Ani agustina (a1 c011007) polynomial

Polynomial by Ani Agustina

Click here

Welcome to Polynomial
Start

Exit


POLYNOMIAL
MATTERS

EXERCISES

First Slide

QUIZ


POLYNOMIAL
MATTERS

EXERCISES

General Aspects of
Polynomial
Value of a polynomial
Polynomial Dividing

Remainder Theorem
Factorization Theorem
Roots of a Polynomial
Equation

First Slide

QUIZ


POLYNOMIAL
MATTERS

EXERCISES
EXERCISES 1
EXERCISES 2
EXERCISES 3
EXERCISES 4
EXERCISES 5
EXERCISES 6

First Slide

QUIZ


POLYNOMIAL
MATTERS

EXERCISES

QUIZ
QUIZ 1
QUIZ 2

First Slide


General Aspects of Polynomial

1. Definition and Polynomial Component
Definition and
Polynomial Component
Algebraic Operations of
Polynomial

Polynomial equation

Algebraic forms, such as a linier equation
and a quadratic equation are kinds of
polynomial equation of degree 1 and 2.
Consider the following algebraic forms:
a. x3 + 4x2 – 7x + 10
b. 5x4 – 2x3 + 3x2 – x – 5
c. x5 + 6x4 – 3x3 + 2x2 + 10x – 3
the algebraic forms above are called
polynomial in variabel x that accomadate in
the polynomial. They consecutively have
degree of 3, 4, and 5.
1

2



Definition and
Polynomial

Polynomial equation

Definition
The general form of polynimial in variabel x
of degree n is:
P(x) = anxn + an-1 xn-1 + an-2 xn-2 + . . . + a1x + a0
The polynomial are arranged based on
descending order of the exponent of x,
where:
an , an-1 , . . . , a1 = real coefficients of the
polynomial and an ≠ 0,
a0 = real constant,
n = degree of the polynomial, n is whole
number.
1

2



Definition and
2.
Polynomial

Polynomial equation

Algebraic Operations of Polynomial

Algebraic operations of polynomial such as
addition, subtraction, multiplication, and
divison have similar characteristic and rulers
as the algerbraic operations on real
numbers.

1

2



Definition and
Polynomial

Polynomial equation

Example :
Given f(x) = x3 + 2x + 3 and g(x) = x5 + 3x4 – 7x2 –
3x + 1
Determine:
a. f(x) + g(x)
b. f(x) – g(x)
Answer:
a. f(x) + g(x) = (x3 + 2x + 3) + (x5 +3x4 – 7x2 – 3x + 1)
= x5 + 3x4 + x3 – 7x2 – x + 4
b. f(x) – g(x) = (x3 + 2x + 3) - (x5 +3x4 – 7x2 – 3x + 1)
= -x5 - 3x4 + x3 + 7x2 + 5x + 2

1

2



Definition and
Polynomial

Polynomial equation

3. Polynomial equation
Definition
Suppose given polynomial of f(x) and g(x),
where
f(x) = anxn + an-1 xn-1 + an-2 xn-2 + . . . + a1x + a0
g(x) = bnxn + bn-1 xn-1 + bn-2 xn-2 + . . . + b1x + b0
f(x) is equal to g(x) (written f(x) ≡ g(x)) if
satisfies:
an = bn , an-1 = bn-1 , an-2 = bn-2 , . . . , a2 = b2 , a1
= b1 , a0 = b0
1

2



Definition and
Polynomial

Polynomial equation

Find the value of a and b of equation x3 + 4x2
– 7x + a = (x – 2) (x + 1) (x + b).
Answer :
X3 + 4x2 – 7x + a = (x – 2) (x + 1) (x + b)
= (x2 – x – 2) (x + b)
= x3 + (b – 1) x2 – (b + 2)x – 2b
b – 1 = 4 → coefficient of x2
b + 2 = 7 → coefficient of x2
Based on the equationa above b = 5
a = -2b → a constant
↔ a = -2(5) = -10
Thus, the value of a = -10 and b = 5
1

2



Determining the value of
polynomial by substitution
method

Determining the value of a
polynomial by Scheme
(chart)

1. Determining the value of polynomial by
substitution method
Using this method, the value of the
polynomial of P(x) for x = k, written P(k), is
determined by substituting the value of k
into the variable of x on the polynomial.
Example
Determine the value of polynomial P(x) =
x5 – 2x4 + 3x3 + 4x2 – 10x + 3 for x = 1
Answer:
P(1) = (1)5 – 2(1)4 + 3(1)3 + 4(1)2 – 10(1) + 3
= 1 – 2 + 3 + 4 – 10 + 3 = -1



method

(chart)

2.

polynomial by Scheme (chart)

The first step of using scheme method is
to write a polynomial from the right to
the left starting from the highest
exponent of the variable. To solve P(x) =
x5 – 2x4 + 3x3 + 4x2 – 10x + 3 for x = 1, you
must write each coefficient each term of
the polynomial inthe chart.
1 2



Consider the following chart.
method

(chart)

Thus, the value of P(1) = -1
Description
Symbol ↗ means multiply by the input
number (in this case input number = 1
since we will find the value of the
polynomial, P(x), with x = 1).
1 2


Polynomial Dividing

Definition
Long Division
Synthetic Division (Horner
Method)

Definition
Consider the following expression.
P(x) = Q(x) . H(x) + S(x)
The expression above is a division of
polynomials, where
P(x) = the divided, Q(x) = the divisor,
H(x) = the quotient, and S(x) = the
remainder.


Polynomial Dividing

1.
Definition
Long Division
Method)

Long Division

The procedure of dividing a polynomial
using long division is similar to the
procedures of dividing integers.
Example
Determine the quotient and the
remainder of the division of P(x) = x3 +
5x2 – 4x – 20 by x + 3
Answers:
P(x) has degree of 3, Q(x) has degree of 1,
and the quotient of H(x) has degree of 3 –
1 = 2, the remainder of division S(x) has
1 2
degree of 1 – 1 = 0


Polynomial Dividing

Definition
Long Division
Method)
Using this method, we get the quotient of
H(x) = x2 + 2x – 10, and the remainder,
S(x) = 10.
1 2


Polynomial Dividing

Definition

2.

Long Division

Horner method is a short way to solve
the division of polynomial. This method
can solve several forms of division of
polynomial.
a. Dividing polynomial by (x – k)
The principle of horner method is derived
from the konsep of long division.
P(x) = ax2 + bx + c

Method)

Synthetic division (Horner Method)

1 2 3 4 5 6 7


Polynomial Dividing

Definition
Long Division
Method)
the qoutient is H(x) = ax + (b + ak) with
S(x) = ak2 + bk + c

1 2 3 4 5 6 7


Polynomial Dividing

Definition
Long Division
Method)

b. Dividing a polynomial by (ax – b)
The division of a polynomial by (ax – b) is
the expansion of the division of
polynomial by (x – k). The general form of
dividing a polynomial by (ax – b) with the
quotient H(x) and the remainder S(x) is:
P(x) = (ax – b) . H(x) + S(x)

1 2 3 4 5 6 7


Polynomial Dividing

Definition
Long Division
Method)

1 2 3 4 5 6 7


Polynomial Dividing

c.
Definition
Long Division
Method)

Dividing polynomial by (ax2 + bx + c)
with a ≠ 0

The division of the polynomial of P(x) by
ax2 + bx + c using the horner method can
be done if the divisor of (ax2 + bx + c) can
be factorized. The general form of this
division is:
P(x) = (ax2 + bx + c) . H(x) + S(x)
With H(x) and S(x) are consecutively the
quotient and the remaider of the
division.
1 2 3 4 5 6 7


Polynomial Dividing

Definition
Long Division
Method)

Suppose (ax2 + bx + c) can be factorized to
be P1 . P2 therefore ax2 + bx + c = P1 . P2 ,
then :
P(x) = (ax2 + bx + c) . H(x) + S(x) = P1 . P2
H(x) + S(x)

1 2 3 4 5 6 7


Polynomial Dividing

Definition
Long Division
Method)

The steps to determine the qoutient and the
remainder of the divison of a polynomial are:
1. Divide the polynomial of P(x) by P1 to get
H0(x) and S1 as the quotient and the
remaider.
2. Divide H0(x) by P2 to get H(x) and S2 as
the quotient and the remainder.
3. The quotient of dividing P(x) by (Q(x) = P1
. P2) is H(x), and the remainder is S(x) = P1
. S2 + S1. Consider if P1 or P2 is in the form
of (ax – b), a ≠ 0, then you need to divide
H0(x) or H(x) by a to get the quotient.
Reconsider the division of a polynomial
by (ax – b) that you have learned.
1 2 3 4 5 6 7


Remainder Theorem
1. Division by (x – k)

Division by (x – k)
Division by (ax – b)
Division by (x – a) (x – b)

Theorem
If polynomial of f(x) with degree of n divided by (x
– k), then the remainder is S = f(k)
Proof:
f(x) = (x – k) H(x) + S
for x – k = 0
x = k → f(k) = (k – k) H(k) + S
f(k) = 0 + S
f(x) = S
then, the remainder of S = f(k), it is proved.
Note:
A polynomial of f(x) is divisible by (x – k) if f(k) = 0


Remainder Theorem


2. Division by (ax – b)
Theorem
If polynomial of f(x) with degree of n divided by
(ax – b), then the remainder is S = f(b/a).


Remainder Theorem

3.


Dividing polynomial of f(x) by (x – a) (x – b) with
the quotient of H(x) and the remainder of S(x), is
written as:
f(x) = (x – a) (x – b) H(x) + S(x)
since the degree of the divisor is 2, the degree of S
is at most 1. Let S = (px + q), then the division can
written as follows.
f(x) = (x – a) (x – b) H(x) + (px + q)



f(x) = (x – k) H(x) + f(k), where H(x) = is the quotient.
I. based on the equation above if f(k)n = 0 then f(x)
= (x – k) H(x). In other words (x – k) is a factor of
f(x).
II. if (x – k) is a factor of f(x) then:
f(x) = (x – k) H(x) for any H(x) such that
f(x) = (k – k) H(x) = 0
from (I) and (II) we have f(k) = 0 if and only if (x – k) is
a factor of f(x).
Therefore, the division of a polynomial by one of its
factor resulted a remainder which is equal to zero.
1

2

3

4



There are several aspects that must be considered
when factorizing a polynomial.
1. If the sum of all coefficients is aqual to zero,
then (x – 1) is a factor of the polynomial
2. If the sum of all coeficients of variables with an
even exponent is equal to the sum of all
coefficients of variabel with an odd exponent,
then (x + 1) is a factor of the polynomial.
3. If (1) and (2) are not satisfied, then solve as in
example.
1

2

3

4



Example
Determine the linier factor of f(x) = x3 – 7x2 + 8x
+ 16
Answer :
consider that the sum of all coefficients of
variables with an even exponen is equal to the
sum of all coefficients of variables with an odd
exponent i.e. (-7 + 16 = 1 + 8). Thus (x + 1) is
one of the factors. Next, we use Horner
method to determine the other factors.
1

2

3

4



f(x) = (x + 1) (x2 – 8x + 16)
= (x + 1) (x – 4) (x – 4)
Therefore, the linier factors of f(x) = x3 – 7x2 +
8x + 16 is (x + 1) and (x – 4)
1

2

3

4


Roots of a Polinomial Equation

Definition of Roots of a
Polynomial Equation
Rational Roots of a
Polynomial

1. Definition of Roots of a Polynomial
Equation
Based on the theorem of factorization
we have proved that (x – k) is a factor
of f(x) if and only if f(k) = 0. Similarly,
we can define the root of a polynomial
equation.
Theorem
(x = k) is the root of a polynomial
equation of f(x) = 0 if and only if f(k) = 0


Roots of a Polinomial Equation

Definition of Roots of a
Polynomial Equation
Rational Roots of a
Polynomial

1

2


Exercises 1

1. The polynomial of (4x – 3) (2x2 + 5) (x-4) has degree of ...

a. 2

b. 3

c. 4

d. 5

e. 6

2. The coefficients of x2 in the polynomial (4x -2) (x + 1) (3x + 1) is ...

a. 4

b. 5

c. 6

d. 9

e. 10


Solution

(4x – 3) (2x2 + 5) (x-4)

= (8x3 - 6x2 + 20x - 15) (x – 4)
= 8x4 - 6x3 + 20x2 - 15x – 32x3 + 24x2 – 80x + 60
= 8x4 – 38x3 + 44x2 – 95x + 60

So, this answer is 4 (c)

Back


Solution

(4x -2) (x + 1) (3x + 1)

= (4x2 + 2x – 2) (3x + 1)
= 12x3 + 4x2 + 6x2 + 2x – 6x – 2
= 12x3 + 10x2 -4x – 2

So, this answer is 10 (e)

Back


Good, your answer is
true !!!
Solution


Sorry, your answer is false !!!

Back


Exercises 2

1. The value of polynomial f(x) = 4x3 + 2x2 – 3x + 5 for x = -2 is ...

a. -13

b. -5

c. -10

d. 5

e. 6

2. The value of polynomial g(x) = 9x5 - 5x4 + 7x3 – 8x2 + 6x – 9 for x = -1 is
...
a. -44

b. 43

c. 34

d. -34

e. -33


Solution
Way 1

Way 2

f(x)
f(-2)
So, the value of f(-2) = -13 (a)

= 4x3 + 2x2 – 3x + 5
= 4(-2)3 + 2(-2)2 – 3(-2) + 5
= 4(-8) + 2(4) + 6 + 5
= -32 + 8 + 11
= -13 (a)

Back


Solution

Way 1
So, the value of f(-1) = -44 (a)

g(x)
Way 2

= 9x5 - 5x4 + 7x3 – 8x2 + 6x – 9
= 9(-1)5 – 5(-1)4 + 7(-1)3 – 8(-1)2 + 6(-1) – 9
=-9–5–7–8–6–9
= - 44 (a)

Back


Exercises 3
1. The qoutient and the remainder of division of f(x) = 2x3 + 5x2 – 3x + 6 by (x – 3)
successively are ...
a. 2x2 – 11x + 30 and 96

c. 2x2 + 11x + 30 and 86

b. 2x2 + 11x - 30 and 96

d. 2x2 + 11x + 30 and 96

e. 2x2 + 11x + 30 and 79

2. The qoutient and the remainder if f(x) = x3 – 5x2 + 2 is divisible by (x2 + 4x – 1) are
...
a. x – 9 and 37x – 7

c. x – 9 and 37x + 7

b. x + 9 and 37x - 7

d. x + 9 and 37x + 7

e. x – 9 and 27x - 7


Solution

Using Horner method

The quotient is H(x) = 2x2 + 11x + 30 with S(x) = 96, so the answer is d.

Back


Solution

f(x) has degree of 3, g(x) has degree of 2, and the quotient of H(x) has
degree of 3 – 2 = 1, the remainder of the division S(x) has degree of 2
– 1 = 1.

Using this method, we get the quotient of H(x) = x – 9 and the
remainder, S(x) = 37x – 7 (a)

Back


Exercises 4
1. the polynomial of f(x) divided by (x -1) the remainder is 6, if divided by (x + 1) the
remainder is -4. Determine the remainder if f(x) divided by (x2 – 1) is ...
a. 5x - 1

c. 5x + 1

b. 5x + 2

d. 5x - 2

e. 5x – 3

2. If polynomial of f(x) divided by (x - 1) the remainder is 3, while if it is divided by (x
+ 2) the remainder is 6, determine the remainder if f(x) is divided by (x – 1) (x + 2)
is ...
a. - x + 4

c. x – 4

b. x + 5

d. x – 5

e. x + 6


Solution

f(x) : (x – 1) then remainder = 6 → f(1) = 6
f(x) : (x + 1) then remainder = -4 → f(-1) = -4
f(x) = (x – 1) (x + 1) H(x) + (px+q)

so, the remainder of S(x) = px + q = 5x + 1 (c)

Back


Solution

f(x) : (x – 1) then remainder = 3 → f(1) = 3
f(x) : (x + 2) then remainder = 6 → f(-1) = 6
f(x) = (x – 1) (x + 2) H(x) + (px+q)

so, the remainder of S(x) = px + q = -x + 4 (a)

Back


Exercises 5

1. If (x + 2) is one of the factors of f(x) = -5x5 + ax3 + x2 – 4, then the value of a is ...
a. -4

c. 20

b. -2

d. 4

e. 16

2. The factorization of f(x) = 2x3 + 3x2 – 17x + 12 is ...

a. (x + 1) (x + 4) (2x – 3)

c. (x + 1) (x - 4) (2x – 3)

b. (x - 1) (x + 4) (2x – 3)

d. (x + 1) (x + 4) (2x + 3)

e. (x - 1) (x - 4) (2x + 3)


Solution

using horner method

-8a + 160 = 0
-8a = -160
a = 20
so, the value of a is 20 (c)

Back


Solution

consider that the sum of all coefficients of variables with an even
exponen is equal to the sum of all coefficients of variables with an
odd exponent i.e. (2 + 3 – 17 + 12 = 0). Thus (x - 1) is one of the
factors. Next, we use Horner method to determine the other factors.

f(x) = (x – 1) (2x2 + 5x – 12)
f(x) = (x – 1) (x + 4) (2x – 3) (b)

Back


Exercises 6

1. the integer roots of x3 – 6x2 + 11x – 6 = 0 are ...
a. 1, 2, and -3

c. -1, -2, and 3

b. -1, 2, and 3

d. 1, 2, and 3

e. 1, -2, and -3

2. If f(x) = x3 – 8x2 + ax + 10 = 0 has roots x = 2, the sum of the two other roots is ...

a. -6

c. 7

b. 6

d. 9

e. 10


Solution

consider that the sum of all coefficients of variables with an even exponen is
equal to the sum of all coefficients of variables with an odd exponent i.e. (1 6 + 11 - 6 = 0). Thus (x - 1) is one of the factors. Next, we use Horner method
to determine the other factors.

f(x) = (x – 1) (x2 - 5x + 6)
f(x) = (x – 1) (x -2) (x – 3)
thus, the roots of f(x) = 0 is x = 1, x = 2 and x = 3 (d)

Back


Solution

Back


Quiz 1


Quiz 2
1. Using the remainder theorem, determine the remainder of the following
division.
a. f(x) = (3x2 – 7x + 1) : (x – 2)
b. f(x) = (2x3 – 4x2 + 3x + 7) : (x + 1)
2. determine the value of a and b in the following statements.
a. The polynomial of f(x) = (2x3 + 5x2 + ax + b) if it is divided by (x + 1)
the remainder is 1, whereas if it is divided by (x – 2) the remainder is
43
b. The polynomial of f(x) = (ax3 – 7x + b) if it is divided by (x – 1) and (x
+ 2) successively is 1 and 4.
3. If x2 – 2x – 3 is a factor of f(x) = x4 – 2x3 – 16x2 – px + q, determine the
value of p and q and other factors!
4. If 1 is the root of x3 + ax2 – 2x + 2 = 0, determine the value of a and the
other roots!
5. Given equation of x4 + 4x3 – x2 – 16x – 12 = 0 has roots of a, b, c, and d.
Determine the value of a2 + b2 + c2 + d2.


Thank you very much !!!!

Ani agustina (a1 c011007) polynomial

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (20)

Similar to Ani agustina (a1 c011007) polynomial

Similar to Ani agustina (a1 c011007) polynomial (20)

Recently uploaded

Recently uploaded (20)

Ani agustina (a1 c011007) polynomial