## Tuesday, October 13, 2015

### Factorize $x^5+x^4+1$

Factorize $x^5+x^4+1$.

If you have never come across this problem before, I'll bet you dollars to donuts you might reply resolutely that $x^5+x^4+1$ could not be factorized. :D

But, fact is, it can. And one doesn't have to have a crystal ball to "see" the factors of $x^5+x^4+1$. We can work it out, slowly but surely!

One of the two things that is required is some basic algebraic manipulation skills such as:

[MATH]\color{black}\bbox[5px,orange]{x=x-1+1}[/MATH]

[MATH]\color{yellow}\bbox[5px,green]{x^2=x^2-1+1=(x-1)(x+1)+1}[/MATH], etc

and the other very important area is one has got to be persistent and won't stop trying until success comes.

\begin{align*}x^5+x^4+1&=x^4(x+1)+1\\&=x^2(x^2)(x+1)+1\\&=x^2(x^2-1+1)(x+1)+1\\&=x^2(x^2-1)(x+1)+x^2(1)(x+1)+1\\&=x^2(x+1)(x-1)(x+1)+x^2(x+1)+1\\&=x^2(x+1)^2(x-1)+x^2(x+1)+1\\&=x^2(x+1)^2(x-1)+x(x)(x+1)+1\\&=x^2(x+1)^2(x-1)+x(x-1+1)(x+1)+1\\&=x^2(x+1)^2(x-1)+x(x-1)(x+1)+x(x+1)+1\\&=\color{green}x(x+1)(x-1)\color{purple}{(x(x+1)+1)}\color{black}+\color{purple}(x(x+1)+1)\\&=\color{green}(x(x+1)(x-1)+1)\color{purple}(x(x+1)+1)\\&=\color{green}(x^3-x+1)\color{purple}(x^2+x+1)\end{align*}

Can you see it now that the manipulation can be employed for multiple times in such a pretty way that we eventually did factor a seemingly impossible expression where at first glance, it's what it's and there would be virtually impossible to factor it?

If you believe in yourself, and attempting at the solution for many times, you will eventually taste the success!