Saturday, October 10, 2015

Learn To Think Mathematically

Train new mode of thinking and hence deduce something useful as problem solving trick.

It's a breeze if you're asked to factor


since it's obviously a difference of square that factored as $a^2-b^2=(a+b)(a-b)$.

What if you're asked to factor


Nope, there is no way one can factor that!

What is the story for $a^3-b^3$? Argh! That one is fairly easy too. For those who have excellent memorization or who used it frequent enough to have that difference of cubes engraved in your mind, you can spell it out its factored form immediately:


And unlike the sum of squares, we can also factor the sum of cubes.


In case you have forgotten about the identities (take for example $a^3+b^3=(a+b)(a^2-ab+b^2)$), you could always work it out by


First, recognize that, $a=-b$ is a root for the expression $a^3+b^3$ since $(-b)^3+b^3=-b^3+b^3=0$.


By performing the polynomial long division once we realized $a=-b$ is a root for the expression $a^3+b^3$:

[MATH]\begin{array}{r}a^2-ab+b^2\hspace{-29px}\\a+b\enclose{longdiv}{a^3+b^3} \\ -\underline{\left(a^3+a^2b\right)} \hspace{-20px} \\ -a^2b+b^3 \hspace{-44px} \\ -\underline{\left(-a^2b-ab^2\right)} \hspace{-61px} \\ ab^2+b^3 \hspace{-90px} \\ -\underline{\left(ab^2+b^3\right)} \hspace{-100px} \end{array}[/MATH]

Now, we shift our focus into factoring $a^4-b^4$.

It's still not too hard a task, since we can treat it as follows:


and note that there is no way to factor $a^4+b^4$.

Now, we're told that we could factor $a^{4} + 4b^{4}$. Argh, that is certainly an uphill task!

It's totally reasonable if we do the following without planning:

$\begin{align*}a^{4} + 4b^{4}&=(a^2)^2+(2b^2)^2\\&=(a^2+2b^2)^2-2(a^2)(2b^2)\\&\stackrel{or}=(a^2-2b^2)^2+2(a^2)(2b^2)\end{align*}$

That is not factoring at all because we ended up with either the sum or difference of two terms, a status of the expression just like what it's being given originally.

That suggests we have to come up with something else in order to factor $a^{4} + 4b^{4}$ properly.

Since $a^{4}$ and $4b^{4}$ can be rewritten as $(a^{2})^2$ and $(2b^{2})^2$ respectively, it's perfectly viable a choice to let

$\begin{align*}a^{4} + 4b^{4}&=(\underbrace{(a^{2})^2+(2b^{2})^2}+\underbrace{\color{purple}\fbox{?}}\color{black})(\underbrace{(a^{2})^2+(2b^{2})^2}-\underbrace{\color{purple}\fbox{?}}\color{black})\\&=((a^{2})^2+(2b^{2})^2)^2-\color{purple}\fbox{?}\fbox{?}\\&=a^4+4a^2b^2+4b^4-\color{purple}\fbox{?}\fbox{?}\\&=a^4+4b^4+4a^2b^2-\color{purple}\fbox{?}\fbox{?}\end{align*}$

It's obvious that $4a^2b^2=(2ab)^2=\color{purple}\fbox{?}\fbox{?}$ and so $2ab=\color{purple}\fbox{?}$ and therefore our mission is accomplished, we have found

$a^{4} + 4b^{4}=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$

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