# Factorizing $(x-1)(x-3)(x-5)(x-7)-64$

We need to factorize:

$$(x-1)(x-3)(x-5)(x-7)-64$$

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,

$$(x-1)(x-7)=(x^2)-8x+7$$

$$(x-3)(x-5)=(x^2)-8x+15$$

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

$$p(p+8)-64=p^2+8p-64$$

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

$$(x-1)(x-3)(x-5)(x-7)-65$$

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

$$(p^2)+8p-65=(p^2)+13p-5p-65=(p+13)(p-5)=(x^2-8x+20)(x^2-8x+2)$$

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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

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