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Find the H.C.F. of $\ 2(x^4+9)-5x^2(x+1),\ 2x^3(2x-9)+81(x-1)$

I can't really see any pattern by inspection, so I carry out the multiplication with the brackets: $2x^4+18-5x^3-5x^2,\ 4x^4-18x^3+81x-81$

I tried to divide them into each other to find the common factor (Euler's method or something? I forgot). I did the obvious first step by multiplying $2x^4+18-5x^3-5x^2$ by $2$, then subtracting that from $4x^4-18x^3+81x-81$, but the result is $-8x^3+10x^2+81x-117$, so clearly continuing this way is almost impossible by hand. Now I don't know what to do. The answer is given as $2x^2-9x+9$, so I'm not understanding something here. Thanks for your help.

ronald christenkkson
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    I don't think it is "clearly impossible by hand". Be brave!!! :) Also, since you know the answer is quadratic, it's only going to take another one or two steps. – David Jan 29 '23 at 07:40

2 Answers2

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You are thinking about The Euclidean algorithm applied to polynomials.

Certainly, the algorithm can be done by hand although it is a bit tedious: $$\begin{align} 4x^4 - 18x^3 + 81x - 81 &= 2(2x^4 - 5x^3 - 5x^2 + 18) + (-8x^3 + 10x^2 + 81x - 117) \\ 4(2x^4 - 5x^3 - 5x^2 + 18) &= (-x)(-8x^3 + 10x^2 + 81x - 117) + (-10x^3 + 61x^2 - 117x + 72) \\ 5(-8x^3 + 10x^2 + 81x - 117) &= 4(-10x^3 + 61x^2 - 117x + 72) + 97(-2x^2 + 9x - 9) \\ 97(-10x^3 + 61x^2 - 117x + 72) &= 5x(-2x^2 + 9x - 9) + 776(2x^2 - 9x + 9) \\ -2x^2 + 9x - 9 &= (-1)(2x^2 - 9x + 9) + 0. \end{align}$$

All that remains is to check that $2x^2 - 9x + 9$ does in fact divide both polynomials.

heropup
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Then $$f:=2x^4+18-5x^3-5x^2=(2x-3)(x-3)(x^2+2x+2)$$ and $$g:=4x^4-18x^3+81x-81=(2x-3)(x-3)(2x^2-9)$$ Hence, $${\rm gcd}(f,g)=(2x-3)(x-3)$$ as desired.

Note: For $f$ we have the candidate rational root are $$\pm\left\{\frac{1}{2},\frac{3}{2},\frac{9}{2}, 1,2,3,6,9,18\right\} $$ But the these we have roots $\frac{3}{2}$ and $3$, then linear factor $(2x-3)$ and $(x-3)$ so $f=(2x-3)(x-3)(x^2+2x+2)$. Similar for $g$.

A. P.
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