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Let $ Z $ be a complex number with nonzero imaginary part such that $$ (2Z + 1)(3Z + 1)(5Z + 1)(30 Z + 1) = 10 $$

Then compute $$ \frac{\text {sum of all values of Z}}{\text {product of all values of Z} } $$

lulu
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Fghj
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    Note: the non-zero imaginary part complicates matters, as the quartic actually has two real roots. As it stands, I think you have to factor the quartic as a product of two quadratics. – lulu Aug 12 '17 at 18:28
  • If you know the two real root of the polynomial, then you can finish quickly. I think it can be decomposed or it has two rational roots. – Exodd Aug 12 '17 at 18:29

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Hint: $1/2+1/30=1/3+1/5 = 2 \cdot 4/15\,$, which suggests the substitution $z = x - 4/15\,$, which gives a biquadratic in $x\,$:

$$ 2\cdot3\cdot5\cdot30\left(z+\frac{1}{2}\right)\left(z+\frac{1}{3}\right)\left(z+\frac{1}{5}\right)\left(z+\frac{1}{30}\right) \\ = 900\left(x+\frac{7}{30}\right)\left(x+\frac{1}{15}\right)\left(x-\frac{1}{15}\right)\left(x-\frac{7}{30}\right) \\ = 900\left(x^2-\frac{7^2}{30^2}\right)\left(x^2-\frac{1}{15^2}\right) \\ = \frac{1}{225}(900x^2-49)(225x^2-1) $$

Then $225(\text{LHS} - 10)$ factors into two qadratics, one of which has the complex roots being sought:

$$ (900x^2-49)(225x^2-1)-2250=(900 x^2 + 71) (225 x^2 - 31) $$

dxiv
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  • you're a wizard harry – Saketh Malyala Aug 12 '17 at 19:04
  • @SakethMalyala Thanks. The technique is fairly well known, and it was probably expected to be used in this case since it's no coincidence that the coefficients were chosen as they did. The symmetry is more easily recognizable and exploitable when the factors on the LHS are monic polynomials (for example this problem), but also applies to problems like the above (or this other one) once you get the gist of it. – dxiv Aug 12 '17 at 22:47