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I need to solve the following polynomial: $$(12x-1)(6x-1)(4x-1)(3x-1)=15$$

I tried do the multiplication and ended up with an even worse expression. There are a few other questions like this one on the list of exercises I'm trying to solve, and I just wanted an easier method, without having to do the "brutal" work.

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    Hint: $\frac{1}{12}+\frac{1}{3}=\frac{1}{6}+\frac{1}{4}\,$. – 2017-02-09
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    What sort of class is this for? – 2017-02-09
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    @dxiv can you develop? Thx – 2017-02-10
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    @Maczinga I posted an extended hint as an answer. – 2017-02-10
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    @dxiv wow very clever idea ! Thx for sharing – 2017-02-10

2 Answers 2

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Hint: write it as: $$12\cdot 6\cdot 4 \cdot 3 \cdot \left(x-\frac{1}{12}\right)\left(x-\frac{1}{6}\right)\left(x-\frac{1}{4}\right)\left(x-\frac{1}{3}\right)=15$$

Note that the symmetric terms have equal sums $\left(x-\frac{1}{12}\right)+\left(x-\frac{1}{3}\right)=\left(x-\frac{1}{6}\right)+\left(x-\frac{1}{4}\right)$ $=2x - \frac{5}{12}$. This suggests the substitution $y=x-\frac{5}{24}\,$ which "shifts" the center of symmetry to $0\,$:

$$ 12\cdot 6\cdot 4 \cdot 3 \cdot \left(y+\frac{1}{8}\right)\left(y+\frac{1}{24}\right)\left(y-\frac{1}{24}\right)\left(y-\frac{1}{8}\right)=15 \\[5px] \iff \quad 12\cdot 6\cdot 4 \cdot 3 \cdot \left(y^2-\frac{1}{8^2}\right)\left(y^2-\frac{1}{24^2}\right) = 15 $$

The latter is a biquadratic which can be solved for $y^2\,$, which then gives $y\,$, then $x$.

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    You should give more details after "the symmetric sums ..." It will be very hard to follow by many readers as it is. Pretty symmetry deserves a pretty exposition. – 2017-02-10
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    But beautiful once you wrap your head around it. I have to wonder how you could come up with this in such short time span. – 2017-02-10
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    @BillDubuque Edited to (hopefully) make it easier to follow. – 2017-02-10
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    @SimplyBeautifulArt There aren't many ways in which a quartic becomes "easy" to solve by hand, so it's lucky that this one worked out ;-) – 2017-02-10
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    [+1] Nice solution. – 2017-02-12
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    dear prof. @Dxiv...can you help me for ask Math question in stack.exchange??..Because this is so hard question for ask for me..And I can not turn the question into mathematics language..Math knowledge is very inadequate.(+latex and english language)..Please..can you help me as your Little student.. :( I prepared direk link.. Can you look?? – 2017-08-30
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    https://files.acrobat.com/a/preview/fe32c9bf-d72e-4185-9fb7-72653adae16b – 2017-08-30
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    @MathematicsisLife Sorry, didn't quite follow that past the first few lines. I suggest you take the first step to compose and post the question, then others may help with formatting, or ask for necessary clarifications. You have been around here for a while, and may have noticed that posts with a clear question and enough context tend to be well received even if the English or MathJax are not perfect (you also know where to find the [MathJax cheatsheet](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference)). – 2017-08-30
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    I understood :( – 2017-08-30
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    @MathematicsisLife Once you do post the question, feel free to tag me in a comment to be sure I see it. P.S. Nice user name ;-) – 2017-08-30
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    Thank you for upvoting... – 2017-08-30
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    @dxiv It is extremely difficult to express a question. – 2017-08-31
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    @dxiv https://math.stackexchange.com/questions/2414343/is-this-limit-equal-to-1 – 2017-09-03
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    @dxiv I am grateful to you for supporting me. :) – 2017-09-11
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    @MathematicsisLife Thank you. I realize that the answers you got do not cover the entire question, but then the question itself is not trivial to answer. – 2017-09-11
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    @dxiv thank you again :) – 2017-09-11
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    @dxiv With your support, the question has not been downvoted. – 2017-09-11
11

Expanding, we get

$$864 x^4 - 720 x^3 + 210 x^2 - 25 x - 14 = 0$$

Using the rational roots theorem, one can then verify that two roots are $x=-1/6$ and $x=7/12$. Factoring these out, we get

$$864 x^4 - 720 x^3 + 210 x^2 - 25 x - 14 = (6x+1)(12x-7)(12x^2-5x+2)$$

And of course the last quadratic is not factorable over $\mathbb R$.