Kamis, 15 Januari 2009

HOW DIFFICULT TO COMMUNICATE MATHEMATICS

HOW DIFFICULT TO COMMUNICATE MATHEMATICS
ABOUT POLYNOMIALS IN ENGLISH

Date : Sunday, January 5th 2009
Place : Ardhita Kirana Rukmi’s Home

It’s difficult to communicate mathematics language to the other. That is not as easy as read or understand any topic in mathematics. We have to be smart to find the right words, not so hard word, so that others can understand what we said. The difficulties to communicate this mathematics language become the challenge for university students of Mathematics in Yogyakarta State University to complete an English assignment lectured by Dr. Marsigit, M. A.. We have to communicate any topic about mathematics, absolutely in English, in pairs. I am with Ardhita Kirana Rukmi.
I take The Polynomials from mathematics book of 11th grade Senior High School, second semester, published by Erlangga.
Firstly, I give her the definitions of the polynomials. Then, I show her an example of the polynomials write. She seems not in trouble in understanding what I said. So, I move to give her several questions about the definitions of the polynomials as follows:
Mention the variable, the degree, and the coefficients for each following polynomials:
a. (4+3t-2t^2+t^3+10t^4-2t^5)
b. (2x^3+5x^2-10x+7)
c. (b^8-6b^7+5b^6-16)
She can do it well. I told her the next subsection about the value of the polynomials after the examples. There are two methods to find the value of the polynomial in this subsection. First method is the substitution and the second is the scheme method. From this two methods, Ardhita like the first method better. She said that she always used the substitution method since the first time she got it to find the value of the polynomials. Here’s the example:
Determine the value of the polynomial f(x)=x^3+3x^2-x+5 if x is replaced by x=m (m element of R).
Answer: For x=m we obtain f(m)=m^3+3m^2-m+5.
I get a little bit difficulties for the scheme method because the explanation from the book uses an alphabet so it seems so abstract. I told Ardhita step by step on scheme method. Unfortunately, she’s not so understands about what I give. It maybe caused by my language that difficult to understand. But, she said that she had understands after I show the picture of the scheme.
I only take the subject matter until the second subsection. I give the exercise after finishing explaining her to see how far she knows about the polynomials. I give her three questions:
1. With the substitution method, find f(1,y) if f(x,y)=x^3y^4+xy^3+y+2x^2+3
2. Find f(2,-1) if f(x,y)=x^2y+y^2-2x+10 with the substitution method
3. Find the value of the polynomials from number 1 and two with the scheme method
She did the two first questions well. For the last question using the scheme method she needs a little longer time than the first.
From this activity I can conclude that Ardhita understanding about the polynomials better than before. The first reason is that she had learned the lesson about the polynomials since junior high school. The second reason is the definition of the polynomial reminding her to the very beginning we learned it. Or on the words it will remind all of the things that maybe passed by her.
Scheme method rarely used by Ardhita to solve the problems. She keen on the substitution method better because this method always she use. It becomes one problem to understand the scheme method quickly. She said that she uses the scheme method in factoring polynomials, not to find the value of the polynomials.
The main challenge for me is in communicate this polynomials topic in English. A little number of mathematics vocabularies is the problem that makes our communication heard not so smooth. Mathematics term maybe rarely heard. But it must be an ordinary thing for mathematics student like me.
Other problem that I face is translating this matter. I take it from the book in Indonesian. There are several term that not existing in the dictionary. I ask to my friend for sometimes or browse from the internet to handle it. The problems can be cleared then.
This activity is so good and gives so many advantages for student to know how far our communication capability in English about mathematics. I motivated to be more diligent to learn and learn, English especially, because nowadays we have to being a part of an international society from the globalization.

THE VIDEOS

1. Pre Calculus Graph
Let’s begin by discussing the graph of a rational function which can have discontinuities. A rational function has a polynomial in the denominator which means you are dividing by something that is valuable quantity.
It’s possible that some value of x will meet to division be zero. Example:
If f(x)=(x+2)/(x-1), when x=1, the function value become f(x)=(1+2)/(1-1)=3/0 with 0 in the denominator. For this function, choosing x=1 is a bad idea.
When there is a bad choice of x when it makes the bottom of the rational function, it shows up a break in a function graph. For example, suppose you to finding the graph f(x)=(x+2)/(x-1). Start with inserting 0 for x. So now we have f(0)=(0+2)/(0-1)=2/(-1)=-1. So, you put off point down on the graph at (0,-2).

Next you try x=1. This time you get f(1)=(1+2)/(1-1)=3/0. That is you know is impossible and it means that you can not compute y-value when x=1.
It also mean that the graph of this function will not have any point for x=1. The graph is separated into two disconnected pieces at x=1.

Take the graph of f(x)=1/(x^2+1). No matter what numbers you choose for x, the denominator will never zero. The graph is smooth and unbroken.

Don’t forget that the general rule on the rational function, you must expect the possibility that the denominator may turn out to be zero.
Review: For polynomials, the graph is a smooth unbroken curve. For a rational function, sometimes the value of x may be zero in the denominator. That is an impossible situation because there is no y for that x. At that point, there is no value for the function and there is a break in the graph.
A break can show up in two ways. The simpler type of a break is just a missing point. This function is an example of this type of break: y=(x^2-x-6)/(x-3).

The gate that appear in the graph is at points where x=3. If you try to substitute 3 for x in the equation, the result is y=(3^2-3-6)/(3-3)=0/0. That is not possible, not feasible, and not allowed. So, there is no y for x=3. That is the typical example of the missing point syndrome and conveniently it always goes with the result like 0/0.
When you see the result of 0/0, it also tells you that it should be possible to factor the top and the bottom of the rational function and simplify. For an example: y=(x^2-x-6)/(x-3)=(x-3)(x-2)/(x-3)=(x+2).
For this kind of the break, the missing point is a loop hole. For the original function without simplify, x=3 is a bad point because it leads to the division be zero, y=0/0. But if you simplify first, then there’s no problem with x=3, y=x+2. It can be one idea or key in calculus.
Removable singularity appears simplest missing points on the graph when x leads to 0/0. For this kind of a break, if you factor and simplify the rational function, the division by zero can be avoided.

2. Kata Kerja

Kata kerja menunjukan sebuah kegiatan atau untuk menjelaskan sebuah kejadian. Tentang apa yang sedang dilakukan oleh suatu kalimat.
Dalam sebuah kalimat sederhana:
Dave berlari
Kata kerja
Karena berlari menunjukan apa yang sedang dilakukan Dave.
Dalam tata Bahasa Inggris, perubahan bentuk kata kerja menunjukan siapa yang sedang melakukan kegiatan tersebut.
I do, you do, he does, we do, they do.
Contoh : Dave runs, tetapi kita mengatakan I run
Kata kerja berubah karena orang lain yang nelakukan kegiatan tersebut
I, You, We, They He, She, it

run runs
kata kerja - to be
I am
You are singular sbyek/kata ganti tunggal, diikuti kata kerja tunggal
She is
It is
We are
They are kata ganti jamak, diikuti kata kerja jamak
Contoh :
Nyonya Midori adalah kata ganti tunggal, jadi menggunakan :
Mrs Midori Yodels bentuk yang digunakan untuk kata ganti tunggal.
Saudara-saudara Midori seperti Else, Gretel, Heidi. Mereka adalah kata ganti jamak, maka menggunakan kata kerja jamak :
Midori’s sister yodel.


3. Basic Trigonometry
Trigonometry (from g reek, trigon and metron). Trigonometry is relly study of rectangle and the relationship between the side and the angle of rectangle.

Function
Sin Ǿ = ?
cos Ǿ = ?
tan Ǿ = ?
to solve them, use triy function soh, cah, toa.
Soh : sine is opposite over hypotenuse
Cah : cosine is adjacent over hypotenuse
Toa : tangent is opposite over adjacent

- Sin Ǿ = 4/5 the other trigfunctions : 1. Cosecant Ǿ

- Cos Ǿ = 3/5 2. Secant Ǿ

- Tan Ǿ = 4/3 3. Catangent Ǿ

If the angle is x, so tan x = ¾ (the invers of tan Ǿ)

4. Kalimat Majemuk

Contoh :
Klausa 1 : It’s the end of the world as we know it, and
Klausa 2 : I feel fine
Ada 2 klausa yang dihubungkan dengan 1 kata penghubung, yaitu “and” ketika satu kalimat digunakan sebagai bagian dalam kelompok yang lebih besar, kalimat yang lebih kecil disebut klausa.
Ketika sebuah klausa dpaat berdiri sendiri dalam sebuah kalimat, maka disebut “independent clause”. Jika memiliki 2 klausa, maka disebut dengan kalimat majemuk. Untuk menggabungkan 2 independent klausa menggunakan tanda titik dua (:), klausa yang kedua menjelaskan klausa yang pertama.
- I love my two sisters. They bake me pie.
Untuk menggabungkan kedua kalimat tersebut, menggunakan tanda titik dua
- I Love my two sister : they bake me pie
• Tanda titik koma (;)
Contoh kalimat :
It’s the end of the world as we know it and I feel fine
Dapat disingkat menjadi :
It’s the end of the world as we know it, I Feel fine.
• Garis penghubung (-)
Terdiri dari beberapa elemen yang mengejutkan. Kita menggunakan garis penghubung karena klausa kedua dihubungkan dengan klausa pertama
Untuk menghubungkan kalimat terdapat 4 cara :
- Kata penguhubung
- Titik dua
- Titik koma
- Garis penghubung
• Kalimat Fragmen
Jika kita mendapati suatu kalimat tidak dapat berdiri sendiri sebagai kalimat lengkap. Contoh kalimat lengkap:
My pet Komodo dragon is a lamb
Contoh kalimat fragmen :
Because he has no teeth
Klausa dependent : - tidak bisa berdiri sendiri
- klausa dependent terdapat pada klausa independent
- bukan kalimat lengkap
Contoh :
Although Tom sleeps regulary, he is constantly tired
Klausa dependent klausa independent
Kalimat diatas disebut kalimat kompleks.

5. Limit By Inspection

There are two conditions :
1. X goes to positive or negative infinity
2. Limit involves a polynomial divide by a polynomial
For example :
Lim x3 + 4 = ~
x2 + x+1
X→ ~
(Because highest power of x in number is 3 greater than highest power in denominator) since all the number are positive and x is positive infinity
The key to defermining limits by inspection is in looking at powers of x in the numerator ana the denominator. To apply these rules : - must be divining by polynomial
- x has to be approaching infinity

1. First shortcut rule
If the highest power of x in numerator is 3 greather than highest power in denominator) since all the number are positive and x is going to positive infinity if you can’t tell it the answer is positive or negative:
- substitute a large number for x
- see if you end up with a positive or negative number
- whatever sign you get is the sign of infinity for the limit.

2. Second shortcut rule:
If the highest power of x is in the denominator, then limit is zero
Lim x2 + 3
x → ~ x3 + I , x2 + 3 = highest power of x in numerator is 2
x3 + 1 = highest power of x in denominator is 3

3. Last shortcut rule
Used when : highest power of x in numerator is same as highest power of x in denominator.
Lim = the quotient on the efficient on the two highest powers
x → ~
-Remember : coefisient → the number that goes with a variable
Ex : 2 is the coefficient of 2x2
75 is the coefficient of 75 x 4
Sho that is no way if x = 3

When you see the result and also tell you direction be possible factor top and bottom of rasional function and simplify

6. Trig Function

Trig function : really study of right triangle

To remember you can use : s o h c a h t o a
i p y o d y a p d
n p p s j p n p j
e o o i a g
s t n c e
i e e e n
t n n t
u t
s
e
Trig Function
- ratios of different sides of triangles
- With respect to an angle
- Only need to know values of sides to find measure of an angles figure of all part of triangle
Trig function : 1. Sine 4. Cosecant
2. cosine 5. Secant
3. tangent 6. cotangent


Six basis trig function defined by :
1. Sides of a triangle
2. angle being measured
opp : side opposite to theta
adj : side adjacent to theta
hyp : hypotenuse
Six trig function : 1. Sin Ө 3. Tan Ө 5. secӨ
2. cos Ө 4. Csc Ө 6. Cot Ө
Trid shortcut

According to this rule :
Lim → coefficients of x³ is over cach other
Lim
x 4

7. Kalimat Sederhana

Kebanyakan tipe kalimat adalah kalimat sederhana, sederhana karena semua elemen didalam kalimat adalah bagian dari subjek dan predikat. Subjek menunjukan kegiatan dari kata kerja utama. Subjek sederhana adalah kata benda khusus yang menunjukan sebuah kegiatan.
The happy little child kicked the gnome over the fence
Subyek
Predikat dari sebuah kalimat terdiri dari : main verb kata kerja utama dan apapun yang mengikutinya. Gabungan keduanya disebut predikat yang lengkap. Contoh :
The happy little child kicked the gnme over the fence predikat yang lengkap.
Karena “kecked the gnome over the fence” menunjukan tentang apa yang ditendang dan bagaimana tendangan itu secara lebih jelas lag.
- Kalimat sederhana dapat diperoleh tanpa subjek dan predikat.
- Kalimat perintah adalah kalimat yang ditunjukan langsung kepada orang kedua, yaitu “kamu”, berfungsi untuk memerintahkan seseorang agar melakukan sesuatu.
Kick that gnome over the fence
- Tidak ada subjek
- Siapa yang sebenarnya melakukan kegiatan tersebut?”kamu”
Bisa ditulis : “Hey you, kick that gnome over the fence.”
Tidak perlu ditulis karena sudah tersirat

POLYNOMIALS

POLYNOMIALS

Subject:
5-1 Definitions of polynomials, the value of the polynomials, and polynomials operations
5-2 The division of polynomials
5-3 Remaining Theorem
5-4 Factoring theorem
5-5 Polynomials Equation solution

Algebraic form had been learned at junior High School which is discussing about the definition of a term, factor, coefficient, and constant. Monomial, binomial, and polynomial in same or different variable, solving the operations of the polynomial, solving the division of the same term, and factoring algebraic terms had been studied too. That matter will be discussed again deeper, then will be developed to the division, the polynomials, remaining theorem, and the roots of the polynomials.
The basic competent of the “Polynomial” is using the division algorithm, remaining theorem, and factoring theorem on problem solving Realization of the competent will be show through study result : using polynomials division algorithm to decide the division result and division remain, using the theorem and the factor to solve the problems, and proof the remaining and the factoring theorem.
To support the success of the competent in this chapter, the indicators of the success are, the student can:
- explain the polynomials division algorithm
- decide the degree of division result polynomials remain by linear or quadratic form
- determine the division result and division remain by linear or quadratic form
- determine the polynomials division remain by linear or quadratic form using the remaining theorem
- determine the linear factor of the polynomials using factoring theorem
- proof the remaining theorem and the factoring theorem

5-1 Definitions of polynomials, the value of the polynomials, and polynomials operations

5-1-1 Definition
Look at there algebraic form:
(i) x2-3x+4
(ii) 4x3 + x2-16x+2
(iii) X4+3x3-12x2+10x+5
(iv) 2x5-10x4+2x3+3x2+15x-6
Those algebraic forms are called polynomials in x.
The degree/ exponent of the polynomial in x decided by the highest exponent of x.
For example:
(i) x2-3x+4 is a second degree polynomial because the highest exponent of x is 2
(ii) 4x3 + x2-16x+2 is a third degree polynomial because the highest exponent of x is 3
(iii) X4+3x3-12x2+10x+5 is a fourth degree polynomial because the highest exponent of x is 4
(iv) 2x5-10x4+2x3+3x2+15x-6 is a fifth degree polynomial because the highest exponent of x is 5
So, the polynomial’s degree in x, generally wrote as:

anxn + an-1xn-1 + an-2xn-2+…+a2x2 + a1x + a0

Where:
- an, an-1, an-2, …, a2, a1, a0, are real number with an  0. an is the coefficient of xn, an-1 is the confident of xn-1, an-2 is the coefficient of xn-2, …, and soon, a0 called as a constant.
- n is a counting number which represent the degree of the polynomials
the terms of the polynomials above are beginning by the term that the variable has he highest power ansn, then followed by the terms with decreasing power of x, an-1xn-1, an-2xn-2, …, a2x2, a1x1, and ended by the constant 90. The polynomials written on that way called arranged in decreasing power rule on variable of x. Remember that the variable must not be x, but can be a,b,c…, y and z.
For illustration, look at these polynomials:
a) 6x2-3x2+10x+4 is a third degree polynomial in x. The coefficient of x3 is 6, for x2 15-3, for x is 10, and the constant is 4.
b) 9y4-y3+5y2-2y-13 is a fourth degree polynomial in x. The coefficient of y4 is 9, for y3 is -1, for x is 5, for y is-2, and the constant is -13.
c) (x-1)(x+1)2=x3+x2-x-1 is a third degree polynomial in x. The coefficient of x3 is 1, for x is -1, and the constant is -1.
d) (t+1)2(t-2)(t+3)=t4+3t3-3t2-11t-6 is a fourth degree polynomial in t. The coefficient of t4 is 1, for t3 is 3, for t3 is -3, for t is-11, and the constant is -6.
Polynomials above are polynomials in one variable called a univariable polynomials. The polynomials in more than one variable called a multivariable polynomials. For illustration, look at following polynomials.
a) x3+x2y4-4x+3y2-10 is a polynomial in two variable x and y. this is a third degree polynomial in x or fourth degree polynomial in y
b) a3+b3+c3+3ab+3ac-bc+8 is a polynomial in three variables a, b, and c. this is a third degree polynomial in a, b, or c.

THE DEFINITION OF THE POLYNOMIALS

EXERCISE 1
1. Mention the variable, the degree, and the coefficient for each following polynomials:
a) 2x3+5x2-10x+7
b) x3-4x+2
c) 10-2y
d) 4+3y-5y2
e) 4a7-3a2+3a-10
f) 3-2a5
g) p4+2p-5
h) 4+3t-2t2+t3+10t4-2t5
2. Repeat question number 1 for following polynomials:
a) 3x4+10x3-5x2+ ¼ x+4
b) 6x7-4x6+2x5-10x4+x3-4x2-5x+8
c) 4-y+3y2-5y3+10y4-y5
d) 3+y-10y2+3y3
e) b8-6b7+5b6-16
f) 4+2b2+3b3-5b4+10b5+6b6-9b+2b10
g) q12+4q8-q6+q4-5q2+q+6
h) 10-q+2q3+4q5- 7q7+10 q9+7 q11+ q13
i) 6-4s+s2-10s3+6s7+s16+3s15
j) z20-4z18+5z16-13z14+4z12+10z7-5z2-z+3
3. Determine the number of variables for each polynomial, then determine the degree (according to the variable)
a) x5y+xy3+4x-5y+12
b) a5b5-a4b3+a3b3+a+b2-6
c) (2x-y)4-(2y+3z)3+(2z-x)6
d) P4+q4+r4+3pq-5pr+6qr+4
4. Repeat question number 3 for following polynomials:
a) x5y5-2x4y3+3x2y2-10xy2-4x+3y-10
b) a7b6-a6b5-6a5b3-8a4+3a2b+5ab2+5a2-6b2+10
c) (2x-y)3-4(y+3z)4+(2z-x)2
d) 3a5+2b5-5c5+3a4-b4+3c4+2abc+4ab-8c+9
5. Multiply these polynomial product (state the result in decreasing power rule then, determine the degree and coefficients
a) (x-4)(x+2)
b) (x2-1)(x+4)
c) (y2-3)(y2+3)
d) (y2-4)(y2+2y+1)
e) (a2+2)(a-4)2
f) (a+4)2(2a-1)
g) (b-1)2(b-2)2
h) (2b-1)2(2-b)2
6. Find the coefficient of the following statements:
a) x2 in (x-1)2(x+2)(x+1)
b) y3 in (2y+y2)(4y2-2y+1)
c) z in z(z-1)(z-2)(2+3)
d) t4 in (t2+2t-1)3


5-1-2 The value of the polynomials
The polynomials can be written in the form of function of the variable, based on the fact that polynomial is an algebraic form in a variable. Polynomials in x can be written as a function of x. for example, general form of the polynomials can be state in a function form as:

f(x)=anxn+ an-1xn-1+ an-2xn-2+…+a2x2+a1x+a0

Notes:
Polynomial function above is state in f (x), sometimes state as:
- S(x) that shows polynomial function in x, or
- P(x) that shows polynomial function in x

Denote the polynomials as a function in x, then the polynomials value can be determined. Generally, polynomials value of f(x) for x = k is f(k) where k is a real number. Then, the value of f(k) can be found with two methods:

Substitution method, or

Scheme method

A. Substitution Method
To explain the way to find the polynomials value with substitution, look at this third degree polynomial in x:
f(x) = x3+3x2-5x+2
- The value of f(x) for x =-1 is:
f(-1) = (-1)3+3 (-1)2-5(-1)+2=-1+3+5+2=9
- The value of f(x) for x =0 is:
f(0) = (0)3+3 (0)2-5(0)+2=0+0-0+2=2
- The value of f(x) for x =1 is:
f(1) = (1)3+3 (1)2-5(1)+2=1+3-5+2=1
- The value of f(x) for x =2 is:
f(2)= (2)3+3 (2)2-5(2)+2=8+12-10+2=12

Based on the description above, the value of the polynomials for a certain variable can be found by the rule of substation method as follows:
The polynomials value of f(x)=anxn+ an-1xn-1+ an-2xn-2+…+a2x2+a1x+a0 for x = k
(k  real numbers) determined by:
f(x)=an(k)n+ an-1(k)n-1+ an-2(k)n-2+…+a2 (k)2+a1 (k)+a0

The value of f(x) for x = k obtained by substitute the value of k to the variable of x on the polynomials of f(x). Hence, counting the polynomials value as above called as a substitution method. Look at these example to understanding the way to counting polynomials value with substitution method.

Example 1:
Determine the value of the polynomial f(x) = x3+3x2-x+5 if x is replaced by:
a) x = 0
b) x = 1
c) x =-1
d) x =2
e) x =-2
f) x = m(m element of R)
g) x = m-2(m element of R)
h) x = m+1(m element of R)
Answer:
f(x) = x3+3x2-x+5, hence
a) for x = 0, we obtain:
f(0) = (0)3+3(0)2-(0)+5=0+0-0+5=5
so, the value of f(x) for x = 0 is f (0) = 5
b) for x = 1, we obtain
f(1) = (1)3+3 (1)2-(1)+5=1+3-1+5=8
so, the value of f(x) for x = 1 is f (1) = 8
c) for x =-1, we obtain
f(-1) = (-1)3+3 (-1)2-(-1)+5=-1+3+1+5=8
so, the value of f(x) for x = -1 is f (-1) = 8
d) for x = 2, we obtain
f(2) = (2)3+3 (2)2-(2)+5=8+12-2+5=23
so, the value of f(x) for x = 2 is f (2) = 8
e) for x = -2, we obtain
f(-2) = (-2)3+3 (-2)2-(-2)+5=-8+12+2+5=11
so, the value of f(x) for x = -2is f (-2) = 8
f) for x = m(mR), we obtain
f(m) = (m)3+3 (m)2-(m)+5=m3+3m2-m+5
so, the value of f(x) for x = m is f (m) = m3+3m2-m+5
g) for x = m-2(mR), we obtain
f(m-2) = (m-2)3+3 (m-2)2-(m-2)+5=m3-3m2-m+11
so, the value of f(x) for x = m-2 is f (m-2) = m3-3m2-m+11
h) for x = m+1(mR), we obtain
f(m+1) = (m+1)3+3 (m+1)2-(m+1)+5=m3+6m2+8m-8
so, the value of f(m+1) for x = m+1 (mR)is f (m+1) = m3+6m2+8m+2

Example 2:
Given a polynomial in 2 variables x and y
f(x,y)=x2y+xy2+3x-4y+2
find
a) f(4,y)
b) f(-1,y)
c) f(x,3)
d) f(x,-2)
e) f(4,2)
f) f(-2,3)
Answer:
f(4,y)=x2y+xy2+3x-4y+2
a) f(4,y) means variable of x = 4 and variable of y constant
f(4,y)= (4)2y+(4)y2+3(4)-4y+2
=16y+4y2+12-4y+2
=4y212y+14
So, f(4,y)=x2y+xy2+3x-4y+2 is polynomial in y
b) f(-1,y) means variable of x = 4 and variable of y constant
f(4,y)= (-1)2y+(-1)y2+3(-1)-4y+2
=y-y2-3-4y+2
=-y2-3y-1
So, f(-1,y)= -y2-3y-1 is polynomial in y
c) f(x,3) means variable of x is constant and variable of y is 3
f(x,3)= x2(3)+x(3)2+3x-4(x)+2
=3x+9x+3x-12+2
=3x2+12x-10
So, f(x,3)= 3x2+12x-10 is a polynomial in x
d) f(x,-2) = x2(-2)+x(-2)2+3x-4(-2)+2
=-2x+4x+3x+8+2
=-2x2+7x+10
So, f(x, -2)= -2x2+7x+10 is a polynomial in x
e) f(4,2) means variable of x = 4 and y = 2
f(4,2)= (4)2(2) +(4) (2) 2+3(4)-4(2) +2
=32+16+12-8+2
= 54
So, f(4,2)= 54 is a real number
f) f(-2,3) means variable of x = -2 and y = 3
f(-2,3)= (-2)2(3)+(-2)(3)2+3(-2)-4(3)+2
= 12-18-6-12+2
= -22
So, f(-2,3)= -22 is a real number
What conclusions can you get based on the results above?


B. Scheme Method
To describe the way to find the polynomials value by scheme method, look at this fourth degree polynomial.
f(x) = a4x4+a3x3+ a2x2+a1x+a0
With substitution method, the polynomial value of f(x) for x = k determine by:
f(x) = a4k4+a3k3+ a2k2+a1k+a0
That f(x) can be arranged with the operation of multiplication addition as follows:
f(x) = a4k4+a3k3+ a2k2+a1k+a0
f(x) = (a4k4+a3k3+ a2k2+a1)k+a0
f(x) = ((a4k4+a3k3+ a2)k+a1)k+a0
f(x) = [{(a4k4+a3)k3+ a2}k2+a1]k+a0

Based on the last equation, we can see that the value of f(k) can be found step by step with an algorithms as follow:
- First step:
Multiply a4 with k, then sum the result with a3
a4k + a3
- Second step
Multiply the result of the first step (a4k + a3) with k, then sum the result with a2
(a4k + a3)k + a2 = a4k2 + a3k + a2
- Third step
Multiply the result of the first step (a4k+a3k+a2 ) with k, then sum the result with a1
(a4k+a3k+a2 )k+a1= a4k2 + a3k + a2k+a1
- Fourth step
Multiply the result of the first step (a4k+a3k+a2k+a1) with k, then sum the result with a0
(a4k+a3k+a2k+a1)k+a0= a4k+a3k+a2k+a1k+a0
The result of this fourth step is the value of f(x)= a4k+a3k+a2k+a1k+a0 for x=k
The processes of the algorithms above can be displayed on a scheme x = k is wrote in the first row of the scheme, then followed by polynomials coefficients. These coefficients arranged from the highest exponent coefficients to the smallest exponent. Look at this scheme:

x=k a4 a3 a2 a1 a0
+ + + +
a4k a4k2+a3k a3k3+a3k2+a2k a4k4+a3k3+a2k2+a1k
a4k a4k2+a3k a4k2+a3k3+a2k a3k3+a3k2+a2k+a1k a4k2+a3k3+a2k+a1k+a0
= f(k)

Finding the value of the polynomials as above called as a scheme method. This name is given because of the scheme that used.

Notes:
(1) Substitution method is suitable to find the value of the polynomials with a simple form for a small value and integer of x
(2) Scheme method can be used o find the value of all forms of the polynomials and for arbitrary x  E

Here are the applications examples to find the value of polynomials with scheme method:
Example 3:
Find the value of these polynomials with scheme method
a) f(x) = x4-3x3+4x2-x+10 for x = 5
b) f(x) = x5-x2+4x-10 for x = 2
Answer:
a) The coefficients of f(x) = x4-3x3+4x2-x+10 are a5=1, a4=a3=0,a2=-1,a1=4, and a0 = -10
For x =5, so k = 5.The scheme is shown on picture 5-2a. Based on the scheme, the value of f(x) = x4-3x3+4x2-x+10 for x=5 is f(5) = 355
b) The coefficients of f(x) = x5-x2+4x-10 are a5=1, a4=a3=0,a2= -1,a1=4, and a0 = -10
For x =2, so k = 5.The scheme is shown on picture 5-2a. Based on the scheme, the value of f(x) = x5-x2+4x-10 for x=2 is f(2) = 26

5 a4 a3 a2 a1 a0
+ + + +
5 10 70 345
1 2 14 69 355


2 a5 a4 a2 a1 a0
+ + + +
2 4 8 36
1 2 4 ¬18 26

Example 4:
Find the value of this polynomial with scheme method:
f(x,y)=x2y+x3 y2+x2+3y+2 for x = 2
Answer:
To find the value of f(x,y) for x=2, f(x,y) is considered as a polynomial in x with the form of decreasing exponents:
f(x,y)=y2x3 + (y+1)x2+(3y+2)
The coefficients of f(x,y) are a3=y2, a2 = y+1, a1=0, and a0 =3y+2. x=2 means k=2. Based on the scheme, the value of f(x,y)=x2y+x3 y2+x2+3y+2 for x = 2 is f(2,y)=8y+7y+6
The value of f(x) = x5-x2+4x-10 for x=2 is f(2) = 26

2 a3 a2 a1 a0
y2 y+1 0 3y+2
+ + +
y2 2y2 4y2+2y +2 8y2+7y +6
y2 2y2y+1 4y2+2y +2 8y2+7y +6



THE VALUE OF THE POLYNOMIALS

EXERCISE 2
1. With the substitution method, find:
a) f(1), if f(x) = x2-3x2+4x-2
b) f(-1), if f(x) = 2x2-4x2+5
c) f(2), if f(x) = x4-2x3+5x2+6
d) f(-2), if f(x) = 2x+2x3-5x2+6x-8
e) f(m), if f(x) = x3-4x2+3
f) f(m-3), if f(x)=x3-2
g) f(m+2), if(x) = x3-2x2+4x+3
h) f(1,y), if f(x,y)=x3y4+xy3+y+2x2+3
i) f(-2,y), if f(x,y)=x3y4+xy3+y+2x2+3
j) f(x,3), if f(x,y)=x3y4+xy3+y+2x2+3
k) f(x,-1), if f(x,y)=x3y4+xy3+y+2x2+3
l) f(2,-1), if f(x,y)=x3y4+xy3+y+2x2+3
m) f(-4,-3), if f(x,y)=x3y4+xy3+y+2x2+3
2. Find the value of these polynomials for given variable value with the substitution method:
a) X3-25 for x = 2
b) x4-4x3-x2+5x-1 for x = 4
c) a3+2a2-a+8 for a=1
d) a6-a for a=-1
e) y3+2y2-5y+10 for y = 4
f) y6-70 for y = 2
g) 4x3y3+5xy2+6x2+6x2-8y2+4 for x = 1
h) 4x3y3+5xy2+6x2+6x2-8y2+4 for x = 2 and y = 3
i) x3+y3z3-3xyz for x = y
j) x3+y3z3-3xyz for x = z
k) x3+y3z3-3xyz for x = -(y+z)
3. Find the value of the polynomials from number 1 and 2 above with the scheme method
4. Use the scheme method to find these values:
a) f(1), if (x)=x2-3x+10
b) f(2), if (f(x) = x3+10x2-5x+4
c) f(10), if(x)=x4-10x3+x-8
d) f(2), if f(x)=x5-4x2+x+4
e) f(1,y), if f(x,y) = 4x3y2-5x2y2+6x2-y2+2
f) f(x,3), if (x,y) = 4x3y2-5x2y2+6x2-y2+2
5. a) Use the substitution and the scheme method to find the value of these polynomial
(i) 4x4+10x2-2x2+5 for x = -0,75
(ii) 8x4+4x3+2,55x+5x+2 for x =-0,75
b) Based on the results of a) above, what conclusions you can get about those two methods?

Bibliography:
Wirodikromo,Santoso. Matematika Jilid 4 untuk SMA kelas XI Semester 2. Jakarta: Erlangga, 2001

YOU HAVE TO BELIEVE IN YOURSELF

YOU HAVE TO BELIEVE IN YOURSELF

The boy are try to motivate us to build our confidence. Anyone can do anything, can be anything, and become anything. No matter where we come from, we are just the same each other. We got to trust to ourselves that we can do what we want to do.
When there is no belief in our heart, we will be nothing. The key to be confident, to be survive in our life is be believe in ourselves. We can be success in our life if we do trust that we can do anything and be anything.
No one have no belief. The boy is a good motivator. He can persuade everybody around him to be come in his speech.

Revision:

DO YOU BELIEVE?

There is a boy named Dalton Sherman. He came from Charles Rice Learning Center. He tried to promote the Charles Rice Learning Center to the audience in Dallas ISD.
Firstly, he asked to the audience, do they believe that he can stand up on the stage, fearless, and talk to them. He said that everyone must believe in him because there is some deals that he can do anything, be anything, and become anything.
Then, he asked to the Dallas ISD, do they believe in his classmate and do they believe that everyone of them can graduate for workplace or college.
He said, they better do believe, because next week they will show up the Dallas ISD that they can reach their highest potential, no matter where they come from, they better not give up on them.
The audiences are the one who feed them and who wipe their tears. They are the one who love them when no one does.
Next, he asked do they believe in their colleagues. They came to the audience’s school because they want to be developed also. No matter what they are, he needs them.
He tried to persuade then to trust to their colleagues, so they will do to. What are they doing is not only for his generation, but to the next generation also.
The boy and his classmate need the audience more than ever. So, the audience needs to believe in themselves.
Then he asks do they believe that every child in Dallas needs to be ready for college? They must believe that they can do. They have to believe in their colleagues, in themselves, and in their goals.
Finally, he say thank you to them who does so many things for him and others. They have to believe in him because he believes in himself. The audience helped him get to where he is today.

THE DEFINITION, EXPLANATION, AND THE EXAMPLE OF THE TERMS

The definition, explanation, and the examples of the terms questioning by…

1. Titik Pangkal: Initial Point
Sentence: Graph y = x^2 +1 from the initial point x = 1.

2. Pasangan Berurutan: Consecutive Pair
Sentence: A consecutive pairs from C = (x, y, z) and D = (1, 2, 3) are {(x,1),(y,2),(z,3)}

3. Bilangan Cacah: Counting Number
Sentence: In math, numbers are grouping in several number sets, such as integers, counting number, etc.

4. Bilangan Berpangkat: Exponential Number
Sentence: 2^3 is called as an exponential number with the power or the exponent of 3.

5. Deret Tak Hingga: Infinitive Row
Sentence: 1+2+3+… is an infinitive row.

6. Menggapai Impian: To Reach the Dream
Sentence: We have to work hard to reach our dream.

7. Mendalami Jati Diri: To Identify the Identity
Sentence: No one knows about us better than ourselves, so we have to identify our identity well.

8. Garis Keturunan: Origin Line/ Genetic Line
Sentence: Adam is the very first man in the origin line of human being.

9. Naik Sepeda Berboncengan: To Drive Together In A Bicycle
Sentence: Be careful when you drive together with your friend in a bicycle.

10. Serius: Serious
Sentence: He does the test seriously.

11. Saling Bersalam-Salaman: To Exchange Greetings/ To Shake Hands
Sentence: They shake their hands when they meet at school.

12. Jangan Melirak-Lirik: Do Not Look Aside
Sentence: Do not look aside when you are driving, for your safety.

13. Pohon Itu Melambai-Lambai: The Tree Is Waving
Sentence: I see the trees are waving along the beach.