Monday, August 3, 2015

Optimization Contest Problem: Prove $x^4+x^3-x^2-x+1>0$ for all real $x$.

In one of my previous blog posts(optimization-contest-problem), we want to prove that [MATH]\color{yellow}\bbox[5px,blue]{x^4+x^3-x^2-x+1}[/MATH] is always greater than zero for all real $x$, or more specifically, for $x\gt 1$.

Math educator should ask open questions that permit a greater variety of responses. This problem that is to prove  $x^4+x^3-x^2-x+1\gt 0$ offers us the best chance to achieve our goal to transform students in terms of their ability to think critically and rationally.

We can actually easily prove that $x^4+x^3-x^2-x+1\gt 0$ for all real $x$ if we employ one of the powerful heuristic strategy, by breaking the problem into cases, solve each case, divide and conquer!

Note that we can rewrite [MATH]\color{yellow}\bbox[5px,purple]{x^4+x^3-x^2-x+1}[/MATH]  in such a way that we can easily show it is always greater than $0$ for [MATH]\color{yellow}\bbox[5px,purple]{x\ge -1}[/MATH].


It is obvious that for [MATH]\color{yellow}\bbox[5px,purple]{x\gt -1}[/MATH], both $x^4\ge 0$ and $(x-1)^2(x+1)\gt 0$.

So we can conclude that [MATH]\color{yellow}\bbox[5px,purple]{x^4+x^3-x^2-x+1\gt 0}[/MATH]for [MATH]\color{yellow}\bbox[5px,purple]{x\ge -1}[/MATH].

Now, for the second part where we have to prove [MATH]\color{yellow}\bbox[5px,green]{x^4+x^3-x^2-x+1\gt 0}[/MATH] for [MATH]\color{yellow}\bbox[5px,green]{x\lt -1}[/MATH].


It's obvious that for [MATH]\color{yellow}\bbox[5px,green]{x\lt -1}[/MATH], both $x\lt 0$ and $(x-1)\lt 0$ while $(x+1)^2\gt 0$.

Combining the signs, it must be true that $x(x-1)(x+1)^2\gt 0$ for [MATH]\color{yellow}\bbox[5px,green]{x\lt -1}[/MATH], which means [MATH]\color{yellow}\bbox[5px,green]{x^4+x^3-x^2-x+1\gt 0}[/MATH]for [MATH]\color{yellow}\bbox[5px,green]{x\lt -1}[/MATH].

All in all, we can say it out loud that [MATH]\color{yellow}\bbox[5px,blue]{x^4+x^3-x^2-x+1\gt 0}[/MATH] for all real $x$.

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