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Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$ over $\mathbb{Q}$

What I tried to do is to write $x^n+5=p(x)(x^3+10x^2+25x)+Ax^2+Bx+C$, where $p(x)$ is a polynomial of degree $n-3$. If I set $x=0$ I obtain that $C=5$.

Now $x^3+10x^2+25x=x(x+5)^2$ and by setting $x=-5$, I get that

$5A-B=(-1)^n5^{n-1}$

But I need one more equation to be able to find the coefficients of the remainder, and I can't get one. What should I do?

user3701033
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3 Answers3

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It is a good idea to use the roots of $q(x)=x^3+10x^2+25x$, but yes: because $-5$ is a double root, things get a little bit trickier. (Also, check that setting $x=0$ gives $C=5$, not $C=0$).

HINT: But since $-5$ is a double root of $q$, it is both a root of $q$ and $q'$. So try to take the derivative at both members and then take again $x=-5$.

Alejandro Nasif Salum
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Hint $\ $ Subtracting $\, C = 5\,$ then cancelling $x$ yields

$\qquad f(x) := x^{\large n-1} = B + Ax + (x+5)^{\large 2} p(x) $

Thus $\, B + Ax = f(-5) + f'(-5)(x+5)\,$ by Taylor expansion at $\,x = -5\,$

Bill Dubuque
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I think there is an error in the attempt that is part of the question. Anyway, here are some relevant quotients and remainders:

$$ \left( x^{4} + 5 \right) = \left( x^{3} + 10 x^{2} + 25 x \right) \cdot \left( x - 10 \right) + \left( 75 x^{2} + 250 x + 5 \right) $$ $$ \left( x^{5} + 5 \right) = \left( x^{3} + 10 x^{2} + 25 x \right) \cdot \left( x^{2} - 10 x + 75 \right) + \left( - 500 x^{2} - 1875 x + 5 \right) $$ $$ \left( x^{6} + 5 \right) = \left( x^{3} + 10 x^{2} + 25 x \right) \cdot \left( x^{3} - 10 x^{2} + 75 x - 500 \right) + \left( 3125 x^{2} + 12500 x + 5 \right) $$ $$ \left( x^{7} + 5 \right) = \left( x^{3} + 10 x^{2} + 25 x \right) \cdot \left( x^{4} - 10 x^{3} + 75 x^{2} - 500 x + 3125 \right) + \left( - 18750 x^{2} - 78125 x + 5 \right) $$ $$ \left( x^{8} + 5 \right) = \left( x^{3} + 10 x^{2} + 25 x \right) \cdot \left( x^{5} - 10 x^{4} + 75 x^{3} - 500 x^{2} + 3125 x - 18750 \right) + \left( 109375 x^{2} + 468750 x + 5 \right) $$ $$ \left( x^{9} + 5 \right) = \left( x^{3} + 10 x^{2} + 25 x \right) \cdot \left( x^{6} - 10 x^{5} + 75 x^{4} - 500 x^{3} + 3125 x^{2} - 18750 x + 109375 \right) + \left( - 625000 x^{2} - 2734375 x + 5 \right) $$ $$ \left( x^{10} + 5 \right) = \left( x^{3} + 10 x^{2} + 25 x \right) \cdot \left( x^{7} - 10 x^{6} + 75 x^{5} - 500 x^{4} + 3125 x^{3} - 18750 x^{2} + 109375 x - 625000 \right) + \left( 3515625 x^{2} + 15625000 x + 5 \right) $$

Will Jagy
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