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- Feb 14, 2012

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Solve the equation $x^3(x+1)=2(x+a)(x+2a)$ where $a$ is a real parameter.

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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,894

-----

Solve the equation $x^3(x+1)=2(x+a)(x+2a)$ where $a$ is a real parameter.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Feb 14, 2012

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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?

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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