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Once I came across the following problem: find the roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$. Here it is how I proceeded:

\begin{align*} (x+1)(x+3)(x+5)(x+7) + 15 & = [(x+1)(x+7)][(x+3)(x+5)] + 15\\ & = (x^2 + 8x + 7)(x^2 + 8x + 15) + 15\\ & = (x^2 + 8x + 7)[(x^2 + 8x + 7) + 8] + 15\\ & = (x^2 + 8x + 7)^2 + 8(x^2 + 8x + 7) + 15 \end{align*} If we make the substitution $y = x^2 + 8x + 7$, we get \begin{align*} y^2 + 8y + 15 = (y^2 + 3y) + (5y + 15) = y(y+3) + 5(y+3) = (y+5)(y+3) = 0 \end{align*} From whence we obtain that: \begin{align*} y + 5 = 0\Leftrightarrow x^2 + 8x + 12 = 0 \Leftrightarrow (x+4)^2 - 4 = 0\Leftrightarrow x\in\{-6,-2\}\\ \end{align*} Analogously, we have that \begin{align*} y + 3 = 0\Leftrightarrow x^2 + 8x + 10 = 0\Leftrightarrow (x+4)^2 - 6 = 0\Leftrightarrow x\in\{-4-\sqrt{6},-4+\sqrt{6}\} \end{align*} Finally, the solution set is given by $S = \{-6,-2,-4-\sqrt{6},-4+\sqrt{6}\}$.

Differently from this approach, could someone provide me an alternative way of solving this problem? Any contribution is appreciated. Thanks in advance.

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  • $\begingroup$ +1 for a well asked question. Grouping the symmetric factors exploits essentially the same symmetry used in my answer. Deriving the two equations which both reduce to $(x+4)^2 - \,\text{something}\,=0$ is another way to "see" (or justify) the substitution $\,y=x+4\,$. $\endgroup$ – dxiv Jul 3 '17 at 23:47
  • $\begingroup$ Reminds me of math.stackexchange.com/questions/1528187/… $\endgroup$ – user223391 Jul 4 '17 at 0:20
  • $\begingroup$ @ZacharySelk Another variation on the same idea here. $\endgroup$ – dxiv Jul 4 '17 at 0:36
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Hint: let $x+4=y$ then the equation writes as:

$$0 = (y-3)(y-1)(y+1)(y+3)+15=(y^2-1)(y^2-9)+15=y^4-10y^2+24$$

The latter is a biquadratic with solutions $y^2 \in \{4, 6\}\,$.

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    $\begingroup$ $$4y=x+1+x+3+x+5+x+7$$ $\endgroup$ – lab bhattacharjee Jul 4 '17 at 1:48
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One way to solve this problem would be to use the Rational Roots Theorem to find the roots $-2$ and $-6$, then use polynomial division to get a quadratic which is easily solved. However this method will not work in general, as a polynomial does not need to have any rational roots at all

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HINT.-Looking about integer solutions for $f(x)=(x+1)(x+3)(x+5)(x+7) + 15 = 0$ possible values should be even and negative so the only candidates are $-2,-4$ and $-6$. We verified that $-2$ and $-6$ are roots. The other two roots are solutions of $$\frac{x^4+16x^3+86x^2+176x+120}{x^2+8x+12}=x^2+8x+10$$

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