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We need to factorize:


We can, by the rational root theorem, see that there are no roots of this polynomial.Next observation is that $64=(8)^2$. So this means that if the first part of the polynomial is a square,we can rewrite the whole polynomial as the difference of two squares.But it turns out that the first part of the polynomial is not a square. However,we can note that,



Therefore,letting $(x^2)-8x+7=p$,we can rewrite the given polynomial as


which I thought would be factorizable, but I seem to be wrong. So how can I factorize this polynomial? I would appreciate a small hint.

EDIT: The original polynomial seems to have been


But there is no point in changing the whole question.To factorize this polynomial,we would do the exact same things as before and find that


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You can factor it into two irreducible quadratics. – ncmathsadist Nov 27 '13 at 14:55
@ncmathsadist,the same thought has struck me.However,I am at a loss to find any practical way of factoring it into 2 irreducible quadratics. – rah4927 Nov 27 '13 at 14:57
Basically, the factorization of a polynomial $P$ does not give you any hint about the factorization of $P+n$, except in special cases. Same when $P$ is an integer. Here, you can't factor into rational quadratics, see WolframAlpha for a complete factorization in $\mathbb{C}$. – Jean-Claude Arbaut Nov 27 '13 at 15:05
arbautjc's comment confirms my worst fears-The textbook has made yet another typo.Does anyone have any idea what the original factorizable polynomial would look like? – rah4927 Nov 27 '13 at 15:14
$ (x-1)(x-3)(x-5)(x-7)-33$ seems to be factorizable.However,33 is way too far from 64 for it to be the original polynomial. – rah4927 Nov 27 '13 at 15:21
up vote 3 down vote accepted

$\left(x-1\right)\left(x-7\right)=y-9$ and $\left(x-3\right)\left(x-5\right)=y-1$ for $y=\left(x-4\right)^{2}$ leading to a factorization of $\left(y-1\right)\left(y-9\right)-64=y^{2}-10y-55$



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