Find the Roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$

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 $y^2 \in \{4, 6\}\,$.


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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Soft Question

Alternative Proof

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