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I need to solve the following polynimial: $$(12x-1)(6x-1)(4x-1)(3x-1)=15$$ I tried to 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: 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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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 $\mathbb R$. |
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Wolfy says the roots are $-1/6, 7/12, (5\pm i\sqrt{71})/24)$. The rational ones I could manually verify; not the complex ones. |
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