Quod Erat Demonstrandum

2008/01/16

[初中] Factorization, useful or not?

Filed under: Junior Form Mathematics,Teaching — johnmayhk @ 9:44 下午
Tags: ,

Is the topic “factorization" in junior form useful? Urm, apart from simplification, solving equation etc, will factorization play a critical role in advanced mathematics? No idea, but at least, I could tell you that we may generalize the idea of factorization in set theory. Take it easy. I just wanna say something in my daily teaching. This topic may have no application (?) in further study in mathematics, however, it may be useful as one of the tools of training of students’ mind. When letting my students play with some challenging (at least, not-straight-forward) problems, nealy all of them are willing to do and ask for help. Well, it is a good moment of promoting the beauty of mathematics!

The following are questions asked from one of my students, Lo, in class. Factorize

1. 2x^2 + 6x - 15z - 2xy + 5yz - 5xz
2. x^4 + 2x^3 + 3x^2 + 2x + 1

Questions above are not difficult, but at least, we could not just apply identities directly without doing some grouping beforehand. The art (art? yes, sometimes, it is more than ‘technique’, it may be an ‘art’) is: how?

For Q.1

2x^2 + 6x - 15z - 2xy + 5yz - 5xz
= 2x^2 - 2xy + 6x - 15z + 5yz - 5xz
= 2x(x - y) + 3(2x - 5z) + 5z(y - x)
= 2x(x - y) + 3(2x - 5z) - 5z(x - y)
= (x - y)(2x - 5z) + 3(2x - 5z)
= (2x - 5z)(x - y + 3)

Teachers can easily set up factorization problems by just expanding polynomials, i.e., starting from (2x - 5z)(x - y + 3), we can come up with 2x^2 + 6x - 15z - 2xy + 5yz - 5xz and ask students to factorize it back. However, if we change z (say) into number, the problem may not be easy to solve. That is, considering

(2x - 5)(x - y + 3)
= 2x^2 - 2xy + 6x - 5x + 5y - 15
= 2x^2 - 2xy + x + 5y - 15

Now, may be students find it difficult to factorize 2x^2 - 2xy + x + 5y - 15.

Further, if we consider the product of trinomials (three terms) with 3 variables x,y,z, it may already be a nightmare, say, could you factorize the following

3x^2 - 4y^2 - 3z^2 - 4xy + 8yz - 8zx ?

For Q.2

x^4 + 2x^3 + 3x^2 + 2x + 1
= x^4 + 2x^3 + (x^2 + 2x^2) + 2x + 1
= x^2(x^2 + 2x + 1) + 2x^2 + 2x + 1
= x^2(x + 1)^2 + 2x(x + 1) + 1
= (x(x + 1))^2 + 2x(x + 1) + 1
= (x(x + 1) + 1)^2
= (x^2 + x + 1)^2

Following this idea, we can create many, like

(a) x^8 + 2x^7 + 3x^6 + 4x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1
(b) x^{10} + 2x^9 + 3x^8 + 4x^7 + 5x^6 + 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1
(c) x^9 + 3x^8 + 6x^7 + 10x^6 + 12x^5 + 12x^4 + 10x^3 + 6x^2 + 3x + 1

It is extremely easy to set up questions above, while, it may not be easy to solve them immediately.

Just help you a bit, observe the following

111^2 = 12321
1111^2 = 1234321
11111^2 = 123454321
111111^2 = 12345654321

Well, it is also an art to strike a balance: give some challenging questions to students without frightening them.

When I was in F.2, I found an old little mathematics book of only one single theme: factorization. I’d lost it long time ago. It taught me many techniques in factorization, as for example, I knew how to factorize something like a^3 + b^3 + c^3 - 3abc (read my old post if you want to). But, it is quite demanding to ask a F.2 student to factorize the following

x^{27} - 1

though it is just a piece of cake for F.6 or F.7 students to use the techniques in complex numbers and obtain

x^{27} - 1
= (x-1)(x^2 - 2x\cos(\frac{2\pi}{27}) + 1)(x^2 - 2x\cos(\frac{4\pi}{27}) + 1)(x^2 - 2x\cos(\frac{6\pi}{27}) + 1)\dots(x^2 - 2x\cos(\frac{26\pi}{27}) + 1)

 There was a funny problem called Tschebotarev problem concerning the factorization of x^n - 1, see if I can find more and share with you next time.

– – – – – –

Just wanna add something below.

When introducing the identity

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

in class, one student, Choi, puzzled that, ‘why can’t we write something like

a^3 - b^3 = ((a^{1.5})^2 - (b^{1.5})^2) = (a^{1.5} - b^{1.5})(a^{1.5} + b^{1.5})

Good question! When putting concrete numbers, it works fine, e.g.

16^3 - 9^3 = (16 ^{1.5})^2 - (9 ^{1.5})^2 = (16 ^{1.5} - 9 ^{1.5})(16 ^{1.5} + 9 ^{1.5})

But, what exactly is the question asking? Factorization of polynomials.

The mathematics object on the L.H.S., a^3 - b^3 is a polynomial in a, b; however, the thing on the R.H.S., like (a ^{1.5} - b ^{1.5}) is NOT a polynomial in a, b. It is because the indices involved are NOT non-negative integers. Hence, the suggestion by the student should not be acceptable.

This is one of the rules. The other is the kind of coefficients.

Also read

https://johnmayhk.wordpress.com/2007/12/02/%e5%88%9d%e4%b8%ad-factorization/
 

1 則迴響 »

  1. Mr
    I respect your opinion , but I can not understand how You are teaching math
    why?
    Are you know how it is obtained a polinomy in the reality?
    You can obtain thousand of polinomies .
    How can you that?
    Relation and correlation of variables-
    It is easy
    Of course, you ought to be in touch with the reality.
    It is sad to teach with knowing applications
    Bye

    迴響 由 JOSE TOMAS OLIVA — 2009/02/10 @ 1:44 上午 | 回覆


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