## Tag Archives: Basic factorization tricks

### Some non trivial factorization examples

I hope to give you a flavour of some non-trivial factorization examples using the following identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab -bc-ca)$ Example 1. Let n be a positive integer. Factorize $3^{3^{n}}(3^{3^{n}}+1) +3^{3^{n}+1}-1$

Solution. Observe that $3^{3^{n}}(3^{3^{n}}+1) +3^{3^{n}+1}-1 = a^{3}+b^{3}+c^{3}-3abc$ where $a=3^{3^{n-1}}$, $b=9^{3^{n-1}}$, and $c=-1$.

Thus, using the above factorization identity, we get the following factorization: $(3^{3^{n-1}}+9^{3^{n-1}}-1)(9^{3^{n-1}}+81^{3^{n-1}}+1-27^{3^{n-1}}+3^{3^{n-1}}+9^{3^{n-1}})$

Example 2. Let a, b, c be distinct positive integers and let k be a positive integer such that $ab+bc+ca \geq 3k^{2}-1$.

Prove that $(1/3)(a^{3}+b^{3}+c^{3})-abc \geq 3k$.

Solution. The desired inequality is equivalent to $a^{3}+b^{3}+c^{3}-3abc \geq 9k$.

Suppose without loss of generality, that $a>b>c$.

Then, since a, b, and c are distinct positive integers, we $a-b \geq 1$, and $(b-c) \geq 1$ and $a-c \geq 2$.

It follows that $a^{2}+b^{2}+c^{2}-ab-bc-ca = (1/2)((a-b)^{2}+(b-c)^{2}+(c-a)^{2}) \geq (1/2)(1+1+4)=3$

We obtain $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca) \geq 3(a+b+c)$ so it suffices to prove that $3(a+b+c) \geq 9k$ or $(a+b+c) \geq 3k$

But, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca = a^{2}+b^{2}+c^{2} - ab -bc- ca +3(ab+bc+ca) \geq 3+3(3k^{2}-1) = 9k^{2}$,

and the conclusion follows.

More later… Nalin

### Factorization uses and tricks

One of the first concepts of elementary algebra is factorization. Apart from its intrinsic use in Math itself, it is applied in several areas like pole-placement method in Control System Design and design of digital filters for use in digital communications, the latter being the foundation stone of IT. Another use is in error control coding which is used in wireless/mobile/cellular communications. One of the trickiest uses of factorization is in Trigonometry, which of course, has countless applications. Let me present to you some tricks, which are not so obvious in factorizing complicated algebraic expressions.

1) Factorization by guessing! Solve the equation $x^{3}-3x^{2}-6x+16=0$.

Solution: The product of the coefficient of highest degree term and constant is 16. So, think of the factors of 16. You will see that the LHS vanishes when $x=2$. (We have used the factor theorem here).

Hence, $x=2$ is one root of the equation and corresponding to this root, we have a factor $x-2$.

The equation may now be written $x^{2}(x-2)-x(x-2)-8(x-2)=0$ or $(x^{2}-x-8)(x-2)=0$

Removing the factor $x-2$, we have $x^{2}-x-8=0$

hence, the three roots of the above cubic $x = 0.5(1+\sqrt{33}$ or $x = 0.5(1-\sqrt{33})$ or $x=2$.

Moral of the story: Guessing the factor reduces the degree of the equation. There will be some trials and errors though.

2) The Method of Undetermined Coefficients:

Resolve into two factors: $2a^{2}x^{2} - 2(3b-4c)(b-c)y^{2}+abxy$

Solution: This equation might seem not factorizable, at first sight, as the *middle term* involves no *c*. (Note: a, b, and c are parameters, and the variables are x and y). Also, the brute force technique for using the formula method for quadratics is way too cumbersome. So, let’s give a shot to it using the method of undetermined coefficients.

Let $2a^{2}x^{2} - 2(3b-4c)(b-c)y^{2}+abxy=(Px+Qy)(Mx+Ny)$

Hence, equating like coefficients, $PM=2a^{2}$, $NQ= -2(3b-4c)(b-c)$, $MQ+NP=ab$.

Again, you have to guess!

Let’s try: let $N=2(b-c)$, $Q=-(3b-4c)$, $M=a$, $P=2a$

We need to check: $MQ+NP=-a(3b-4c)+2(b-c)(2a)=ab$ Bingo:-)

So, the factors are $(ax+2(b-c)y)(2ax-y(3b-4c))$

3) An important factorization identity is: $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Would you like to try proving this?(It is not so easy!)

More later, — Nalin