Quod Erat Demonstrandum

2016/12/18

因式分解與畢氏數組

Filed under: Junior Form Mathematics — johnmayhk @ 3:40 下午
Tags:

有些二次式 x^2+bx+c,可以(在整數環上)因式分解為

x^2+bx+c\equiv (x+p)(x+q)

但一般來說,以下四式

x^2+bx+c
x^2-bx+c
x^2+bx-c
x^2-bx-c

未必全部可以。

當然有些情況可以,例如

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)

原來 (more…)

2016/07/10

論商餘(三)

Filed under: mathematics,NSS,Teaching — johnmayhk @ 9:45 下午
Tags: , ,

一. 帶餘除法

(a) 多項式與數字

聞說以下是某校的試題:

一看,個心離一離。 (more…)

2016/02/19

hcf and lcm

Filed under: NSS — johnmayhk @ 11:14 上午
Tags: ,

某必修數學題:

若多項式 P4x^2y^2z^2 的 H.C.F. 及 L.C.M. 分別為 2x^2yz^220x^2y^2z^3,求 P

可能一開始已向學生提及,對於任何兩個正整數 a,b,恆有

ab= H.C.F \times L.C.M.

有學生解上題如下: (more…)

2015/08/27

某 monic 多項式

Filed under: NSS,Pure Mathematics — johnmayhk @ 10:09 上午
Tags: , ,

偶見高X論壇某題:

證明 x^4+x^3+x^2+x+1 > 0

回應者用了較麻煩的方法處理。

其實,當 x\ne 1(more…)

2012/09/20

To 4E students

Filed under: mathematics,NSS — johnmayhk @ 11:11 上午
Tags:

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
Tags:

liszenga 留言"why there aren’t answer of factorization Quiz and worksheet!!"

因為

1.我 (more…)

2010/10/21

因式分解二次多項式

Filed under: NSS — johnmayhk @ 5:37 下午
Tags: ,

教 sum and product of roots,設 \alpha, \beta 為二次方程 ax^2 + bx + c = 0 的根(roots),則

ax^2 + bx + c \equiv a(x - \alpha)(x - \beta) ………. (*)

從而得

ax^2 + bx + c \equiv ax^2 - a(\alpha + \beta)x + a\alpha\beta

再由比較係數,知

\alpha + \beta = \frac{-b}{a}
\alpha\beta = \frac{c}{a}

梁同學問:「為何可以比較係數?」

答:「(*) 是恆等式。」

梁:「但為何 (*) 是真的?」 (more…)

2008/12/07

做數雕蟲小技系列:看整體

看整體有時比看局部好。舉例

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
Tags: ,

Let’s start with the following question.

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
Tags: ,

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
Tags: , ,

把授課員眼中認為非常簡單的恆等式習題寫在黑板上,見同學 A 乖乖抄完,之後,他好像有點魂遊丈外,我走到他那裡,問之 (more…)

2008/02/23

Quadratic formula program @ CASIO 3950p/3650p

Filed under: HKCEE,Junior Form Mathematics — johnmayhk @ 11:17 下午
Tags:

驚恐中。我班的中五同學不懂做

Factorize 4x^2 - 4xy + y^2.

(不變金句:教了,不等於學了)

在中二的課,談到 cross method,我教同學用計算機。Casio 3950p (or 3650p) 較以往好,是它顯示的根(roots)是分數,不是點數,這樣可方便同學寫出答案。中五的同學,若你的 Casio 3950p 還未有 quadratic formula,快快看:

http://intranet.sfxc.edu.hk/it-school/homepage/nwc/casio_%20fx_3950p_quadratic_formula_program.doc

(more…)

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/
 

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?

在WordPress.com寫網誌.