# Quod Erat Demonstrandum

## 2017/09/14

### 又因式分解

Filed under: Junior Form Mathematics — johnmayhk @ 12:45 下午
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Factorize $2x^2+7xy+9x+13y+6y^2-5$. (more…)

## 2016/12/18

### 因式分解與畢氏數組

Filed under: Junior Form Mathematics — johnmayhk @ 3:40 下午
Tags: $x^2+bx+c\equiv (x+p)(x+q)$ $x^2+bx+c$ $x^2-bx+c$ $x^2+bx-c$ $x^2-bx-c$ $x^2+5x+6\equiv (x+2)(x+3)$ $x^2-5x+6\equiv (x-2)(x-3)$ $x^2+5x-6\equiv (x-1)(x+6)$ $x^2-5x-6\equiv (x+1)(x-6)$

## 2016/07/10

### 論商餘（三）

Filed under: mathematics,NSS,Teaching — johnmayhk @ 9:45 下午
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(a) 多項式與數字 ## 2016/02/19

### hcf and lcm

Filed under: NSS — johnmayhk @ 11:14 上午
Tags: , $ab=$ H.C.F $\times$ L.C.M.

## 2015/08/27

### 某 monic 多項式

Filed under: NSS,Pure Mathematics — johnmayhk @ 10:09 上午
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## 2012/09/20

### To 4E students

Filed under: mathematics,NSS — johnmayhk @ 11:11 上午
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For Q.8, you may refer to

https://johnmayhk.wordpress.com/2008/11/16/f2-mathematics-factorization-by-cross-method/

For 0.5-mark questions, (more…)

## 2011/01/29

### 無聊因式分解

Filed under: Junior Form Mathematics — johnmayhk @ 6:41 下午
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liszenga 留言＂why there aren’t answer of factorization Quiz and worksheet!!＂

1.我 (more…)

## 2010/10/21

### 因式分解二次多項式

Filed under: NSS — johnmayhk @ 5:37 下午
Tags: , $ax^2 + bx + c \equiv a(x - \alpha)(x - \beta)$ ………. (*) $ax^2 + bx + c \equiv ax^2 - a(\alpha + \beta)x + a\alpha\beta$ $\alpha + \beta = \frac{-b}{a}$ $\alpha\beta = \frac{c}{a}$

## 2008/12/07

### 做數雕蟲小技系列：看整體

1. Factorize $(x + y)^3 + (x - y)^3$. (more…)

## 2008/11/26

### F.2 Mathematics : a minor problem in factorization

Filed under: Junior Form Mathematics — johnmayhk @ 6:11 下午
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Factorize $1 - x^2(1 - 2x)^2$. (more…)

## 2008/11/16

### F.2 Mathematics: factorization by cross method

Filed under: Junior Form Mathematics — johnmayhk @ 6:20 下午
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Factorize $x^2 - 5x - 6$.

By using the cross method, students may give the following two ‘possible answers’.

A. $(x - 2)(x - 3)$
B. $(x + 1)(x - 6)$ (more…)

## 2008/10/25

### 中二數學課：恆等式

Filed under: Junior Form Mathematics,Teaching — johnmayhk @ 6:40 下午
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## 2008/02/23

### Quadratic formula program @ CASIO 3950p/3650p

Filed under: HKCEE,Junior Form Mathematics — johnmayhk @ 11:17 下午
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Factorize $4x^2 - 4xy + y^2$.

（不變金句：教了，不等於學了）

## 2008/01/16

### [初中] Factorization, useful or not?

Filed under: Junior Form Mathematics,Teaching — johnmayhk @ 9:44 下午
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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. $111^2 = 12321$ $1111^2 = 1234321$ $11111^2 = 123454321$ $111111^2 = 12345654321$

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.

– – – – – –

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.

## 2007/12/02

### [初中] Factorization

Filed under: Junior Form Mathematics — johnmayhk @ 3:50 下午
Tags:

The following is a question in the uniform test.
Factorize $12xy - 6x^2$.
One student, Lai, gave the solution as $12xy - 6x^2$
= $12x(y - \frac{x}{2})$
If you were a teacher, will you give marks?
Well, you may have the “correct answer" in your mind, it should be $6x(2y - x)$, right? But why the answer given by Lai was not an answer? More, if somebody gives the following, what is your comment? $12xy - 6x^2$
= $24x^3y(\frac{1}{2x^2} - \frac{1}{3xy})$
Urm, let me further my discussion by giving two more examples.
Many students know how to factorize $x^2 - y^2$, that is $x^2 - y^2$
= $(x + y)(x - y)$
Urm, can I further my calculation in writing something like $x^2 - y^2$
= $(x + y)(x - y)$
= $(x + y)(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$
Do you think the above is an answer to the factorization problem?
Some may say we cannot factorize $x^2 + 1$, but after introducing the complex numbers, can we say $x^2 + 1$
= $(x + i)(x - i)$ where $i^2 = -1$
is a process of factorization?
All in all, the above may force us to think about what do we mean by factorization exactly?