Download Division of Polynomials

Dividing Polynomials
First a quick look at dividing monomials:
Example 1: (a)
x
5

x 2  x 3  x 3  x3
x 2  1 1
2
x
Law of exponents;
Identity property
of multiplication
(b)
Cancellation
property
 
 
2
4
6
6
3x 6 x
18 x
6x
4


 6x
2
2
1
3x
3x 1
Example 1 demonstrates a general rule:
If m and n are nonnegative integers, and m > n, then
a
m
a
n a
mn
if a  0.
Dividing two polynomials
To divide two polynomials, we first must write each polynomial in standard form. The process then
follows a pattern similar to that of Example 4. The next example illustrates the procedure.
Example 2: Find the quotient and the remainder when
3x3 + 4x2 + x + 7
is divided by
x2 + 1
Solution: Each of the given polynomials is in standard form. The dividend is 3x3 + 4x2 + x + 7, and the
divisor is x2 + 1.
STEP 1: Divide the leading term of the dividend, 3x3, by the leading term of the divisor,
x2. Enter the result, 3x, over the term 3x3, as shown below.
3x
x2 + 1)3x3 + 4x2 + x + 7
STEP 2: Multiply 3x by x2 + 1 and enter the result below the dividend.
3x
x2 + 1)3x3 + 4x2 + x + 7
3x3
+ 3x
Notice that the 3x term is aligned under the x to make the
next step easier.
STEP 3: Subtract and bring down the remaining terms.
3x
x2 + 1)3x3 + 4x2 + x + 7
3x3
+ 3x
4x2 - 2x + 7
Subtract.
Bring down the 4x2 and the 7.
10
STEP 4: Repeat STEPS 1 – 3 using 4x2 - 2x + 7 as the dividend.
3x + 4
x2 + 1)3x3 + 4x2 + x + 7
3x3
+ 3x
4x2 - 2x + 7
Divide 4x2 by x2 to get 4.
Multiply x2 + 1 by 4; subtract.
Since x2 does not divide -2x evenly (that is, the result is not a monomial), the process ends. The
quotient is 3x + 4, and the re is -2x + 3.
Check: (Quotient)(Divisor) + Remainder = (3x + 4)(x2 + 1) + (-2x + 3)
= 3x3 + 4x2 + 3x + 4 + (- 2x + 3)
= 3x3 + 4x2 + x + 7 = Dividend
Thus,
3
2
3x  4 x  x  7
 2x  3
 3x  4 
2
2
x 1
x 1
The next example combines the steps involved in long division.
Example 2: Find the quotient and the remainder when
x4 - 3x3 + 2x - 5 is divided by x2 - x + 1
Solution: In setting up this division problem, it is necessary to leave a space for the
missing x2 term in the dividend.
Divisor
2
x -x+1
Subtract
Subtract
Subtract
)
x2 - 2x – 3
Quotient
x4 - 3x3
+ 2x -5
x4 - x3 + x2
- 2x3 - x2 + 2x -5
- 2x3 + 2x2 - 2x
- 3x2 + 4x - 5
- 3x2 + 3x - 3
x-2
Dividend
Remainder
Check: (Quotient)(Divisor) + Remainder
= (x2 - 2x - 3)(x2 – x + 1) + x - 2
= x4 - x3 + x2 - 2x3 + 2x2 - 2x - 3x2 + 3x - 3 + x - 2 = x4 - 3x3 + 2x - 5 = Dividend
Thus,
4
3
x  3x  2 x  5
x2
2
 x  2x  3 
2
2
x  x 1
x  x 1
11
Example 7: Find the quotient and remainder when
2x3 + 3x2 - 2x - 3
is divided by
2x + 3
x2 - 1
2x + 3 ) 2x3 + 3x2 - 2x - 3
2x3 + 3x2
Solution:
-2x - 3
-2x - 3
0
Check: (Quotient)(Divisor) + Remainder
= (x2 - 1)(2x + 3) + 0
= 2x3 + 3x2 - 2x - 3 = Dividend
Thus,
3
2
2 x  3x  2 x  3
2x  3
x
2
1
0
2x  3
x
2
1
The process for dividing two polynomials leads to the following result:
Theorem: The remainder after dividing two polynomials is either the zero polynomial or a polynomial of degree
less than the degree of the divisor.
Practice Exercise 2:
Find the quotient and remainder when:
1.
2.
3.
Answers to Practice Exercises:
1.
2.
3.
x3 + 3x2 – 7x + 4 is divided by x + 2
x4 – 81 is divided by x2 + 9
2x3 – 2x2 + 7x is divided by 2x – 2
Quotient: x2 + x – 9;
Quotient: x2 – 9;
remainder: 22
remainder: 0
2 7
Quotient: x 
2
remainder: 7
12
Exercise 2.6
In Problems 21-50, find the quotient and the remainder. Check your work by verifying that
(Quotient)(Divisor) + Remainder = Dividend
21.
23.
4x3 – 3x2 + X + 1
4x3 – 3x2 + X + 1
x
divided by
divided by x + 2
22.
24.
25.
4x3 – 3x2 +
divided by x - 4
26.
3x3 – x2 +
27.
4x3
28.
3x3 – x2 +
29.
31.
33.
35.
+1
–
+X+1
4x3 – 3x2 + X + 1
4x3 – 3x2 + X + 1
4x3 – 3x2 + X + 1
4x3 – 3x2 + X + 1
3x2
X
divided by
x2
divided by x2 + 2
divided by x3 – 1
divided by x2 + X + 1
divided by x2 – 3x – 4
37. x4 – 1 divided by x – 1
38. x4 – 1 divided by x + 1
39. x4 – 1 divided by x2 – 1
40. x4 – 1 divided by x2 + 1
41. -4x3 + x2 – 4 divided by x – 1
42. -3x4 – 2x – 1 divided by x – 1
43. 1 – x2 + x4 divided by x2 + x + 1
3x3 – x2 + X – 2
3x3 – x2 + X – 2
divided by x
divided by x
– 2 divided by x – 4
– 2 divided by x2
30.
3x3 – x2 + X – 2 divided by x2 + 2
32.
3x3 – x2 + X – 2 divided by x3 – 1
34. 3x3 – x2 + x – 1
divided by x2 + X + 1
3
2
36. 3x – x + X – 2 divided by x2 – 3x – 4
44. 1 – x2 + x4 divided by x2 – x + 1
45. 1 – x2 + x4 divided by 1 – x2
46. 1 – x2 + x4 divided by 1 + x2
47. x3 – a3 divided by x – a
48. x3 + a3 divided by x + a
49. x4 – a4 divided by x – a
50. x5 – a5 divided by x – a
X
X
13
Reducing Rational Expressions to Lowest Terms
If we form the quotient of two polynomials, the result is called a rational expression. Some examples of rational
expressions are:
2
a)
x 1
2
b)
3x  x  2
x
x
c)
2
x 5
x
2
1
d)
xy
 x  y 2
Expressions (a), (b), and (c) are rational expressions in one variable, x, whereas (d) is a rational expression in two
variables, x and y.
Rational expressions are described in the same manner as fractions.
Thus, in expression (a), the polynomial x3 + 1 is called the numerator, and x is called the denominator. When
the numerator and denominator of a rational expression contain no common factors (except 1 and -1), we say the
rational expression is reduced to lowest terms, or simplified.
A rational expression is reduced to lowest terms by completely factoring the numerator and the denominator and
canceling any common factors by using the cancellation property,
ac a
 if b . 0 and c  0
bc b
Example 1: Reduce to lowest terms:
2
x  4x  4
2
x  3x  2
We begin by factoring the numerator and the denominator:
x2 + 4x + 4 = (x + 2)(x + 2)
x2 + 3x + 2 = (x + 2)(x + 1)
Since a common factor, x + 2, appears, the original expression is not in lowest terms. To reduce it to lowest terms,
we use the cancellation property.
(x  2)(x  2)
(x  2)(x  1)

x2
x 1
Warning: Apply the cancellation property only to rational expressions written in factored form. Be sure to cancel
only the common factors.
3
Example 2: Reduce to lowest terms:
x  5x
2
x x
Solution: We proceed directly to rewriting the rational expression so that both the numerator and
denominator are factored completely:
x3 + 5x2 = x2(x + 5) = x(x + 5)
x2 + x
x(x + 1)
x+1
14
Once a rational expression has been reduced to lowest terms, it may be left in factored form or multiplied
out. Thus, the simplified form of the solution to Example 2 may be written as:
x(x + 5)
x+1
x2 + 5x
x+1
or as
Later, we will see that leaving a rational expression in factored form is generally preferable.
4
x  8x
2
x  2x
Example 3: Reduce to lowest terms:
Solution:
x
x
4
2
 8x
 2x


x x3  8

x x  2 


x x  2  x 2  2 x  4
x x  2 

Monomial
Difference of
Factors
two cubes
2
x  2x  4
2
=
 x  2x  4
1
If the leading coefficient of a polynomial in standard form is negative, it is usually easier to rewrit e the
polynomial as the product of -1 and its additive inverse before factoring. For example,
- x2 – x + 6 = -1 (x2 + x – 6) = (-l)(x + 3)(x - 2)
When factoring out -1, remember to change the sign of each term in the polynomial when writing its additive
inverse.
The next example illustrates this procedure as it applies to reducing rational expressions.
Example 4: Reduce to lowest terms:
Solution:
6 -– x – x2
x2 – x – 12 =
2
6xx
2
x  x  12
x2 -– x + 6
x – x – 12
=
-–
2
Factor out –1
Write numerator
In standard form
=
(-1)(x2 + x – 6)
x2 – x –12
 1x  3x  2   1x  2
x  4x  3
x4
Factor
Cancel
The solution to Example 4 may be written in any of the following equivalent ways:
 1x  2 or
x4
x2
2 x
x2
or
or 
x4
x4
x4
15
Evaluating Rational Expressions
To evaluate a rational expression means to evaluate the polynomial in the numerator and the polynomial in the
denominator. However, the polynomial in the denominator of a rational expression cannot have a value equal to 0,
since division by 0 is not defined. In other words, the domain of the variable of a rational expression must exclude
any values that cause the polynomial in the denominator to have a value equal to 0.
Example 5 Evaluate the rational expression below for the given values of x.
x2  2x 1
x2  9
(a) x = 2
(b) x = -1
(c) x = 0
(d) x = 1
Solution:
(a) We substitute 2 for x in the rational expression to obtain
x 2  2 x  1 2  22   1 4  4  1 1
1




2
2
x 9
49
5
5
2  9
2
(b) We substitute -1 for x in the rational expression to obtain
x 2  2 x  1  1  2 1  1 1  2  1 4
1




2
1

9

8
2
x2  9
 1  9
2
(c) We substitute 0 for x in the rational expression to obtain
x 2  2 x  1 0  20  1 1
1



2

9
9
x2  9
0  9
2
(d) We substitute 1 for x in the rational expression to obtain
x 2  2 x  1 1  21  1 1  2  1 0



0
1 9
8
x2  9
12  9
2
The domain of the variable x in the ration expression given in Example 5 is any real number except 3 and 3, since these values would cause the denominator, x2 – 9, to equal 0.
Exercise 3.2
In problems 1 – 20, reduce each rational expression to lowest terms.
2
2
2x  4
3x
x x
1.
2.
3.
4.
2
x
9 x  18
x x
2
2
2
x  3x  2
x 9
5. x  4 x  4
6.
7.
8.
2
2
2
x

4
x

6
x

9
x  6x  8
9.
13.
17.
2
x  x2
3
2
x  4x  4x
2
x  2x
3x  6
2
y  25
2 y  10
10.
14.
2
x  6x  9
2
3x  9 x
2
15 x  24 x
3x
18.
2
3y  2
2
3y  5y  2
11.
5 x  10
2
12.
x 4
15.
24 x
12 x
19.
2
16.
 6x
2
x  4x  5
x 1
20.
3
x x
3
2
x x
2
x  4x  4
2
x  2x
4x  8
12 x  24
3 x  12
2
x  16
2
xx
2
x x2
16
In problems #’s 21 – 24 evaluate each rational expression for the given value of the variable.
2
x 1
6x
21.
for x = 3
22.
for x = 2
2
3x
x 1
23.
2
x  4x  4
for x = -4
2
x  25
24.
x
2
x
 6x  9
2
for x = -5
 16
Multiplication and Division of Rational Expressions
The rules for multiplying and dividing rational expressions are the same as the rules for multiplying and dividing
fractions.
If
a
c
are two rational expressions, their product is given by the rule:
and
b
d
Equation (1)
a c ac


if b  0 and a  0
b d db
In using equation (1), be sure to factor each polynomial completely so that common factors can be cancelled.
We will follow the practice of leaving our answers in factored form.
Example 1:
x 1

6x
3
x  12 x 3
2x

2
2
x  1 6x x  1


Rule for multiplication

x  12 x  x  x  x 2
2  3  x x  1x  1 3 x  1
Factor
Example 2:



2
2
4 x  8 x  1 4 x  8 x  1


2
2
x  x 3x  6
x  x 3 x  6 

Rule for multiplication

4 x  2  x  1 x  1
x x  13 x  2 

4 x  2  x  1
3 x x  2 
Factor
Notice in Example 2 that we wrote the factors of the answer so that the numerical factors appear first. In this
way, the factor 3 in the denominator, for example, will not be mistaken for an exponent.
2
2
2
2
x  2x  1 4x  4
x  2x  1 4x  4
Example 3:


3
2
3
2
x x
x x2
x x x x2






2

x  1x  14x  1 4x  1


2
x x  1x  2 x  1 xx  2 
17
Example 4:
2
6x


2
6x
6x x 2  1
x2  1
x 1



2
x  1 3 x2
3x
x  1 3x 2
2
x 1
Rule for division
Rule for multiplication
2  3  xx  1x  1

3  x  xx  1
x3
2
x3
x 4

2
2
x  x  12
x 4
3
x 8
Example 5:

x

2x  1
x
x38

x 2  x  12

x  2 x 2  2 x  4
x  4x  3
 2 x  2
x3




x  3x  2 x 2  2 x  4  x 2  2 x  4
x  2x  2x  4x  3 x  2x  4

Exercise 3.3
In problems 1 – 21, perform the indicated operation and simplify the result. Leave your answer in factored form.
2
3 x 12
2
2
5x
9
8x
x 1

1.
2.
3.


4
x
18
2x
x 1
2x
2
2
x6
2x
4. x  4  9 x
5. x  5  x
6.

2 3 x  18
2
6x
2 x  10
x2
6x
3x
7.
10.
4x  8
3x  6
2x  8

3
2x  4
3x

6x
8.
x
2
2
11.
16.
18.
5x
x
x
2
2

x
x
2
2
2
x

x
17.
x
x6
2
2
2
 25
 2 x  15
19
2x  4
2x  2
1
4x  8
15.
 3x
4x  2
x
x
x
2
 3 x  10
2
 2 x  35
2
 x  12
2
 x  12


x
x
x
x
x

9 x  15
12.
4
x
x

x

3x  9
9.
10 x
12
14.
4 x  18
 4x  5
3x  6
4x  1

5x
2
x6
8x  2
 16
2
2
13. 4 x  1  x  4 x
2
2x  1
x  16
6 x  27
5
2
4
6x


1 x
2
6 x  10
12
12  6 x
2
 4 x  21
2
 9 x  14
2
 7 x  12
2
 7 x  12
18
20.
x
2
x
 7x  6
2
x

x6
x
2
2
 5x  6
21.
 5x  5
1 x
2
1 x
2

3
x x
3
x x
Addition and Subtraction of Rational Expressions
The rules for adding and subtracting rational expressions are the as the rules for adding and subtracting fractions.
Thus, if the denominators of two rational expressions to be added (or subtracted) are equal, we add (or subtract) the
numerators and keep the common denominator. That is, if a/b and c/b are two rational expressions, then
a
Equation (1):
c

b

ac
a
b
b
b

c

ac
b
if b  0
b
In this section, we will again follow the practice of leaving our answers in factored form.
x
5
x5


2
2
2
x 1 x 1 x 1
Example 1:
2x
Example 2:
2
4
2x  5

2x
2

x3
2x  5

2 x 2  1  x  3
2x  5
 x 1
2x  5
2x
2
2x  2
2 x  1
2




2
2
2



x

1
x

1
x
1
x 1 x 1 x 1
Example 3:
Example 4:
2x
Find :
x3

5
3 x
Solution: Notice that the denominators of the two rational expressions to be added are different. However, the
denominator of the second expression is just the additive inverse of the denominator of the first. That is
3 – x = - x + 3 = -1(x – 3) = - (x – 3)
2x
Thus
x3
Example 5:

5
3 x
x
x3


2x
x3
3x  2
x3


5
  x  3

x  3 x  2 
x3

2x
x3

5
x3

2 x   5
x3

2x  5
x3
x  3 x  2  2 x  2  2 x  1


x3
x3
x3
Notice in Example 5 that we subtracted the quantity (3x + 2) from the x in the first step. When subtracting rational
expressions, be careful to place parentheses around the numerator of the faction being subtracted to ensure that the
entire numerator is subtracted.
19
If the denominators of two rational expressions to be added or subtracted are not equal, we can use the following
general rules for adding and subtracting quotients:
Equation (2):
a

b
Equation (3):
a
b
c

d

c
if b  0 and d  0
bd

ad  bc
d
x3
Example 6:
ad  bc
x4
if b  0 and d  0
bd

x
x2

x  3x  2  x  4 x 
x  4x  2
x 2  5x  6  x 2  4x
2x 2  x  6

x  4x  2
x  4x  2 
Least common Multiple
If the denominators of two rational expressions to be added (or subtracted) have common factors, we usually do not
use the general rules given by equations (2) and (3), since, in doing so, we make the problem more complicated than
it needs to be. Instead, just as with fractions, we apply the least common multiple (LCM) method by using the
polynomial of least degree that contains each denominator polynomial as a factor. Then we rewrite each rational
expression using the LCM as the common denominator and use equation (1) to perform the addition (or subtraction).
To find the least common multiple of two or more polynomials, first factor completely each polynomial. The LCM
is the product of the different prime factors of each polynomial, each factor appearing the greatest number of times
it occurs in each polynomial. The next two examples will give you the idea.
Example 7: Find the least common multiple of the following pair of polynomials:
2
xx  1 x  1
and
3
4x  1x  1
Solution: The polynomials are already factored completely as
x(x - 1)2(x + 1)
and
4(x - 1)(x + 1)3
Start by writing the factors of the left-hand polynomial. (Alternatively, you could start with the one on the right.)
x(x - 1)2(x + 1)
Now look at the right-hand polynomial. Its first factor, 4, does not appear in our list, so we insert it:
4 x(x - 1)2(x + 1)
The next factor, x-1, is already in our list, so no change is necessary. The final factor is (x + 1) 3. Since our list has x
+ 1 to the first power only, we replace x + 1 in the list by (x + 1) 3. The LCM is
4 x(x - 1)2(x + 1)3
Notice that the LCM is, in fact, the polynomial of least degree that contains x(x - 1)2(x + 1) and 4(x - 1)(x + 1)3 as
factors.
Example 8: Find the least common multiple of the following pair of polynomials:
2x2 - 2x - 12
and
x3 - 3x2
20
Solution: First, we factor completely each polynomial:
2x2 - 2x - 12 = 2(x2 - x - 6) = 2(x- 3)(x + 2)
x3 - 3x2 = x2(x - 3)
Now we write the factors of the first polynomial:
2(x - 3)(x + 2)
Looking at the factors that appear in the second polynomial, we see that the first factor, x 2, does not appear in our
list, so we insert it:
2x2(x - 3)(x + 2)
The remaining factor, x - 3, is already on our list.
Thus, the LCM of 2x2 - 2x – 12 and x3 - 3x2 is 2x2(x - 3)(x + 2).
Exercise 3.4
In problems 1 – 24, perform the indicated operation(s) and simplify the result. Leave your answer in factored form.
1.
x

2
4.
3
2.
x
x5
x

3

13.
16.
19.
22.
5.
8.
x
x 1
x
x 1
2x
x 1
x


3x  1
x
1
11.
2x  3
14.
x 1
2x  1
2

17.
x 1
3
x  1
x


2
x 1
20.

x
x 1

23.
x 1
3x
x4

x
2
x 4
3x
1
x
6.
9.
x
2

4
2x
x 1
4
2x  1


1
x2
2x
x3
1

12.
15.
18.
x
x
x 1
x2
3
x 1
2
x5
x2
x2
x 1
3


1
x 1
21.
24.
3
x
x 1
1
x

4
x
x
x
2
x 1
1
x3
4
3
4
x
3.
x 1
4
x

x
x
x 1
6

x3
4
3
10.
2
3
x
7.
x

1
2





4
x2
3
x5
x2
x2
1
2
x 1
x2
x 1
1
x 1


x 1
x2
1
x2
21
In problems 25 – 32, find the LCM of the given polynomial.
25.
x2 – 4;
x2 – x – 2
28.
3x2 – 27;
30.
x– 3;
32.
x2 + 4x + 4; x3 + 2x2; (x + 2)3
26.
x2 – x – 12;
x2 – 8x + 16
2x2 – x – 15
29.
4x3 – 4x2 + x;
x2 + 3x; x3 – 9x
31.
x3 – x;
27.
x3 – x;
x2 – x
2x3 – x2; x3
x3 – 2x2 + x;
x3 – 1
22