How do I solve $(x-1)(x-2)(x-3)(x-4)=3$ How to solve $$(x-1) \cdot (x-2) \cdot (x-3) \cdot (x-4) = 3$$
Any hints?
 A: Look at $(x-1)(x-4)$ and $(x-2)(x-3)$ they multiply as $(x^{2}-5x+4)$ and $x^{2}-5x+6$. Now put $t= x^{2}-5x$ and reduce it to a quadratic equation.
A: I assume the hints would already have given you the answer. If not, here is the full answer:
Let y = x-2.5
(y+1.5)(y-1.5)(y+0.5)(y-0.5) = 3.
so $(y^2-2.25)(y^2-0.25) = 3$.
Let $z = y^2-1.25$.
(z-1)(z+1) = 3.
So $z^2-1 = 3$.
Hence $z^2 = 4$.
z = -2 gives $y^2 = z + 1.25 = -0.75$.
So $y = \pm \sqrt{0.75}i$. Clearly, this should be ignored if you only want real roots.
$x = 2.5 + y = 2.5 \pm \sqrt{0.75}i$.
z = 2 gives $y^2 = z + 1.25 = 3.25$.
So $y = \pm \sqrt{3.25}$. Clearly, this should be ignored if you only want real roots.
$x = 2.5 + y = 2.5 \pm \sqrt{3.25}$.
If you want all the terms in the product to be positive, then obvious ly $x = 2.5 + \sqrt{3.25}$ is the only one that works. This is roughly 4.30277564.
A: Hint $\ $ The LHS is a difference of squares $\rm\:y^2\!-\!1,\:$ hence so too is $\rm\:(y^2\!-\!1)-3\: =\: y^2\!-\!2^2,\:$ viz.
$\rm\qquad\ \:\! (x\!-\!1)(x\!-\!4) (x\!-\!2)(x\!-\!3)\ =\ (x^2\!-\!5x+4)(x^2\!-\!5x+6)\ =\ (x^2\!-\!5x+5)^2 \!-\! 1^2 $
$\rm\ \ \Rightarrow\ (x\!-\!1)(x\!-\!4) (x\!-\!2)(x\!-\!3)\!-\!3\ =\ (x^2\!-\!5x+5)^2 \!-\! 2^2\ =\ (x^2\!-\!5x+3)(x^2\!-\!5x+7)$
