac{x+3}{-2+3}-1\cdot\frac{x+2}{-3+2}=2x+5.$$

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  • $\begingroup$ The answer is correct but is there any simpler way to solve? My level isn't there yet. Thanks. $\endgroup$ – Gunners98 Aug 6 '15 at 18:45
  • $\begingroup$ Of course, for a linear polynomial you can solve equations to find the coefficients of $r(x)$. I just wanted to show the implications (all the more so as the general Lagrange's interpolation polynomial involves ideas from linear algebra. Besides, there are ceratain formulae that should be known: one cannot always reinvent the wheel. $\endgroup$ – Bernard Aug 6 '15 at 18:53
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HINT. Set $P(x)=Q(x)(x^2+5x+6)+ax+b$ and choose two suitable values of $x$ to make a simultaneous equation for $a$ and $b$

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Hint :

$$(x+2)^{2n+1}=(x+3-1)^{2n}(x+2)\equiv(-1)^{2n}(x+2)\pmod{(x+2)(x+3)}$$

and $$(x+3)^{m+1}=(x+3)(x+2+1)^m\equiv(x+3)(1)^m\pmod{(x+2)(x+3)}$$

where $m,n$ are non-negative integers

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Set the polynomial x^2 + 5x + 6 to zero then find x by quadratic formula. and put that x in the first polynomial.

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