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