1. Polynomial by Ani Agustina Click here Welcome to Polynomial
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2. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES
First Slide QUIZ
3. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES
First Slide QUIZ
4. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES
First Slide QUIZ
5. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES
First Slide QUIZ
6. Polynomial by Ani Agustina 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
7. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES
EXERCISES 1 EXERCISES 2 EXERCISES 3 EXERCISES 4 EXERCISES 5
EXERCISES 6 First Slide QUIZ
8. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES QUIZ
QUIZ 1 QUIZ 2 First Slide
9. Polynomial by Ani Agustina 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
10. Polynomial by Ani Agustina General Aspects of Polynomial
Definition and Polynomial Component Algebraic Operations of
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
11. Polynomial by Ani Agustina General Aspects of Polynomial
Definition and Polynomial Component 2. Algebraic Operations of
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
12. Polynomial by Ani Agustina General Aspects of Polynomial
Definition and Polynomial Component Algebraic Operations of
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
13. Polynomial by Ani Agustina General Aspects of Polynomial
Definition and Polynomial Component Algebraic Operations of
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
14. Polynomial by Ani Agustina General Aspects of Polynomial
Definition and Polynomial Component Algebraic Operations of
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
15. Polynomial by Ani Agustina Value of a polynomial
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
16. Polynomial by Ani Agustina Value of a polynomial
Determining the value of polynomial by substitution method
Determining the value of a polynomial by Scheme (chart) 2.
Determining the value of a 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
17. Polynomial by Ani Agustina Value of a polynomial Consider
the following chart. Determining the value of polynomial by
substitution method Determining the value of a polynomial by Scheme
(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
18. Polynomial by Ani Agustina 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.
19. Polynomial by Ani Agustina Polynomial Dividing 1.
Definition Long Division Synthetic Division (Horner 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
20. Polynomial by Ani Agustina Polynomial Dividing Definition
Long Division Synthetic Division (Horner Method) Using this method,
we get the quotient of H(x) = x2 + 2x 10, and the remainder, S(x) =
10. 1 2
21. Polynomial by Ani Agustina 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 Synthetic Division (Horner Method) Synthetic division (Horner
Method) 1 2 3 4 5 6 7
22. Polynomial by Ani Agustina Polynomial Dividing Definition
Long Division Synthetic Division (Horner Method) the qoutient is
H(x) = ax + (b + ak) with S(x) = ak2 + bk + c 1 2 3 4 5 6 7
23. Polynomial by Ani Agustina Polynomial Dividing Definition
Long Division Synthetic Division (Horner 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
24. Polynomial by Ani Agustina Polynomial Dividing Definition
Long Division Synthetic Division (Horner Method) 1 2 3 4 5 6 7
25. Polynomial by Ani Agustina Polynomial Dividing c.
Definition Long Division Synthetic Division (Horner 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
26. Polynomial by Ani Agustina Polynomial Dividing Definition
Long Division Synthetic Division (Horner 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
27. Polynomial by Ani Agustina Polynomial Dividing Definition
Long Division Synthetic Division (Horner 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
28. Polynomial by Ani Agustina 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
29. Polynomial by Ani Agustina Remainder Theorem Division by (x
k) Division by (ax b) Division by (x a) (x b) 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).
30. Polynomial by Ani Agustina Remainder Theorem 3. Division by
(x k) Division by (ax b) Division by (x a) (x b) Division by (x a)
(x b) 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)
31. Polynomial by Ani Agustina Factorization Theorem 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
32. Polynomial by Ani Agustina Factorization Theorem 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
33. Polynomial by Ani Agustina Factorization Theorem 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
34. Polynomial by Ani Agustina Factorization Theorem 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
35. Polynomial by Ani Agustina 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
36. Polynomial by Ani Agustina Roots of a Polinomial Equation
Definition of Roots of a Polynomial Equation Rational Roots of a
Polynomial 1 2
37. Polynomial by Ani Agustina Roots of a Polinomial Equation
Definition of Roots of a Polynomial Equation Rational Roots of a
Polynomial 1 2
38. Polynomial by Ani Agustina 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
39. Polynomial by Ani Agustina 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
40. Polynomial by Ani Agustina 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
41. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
42. Polynomial by Ani Agustina Sorry, your answer is false !!!
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43. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
44. Polynomial by Ani Agustina 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
45. Polynomial by Ani Agustina 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
46. Polynomial by Ani Agustina 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 =-957869 = - 44 (a) Back
47. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
48. Polynomial by Ani Agustina Sorry, your answer is false !!!
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49. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
50. Polynomial by Ani Agustina 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
51. Polynomial by Ani Agustina Solution Using Horner method The
quotient is H(x) = 2x2 + 11x + 30 with S(x) = 96, so the answer is
d. Back
52. Polynomial by Ani Agustina 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
53. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
54. Polynomial by Ani Agustina Sorry, your answer is false !!!
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55. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
56. Polynomial by Ani Agustina 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
57. Polynomial by Ani Agustina 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
58. Polynomial by Ani Agustina 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
59. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
60. Polynomial by Ani Agustina Sorry, your answer is false !!!
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61. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
62. Polynomial by Ani Agustina 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)
63. Polynomial by Ani Agustina Solution using horner method -8a
+ 160 = 0 -8a = -160 a = 20 so, the value of a is 20 (c) Back
64. Polynomial by Ani Agustina 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
65. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
66. Polynomial by Ani Agustina Sorry, your answer is false !!!
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67. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
68. Polynomial by Ani Agustina 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
69. Polynomial by Ani Agustina 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
70. Polynomial by Ani Agustina Solution Back
71. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
72. Polynomial by Ani Agustina Sorry, your answer is false !!!
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73. Polynomial by Ani Agustina Good, your answer is true !!!
Solution
74. Polynomial by Ani Agustina Quiz 1
75. Polynomial by Ani Agustina 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.
76. Polynomial by Ani Agustina Thank you very much !!!!