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?

Solution 1:

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.

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 $

$\rm\ \ \Rightarrow\ (x\!-\!1)(x\!-\!4) (x\!-\!2)(x\!-\!3)\!-\!3\ =\ (x^2\!-\!5x+5)^2 \!-\! 2^2\ =\ (x^2\!-\!5x+3)(x^2\!-\!5x+7)$