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Problem Of The Week #384 Sep 19th, 2019

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anemone

MHB POTW Director
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Feb 14, 2012
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anemone

MHB POTW Director
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Feb 14, 2012
3,894
Congratulations to Opalg for his correct solution(Cool), which you can find below:

If $x^3(x+1) = 2(x+a)(x+2a)$ then $x^4 + x^3 - 2x^2 - 6ax - 4a^2 = 0$. That factorises as $$x^4 + x^3 - 2x^2 - 6ax - 4a^2 = (x^2-x-2a)(x^2+2x + 2a) = 0.$$ If $x^2-x-a = 0$ then $x = \frac12\bigl(1\pm\sqrt{1+8a}\bigr).$ If $x^2+2x + 2a = 0$ then $x = -1\pm\sqrt{1-2a}$.

Trade secret: where did the factorisation $x^4 + x^3 - 2x^2 - 6ax - 4a^2 = (x^2-x-2a)(x^2+2x + 2a)$ come from?
Replace the parameter $a$ by $y$, and get Desmos to graph the curve $x^4 + x^3 - 2x^2 - 6xy - 4y^2 = 0$:

This clearly splits into two separate parabolas, which both go through the origin. The upwards-opening parabola has its vertex at $\bigl(\frac12,-\frac18\bigr)$. The downwards-opening parabola has its vertex at $\bigl(-1,\frac12\bigr)$. From that it is easy to find that the first parabola has equation $y = \frac12\bigl(x-\frac12\bigr)^2-\frac18$, or $x^2 - x - 2y = 0$. And the second parabola has equation $y = -\frac12(x+1)^2 + \frac12$, or $x^2 + 2x + 2y = 0.$ Therefore $$x^4 + x^3 - 2x^2 - 6xy - 4y^2 = (x^2 - x - 2y)(x^2 + 2x + 2y).$$ Now replace $y$ by $a$ to get the required factorisation.
 
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