Chapter 8
Polynomials and Partial Fractions
Vocabulary
Polynomial
Term
Expression
Constant
Variable
Degree of polynomial
First degree
Linear
‫تعبير متعدد الحدود األسية‬
‫حد‬
‫تعبير‬
‫ثابت‬
‫متغير‬
‫درجة التعبير المتعدد الحدود األسية‬
‫من الدرجة األولى‬
‫خطي \ من الدرجة األولى‬
Second degree
Quadratic
Third degree
Long division
Remainder
Partial fractions
Resolve
Decompose
‫تربيعي \من الدرجة الثانية‬
‫تربيعي \من الدرجة الثانية‬
‫من الدرجة الثالثة‬
‫القسمة المطولة‬
‫باقي‬
‫الكسور الجزئية‬
‫يحلل \ يجزئ‬
‫يحلل \ يجزئ‬
8.1 Polynomials
A polynomial is an algebraic expression in which all terms have variables that are raised to whole
number powers. Polynomial terms cannot contain variables which are raised to fractional powers or
terms which contains variables in the denominator. All the following expressions are polynomials:
5x2  3x  4
6x2 y3  4x3 y  xy  8
 5z4
12
xy2  4z
While the following expressions are not polynomials since they do not comply with the rules of
polynomials definition:
3x4  2
3x
3x2 y2  4xy  x  8
4x4  x2 3  3
The degree of a polynomial is equal to the degree of the term having the highest degree. For example,
the degree of the first term of 2x2 y4  3x3 y  x2 y  5 is 6 which is higher than the degree of each of the
other three terms and hence the degree of this polynomial is 6.
Example 1. Determine whether each of the following expressions is a polynomial or not:
(b) zx2  8x  7xy4
(a) 3x2  11x  4
and (f) 2y  3x
2
45
(c) 2y  y  8x  2
(d) 16
(e) x2  4  2xy2
z
.
Chapter 8: Polynomials and Partial Fractions
1
Solution
(a)
3x2  11x  4
is a polynomial
(b)
zx2  8x  7 xy 4
is a polynomial
(c)
2y  y  8x  2
is not a polynomial
(d)
16
is a polynomial
(e)
x2  4  2xy2
z
is not a polynomial
(f)
2y2  3x4 5
is not a polynomial
Example 2. Determine the degree of each of the following polynomials:
(a) 6 x 2 z 2  7 x  8 z 2
(b) x5  3x 2  5x 2 y 2
(c) 6 y
(d) 8
and (e) 6 xyz .
Solution
(a)
6 x2 z 2  7 x  8z 2
is a fourth degree polynomial
(b)
x5  3x 2  5 x 2 y 2
is a fifth degree polynomial
(c)
6y
is a first degree polynomial
(d)
8
is a zero degree polynomial
(e)
6 xyz
is a third degree polynomial
8.2 Division of Polynomials
In division of polynomials, long division is used in the same way it is used in the division of numbers.
The result of division of polynomials may or may not contain a remainder and as illustrated in the
following examples.
Example 3. Divide (3x2  7x  4) by (x  1) .
Solution
Using long division:
3x  4
x  1 3x 2  7 x  4
3x 2  3x
 4x  4
 4x  4
0
Thus (3x 2  7 x  4)  ( x  1)  3x  4
2
Chapter 8: Polynomials and Partial Fractions
Example 4. Determine (4x3  4x2  11x  9)  (2x  1) .
Solution
Using long division:
2x2 + 3x  4
2x  1 4x3  4x2  11x  9
4x3  2x2
6x2  11x  9
6x2  3x
 8x  9
 8x  4
 13
Thus (4x3  4x2  11x  9)  (2x  1)  2x2  3x  4 remainder  13 ,
(4x3  4x2  11x  9)  (2x  1)  2x2  3x  4  13
2x  1
or
5
3
Example 5. Find: x  2x  23x  18
x2
Solution
Using long division:
x4  2x3  2x2  4x  31
x  2 x5  0x4  2x3  0x2  23x  18
x5  2x4
 2x4  2x3  0x2  23x  18
 2x4  4x3
2x3  0x2  23x  18
2x3  4x2
 4x2  23x  18
 4x2  8x
31x  18
31x  62
 80
Hence
(x5  2x3  23x  18)  (x  2)  x4  2x3  2x2  4x  31 remainder  80 ,
or
(x5  2x3  23x  18)  (x  2)  x4  2x3  2x2  4x  31  80
x2
Chapter 8: Polynomials and Partial Fractions
3
8.3 Partial Fractions
5 and 3 is carried out as follows:
x2
x 3
The addition of the two rational expressions
5  3  5(x  3)  3(x  2)
x  2 x 3
(x  2)(x  3)
 5x  15  3x  6
x2  x  6

8x  9
x  x6
2
8x  9 into
x  x6
The process of converting
5  3 is called resolving or decomposing into
x  2 x 3
2
partial fractions
In general, there are three types of partial fractions based on the form of the denominator of the
rational expression which is to be resolved into partial fractions. These types are:
(i)
Denominator contains linear factors:
polynomial
 A  B  C
( x  a)( x  b)( x  c) ( x  a) ( x  b) ( x  c)
(ii) Denominator contains repeated linear factors:
polynomial
( x  a) 3
A


( x  a)
B
( x  a) 2

C
( x  a) 3
(iii) Denominator contains quadratic factors:
polynomial
(ax 2  bx  c)( x  d )
Example 6. resolve

Ax  B
(ax 2  bx  c)
5x  12
x  5x  6
2

C
(x  d )
into partial fractions.
Solution
5x  12
x  5x  6
2

5x  12
(x  2)(x  3)
which belongs to the first type of partial fractions since the
denominator contains two linear factors.
5x  12  A  B  A(x  3)  B(x - 2)
(x  2)(x  3) (x  2) (x  3)
(x  2)(x  3)
Thus 5x  12  A(x  3)  B(x  2)
Let x  3 . Then
4
Chapter 8: Polynomials and Partial Fractions
5  3  12  A(3  3)  B(3  2)
15  12  A(0)  B(1)
3 0B
B3
Let x  2 . Then
5  2  12  A(2  3)  B(2  2)
10  12  A(1)  B(0)
 2  A  0
A2
Hence
5x  12
(x  5x  6)

2
2  3
(x  2) (x  3)
Example 7. Decompose 7x  16 into partial fractions.
(x  3)2
Solution
The rational expression 7x  16 belongs to the second type of partial fractions since the denominator
(x  3)2
contains two repeated linear factors.
7x  16 
A
B

2
(x  3)
(x  3)2
(x  3)

A(x  3)  B
(x  3)2
 Ax  3A  B
(x  3)2
7x  16  Ax  3A  B
7x  Ax
A7
 16  3A  B
 16  3  7  B
 16  21  B
 16  21  B
B5
Thus 7x  16 
(x  3)
2
7
(x  3)

5
(x  3)2
Chapter 8: Polynomials and Partial Fractions
5
Example 8. Resolve
4x2  x  7 into partial fractions.
(x  1)(x2  3)
Solution
Since there is a quadratic factor in the denominator of the rational expression
belongs to the third type of partial fractions.
4x2  x  7  Ax  B  C
(x  1)(x2  3) (x2  3) (x  1)

(Ax  B)(x  1)  C(x2  3)
(x  1)(x2  3)
4x2  x  7  (Ax  B)(x  1)  C(x2  3)
Let x  1 . Then
4  12  1  7  (A  1  B)(1  1)  C(12  3)
12  4C
C3
4x2  x  7  (Ax  B)(x  1)  C(x2  3)
4x2  x  7  (Ax2  Ax  Bx  B)  C(x2  3)
4x2  x  7  Ax2  Ax  Bx  B  Cx2  3C
4x2  Ax2  Cx2
4x2  (A  C)x2
4AC
4A3
A 1
Also
x  Ax  Bx
x  1  x  Bx
x  x  Bx
x  x  Bx
2x  Bx
B2
Hence
6
4x2  x  7  x  2 
3
2
2
(x  1)(x  3) (x  3) (x  1)
Chapter 8: Polynomials and Partial Fractions
4x2  x  7
then it
(x  1)(x2  3)