I'm sure you're pretty familiar with factoring sum and difference of squares or even cubes and recognizing perfect squares at first glance, take for example, when you see
$a^2-b^2$ you know it can be represented as $a^2-b^2=(a+b)(a-b)$
$a^2+b^2$ you know it can be represented as $a^2+b^2=(a+b)^2-2ab\stackrel{\text{or}}=(a-b)^2+2ab$
$a^3-b^3$ you know it can be represented as $a^3-b^3=(a-b)(a^2+ab+b^2)$
$a^3+b^3$ you know it can be represented as $a^3+b^3=(a+b)(a^2-ab+b^2)$
Also, it's perhaps quite obvious to you that $x^2-6x+9=(x-3)^2$, $x^2+10x+25=(x+5)^2$, $x^2-6x+9=(x-3)^2$, and even $x^3-3x^2+3x-1=(x-1)^3$, but, what can you say about the expression below when you're asked to simplify it?
$\large \left(a^4+\dfrac{b^4}{4}\right)$
Can you easily transform it to becomes a product of two factors? Or anything you could do to simplify it from two terms to become a single term?
You might say, this couldn't be too difficult a thing to do:
$\begin{align*} \left(a^4+\dfrac{b^4}{4}\right)&=\left((a^2)^2+\left(\dfrac{b^2}{2}\right)^2\right)\\&=\color{blue}\square^2+\color{red}\square^2\\&=\left(\color{blue}\square\color{black}+\color{red}\square\right)^2-2\color{blue}\square\color{red}\square\\&=\left(a^2+\dfrac{b^2}{2}\right)^2-2ab \end{align*}$
Okay, you have done something, but, wouldn't you agree with me that the end result looks more complicated than the given sum? We're not simplifying it but instead complicating it unfortunately.
We need to devise another plan, if we want to factorize an expression, we could perhaps see if polynomial long division could shed some light...
First we simply try out any divisor that makes a bit of sense, we know letting the first term of the divisor as $a^2$ wouldn't go amiss since $a^2$ times with $a^2$ will give us $a^4$, which is the first term in our quotient. So, let's make the first trial of the divisor as $a^2+ab$ and see where that leads us:
[MATH]\begin{array}{r}a^2-ab+b^2\hspace{130px}\\a^2+ab\enclose{longdiv}{a^4+0a^3b+0a^2b^2+0ab^3+\dfrac{b^4}{4}} \\ -\underline{\left(a^4+a^3b\right)} \hspace{150px} \\ -a^3b+0a^2b^2+0ab^3+\dfrac{b^4}{4} \hspace{18px} \\ -\underline{\left(-a^3b-a^2b^2\right)} \hspace{100px} \\ a^2b^2+0ab^3+\dfrac{b^4}{4} \hspace{20px} \\ -\underline{\left(a^2b^2+ab^3\right)} \hspace{60px} \\ -ab^3+\dfrac{b^4}{4} \hspace{30px} \end{array}[/MATH]
This could not be right, there are remainders at the end of the division process.
Let's try other division, why not we try $a^2+ab+b^2$?
[MATH]\begin{array}{r}a^2-ab+b^2\hspace{130px}\\a^2+ab+b^2\enclose{longdiv}{a^4+0a^3b+0a^2b^2+0ab^3+\dfrac{b^4}{4}} \\ -\underline{\left(a^4+a^3b+a^2b^2\right)} \hspace{100px} \\ -a^3b-a^2b^2+0ab^3+\dfrac{b^4}{4} \hspace{20px} \\ -\underline{\left(-a^3b-a^2b^2-ab^3\right)} \hspace{50px} \\ ab^3+\dfrac{b^4}{4} \hspace{20px} \end{array}[/MATH]
There doesn't sound promising either, so we will try the divisior of $a^2+ab+\dfrac{b^2}{2}$:
[MATH]\begin{array}{r}a^2-ab+\dfrac{b^2}{2}\hspace{120px}\\a^2+ab+\dfrac{b^2}{2}\enclose{longdiv}{a^4+0a^3b+0a^2b^2+0ab^3+\dfrac{b^4}{4}} \\ -\underline{\left(a^4+a^3b+\dfrac{a^2b^2}{2}\right)} \hspace{90px} \\ -a^3b-\dfrac{a^2b^2}{2}+0ab^3+\dfrac{b^4}{4} \hspace{20px} \\ -\underline{\left(-a^3b-a^2b^2-\dfrac{ab^3}{2}\right)} \hspace{50px} \\ \dfrac{a^2b^2}{2}+\dfrac{ab^3}{2}+\dfrac{b^4}{4} \hspace{20px} \\ -\underline{\left(\dfrac{a^2b^2}{2}+\dfrac{ab^3}{2}+\dfrac{b^4}{4}\right)} \hspace{10px} \end{array}[/MATH]
Yes! We did it!
$\large \left(a^4+\dfrac{b^4}{4}\right)= \left(a^2-ab+\dfrac{b^2}{2}\right)\left(a^2-ab+\dfrac{b^2}{2}\right)$
The experimenting with the help of long polynomial division actually only consumed us a bit of time, and see! We could factor a seemingly non-factorable expression perfectly! :D
I hope indeed from now onwards that you could memorize the above identity, it could be of giant help in some circumstances.
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