Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to solve $$(x-1) \cdot (x-2) \cdot (x-3) \cdot (x-4) = 3$$

Any hints?

share|cite|improve this question

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.

share|cite|improve this answer
Alternately, $z = x-\frac{5}{2}$, then the equation becomes $(z^2-\frac{1}{4})(z^2-\frac{9}{4})=3$. But this is basically the same solution.... – N. S. Apr 16 '12 at 16:31
Note that solving said equation for $t$ yields roots $t = -3, -7.\:$ Then you solve $\:x^2-5x = t,\:$ which amounts to solving the two quadratics I gave. Rather than this roundabout way, it's simpler to notice that subtracting $3$ preserves the difference of squares form, as in my answer. – Bill Dubuque Apr 16 '12 at 17:13
@N.S.: Equivalent it may be, but your substitution $z=x-\frac{5}{2}$ is a natural move, in that it brings out the symmetry. – André Nicolas Apr 16 '12 at 17:30
Several solution methods (and some motivation behind them) for the equation $(x-r)(x-2r)(x-3r)(x-4r) = a$ are given in the following sci.math thread:… (Google) (Math Forum) – Dave L. Renfro Apr 16 '12 at 18:11
To elaborate on what N.S. and André are saying: that substitution depresses the quartic, in that it kills the cubic term, and very luckily, the linear term as well... – J. M. Apr 17 '12 at 11:31

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 $

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

share|cite|improve this answer

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.

share|cite|improve this answer
In writing mathematics, once should be encouraged to use rationals 3/2 and not decimals 1.5 ... but of course rationals are less convenient for on-line writing. – GEdgar Apr 17 '12 at 12:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.