# $\huge{"}$unfactorable$\huge{"}$ Alright, I guess you could call it a "pet peev" but when ever I hear the word unfactorable, it just hits a nerve, because in reality, all equations are factorable...technically.

The most stereotypically "unfactorable"... Or should I say "not pretty", equation is $x^2-x-1$. WELL GUESS WHAT you can factor it as $(x-\frac{1}{2}-\frac{\sqrt5}{2})(x-\frac{1}{2}+\frac{\sqrt5}{2})$.

Yes, it sure isn't pretty, but it works. You can use the quadratic equation and match the first coefficients. This will work sometimes, but sometimes, you must use a little more unconventional ways.

Apparently, every teacher has this deadly disease called "unfactorablus polynomium" (lol) because not one teacher at our school taught us this.

I mean, when I say that EVERY polynomial is factorable, I mean EVERY SINGLE ONE.

💭💭💭. "what about $x-5$?"

Not factorable???? Nope, too bad, sorry math, I win, $\textit{it is}$. Just use difference of cubes.

$x-5$

$(^3\sqrt{x}- ^3\sqrt5)(^3\sqrt{x^2}+^3\sqrt{5x}+^3\sqrt{5^2})$

There, it's done being factored.... WRONG... I see a sum of squares and a difference of squares.

$(^6\sqrt{x}-^6\sqrt5)(^6\sqrt{x}+^6\sqrt5)(^6\sqrt{x^2}+i\sqrt{^3\sqrt{5x}+^3\sqrt{5^2}})(^6\sqrt{x^2}-i\sqrt{^3\sqrt{5x}+^3\sqrt{5^2}})$

But you get the point. I could go on forever using these BRILLIANT factoring Lemmas... But turns out, that only one of them actually has a 0.. So there will only be one intercept because the original equation only has one root. In fact, none of the others even have imaginary roots. Thus the original equation is not changed.

So I ask of you all to please spread the knowledge of the Unfactorablus Polynomium disease. And remember:

"I don't care what it is, everything is factorable......unless it's a potato" -Trevor Arashiro 2014.

Don't forget to hashtag your posts

# . #everythingisfactorable #factoringforthewin #fundamentaltheoremofalgebraiswrong #cuztrevorsaidso Note by Trevor Arashiro
6 years, 9 months ago

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Maybe teachers should mention the Fundamental Theorem of Algebra, for the sake of completeness.

- 6 years, 9 months ago

So ... I was watching this new show called Scorpion, about a bunch of high-IQ misfits problem solving think-tank

In a flashback, one of them had a teacher who wrote several numbers on the board, including 703.
Teacher: "Which of these numbers is divisible by 4?"
Student: "All of them."
Teacher: "Wrong! 703 is not ..."
Student: "703 divided by 4 is 175.75. You did not define your terms properly."

Staff - 6 years, 9 months ago

Haha. That is one cool guy.

- 6 years, 9 months ago

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- 2 years, 2 months ago