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
$$(p^2)+8p-65=(p^2)+13p-5p-65=(p+13)(p-5)=(x^2-8x+20)(x^2-8x+2)$$
Best Answer
$\left(x-1\right)\left(x-7\right)=y-9$ and $\left(x-3\right)\left(x-5\right)=y-1$ for $y=\left(x-4\right)^{2}$ leading to a factorization of $\left(y-1\right)\left(y-9\right)-64=y^{2}-10y-55$
Here:
$y^{2}-10y-55=\left(y-5-4\sqrt{5}\right)\left(y-5+4\sqrt{5}\right)=\left(\left(x-4\right)^{2}-5-4\sqrt{5}\right)\left(\left(x-4\right)^{2}-5+4\sqrt{5}\right)=\left(x-4+\sqrt{5+4\sqrt{5}}\right)\left(x-4-\sqrt{5+4\sqrt{5}}\right)\left(x-4+i\sqrt{4\sqrt{5}-5}\right)\left(x-4-i\sqrt{4\sqrt{5}-5}\right)$