16
4b
MODULE 4. INTEGER EXPONENTS AND POLYNOMIALS
Polynomial Arithmetic
Adding and Subtracting Polynomials
In this section we concentrate on adding and subtracting polynomial expressions. Let’s begin with an addition example.
You Try It!
EXAMPLE 1. Simplify:
(a2 + 3ab − b2 ) + (4a2 + 11ab − 9b2 )
Solution: Use the commutative and associative properties to change the order
and regroup. Then combine like terms.
(a2 + 3ab − b2 ) + (4a2 + 11ab − 9b2 )
= (a2 + 4a2 ) + (3ab + 11ab) + (−b2 − 9b2 )
= 5a2 + 14ab − 10b2
!
You Try It!
EXAMPLE 2. Simplify:
(x3 − 2x2 y + 3xy 2 + y 3 ) + (2x3 − 4x2 y − 8xy 2 + 5y 3 )
Solution: This time we will combine all like terms mentally, simply writing
down the answer.
(x3 − 2x2 y + 3xy 2 + y 3 ) + (2x3 − 4x2 y − 8xy 2 + 5y 3 )
= 3x3 − 6x2 y − 5xy 2 + 6y 3
Here is a list of the mental calculations used to combine like terms above.
• x3 + 2x3 = 3x3
• −2x2 y − 4x2 y = −6x2 y
• 3xy 2 − 8xy 2 = −5xy 2
• y 3 + 5y 3 = 6y 3
!
4B. POLYNOMIAL ARITHMETIC
17
Negating a Polynomial
Before attempting subtraction of polynomials, let’s first address how to negate
or “take the opposite” of a polynomial. First recall that negating is equivalent
to multiplying by −1.
Negating. If a is any number, then
−a = (−1)a.
That is, negating is equivalent to multiplying by −1.
We can use this property to simplify −(a + b). First, negating is identical to
multiplying by −1.
−(a + b) = (−1)(a + b)
Next, use the distributive property to distribute the −1 times each term in the
expression a + b.
= (−1)a + (−1)b
Finally, use again the fact that multiplying by −1 is equivalent to negating.
= −a − b
Thus, −(a + b) = −a − b. However, it is probably simpler to note that the
minus sign in front of the parentheses simply changed the sign of each term
inside the parentheses.
You Try It!
EXAMPLE 3. Simplify: −(−3x2 + 4x − 8)
Solution: First, negating is equivalent to multiplying by −1.
−(−3x2 + 4x − 8) = (−1)(−3x2 + 4x − 8)
Next, distribute the −1, then simplify
= (−1)(−3x2 ) + (−1)(4x) − (−1)(8)
= 3x2 − 4x + 8
Alternate solution: As we saw above, a negative sign in front of a parentheses
simply changes the sign of each term inside the parentheses. So it is much more
efficient to write
−(−3x2 + 4x − 8) = 3x2 − 4x + 8,
changing the sign of each term inside the parentheses.
!
18
MODULE 4. INTEGER EXPONENTS AND POLYNOMIALS
Subtracting Polynomials
Now that we know how to negate a polynomial (change the sign of each term
of the polynomial), we’re ready to subtract polynomials.
You Try It!
EXAMPLE 4. Simplify:
(y 3 − 3y 2 z + 4yz 2 + z 3 ) − (2y 3 − 8y 2 z + 2yz 2 − 8z 3 )
Solution: First, distribute the minus sign, changing the sign of each term of
the second polynomial.
(y 3 − 3y 2 z + 4yz 2 + z 3 ) − (2y 3 − 8y 2 z + 2yz 2 − 8z 3 )
= y 3 − 3y 2 z + 4yz 2 + z 3 − 2y 3 + 8y 2 z − 2yz 2 + 8z 3
Now combine like terms mentally.
= −y 3 + 5y 2 z + 2yz 2 + 9z 3
!
Multiplying Polynomials
In this section we will find the products of polynomial expressions and functions. We start with the product of two monomials, then graduate to the
product of a monomial and polynomial, and complete the study by finding the
product of any two polynomials.
The Product of Monomials
As long as all of the operators are multiplication, we can use the commutative
and associative properties to change the order and regroup.
You Try It!
EXAMPLE 5. Simplify: −5(7y).
Solution: Use the commutative and associative properties to change the order
and regroup.
−5(7y) = [(−5)(7)]y
= −35y
Reorder. Regroup.
Multiply: (−5)(7) = −35.
!
19
4B. POLYNOMIAL ARITHMETIC
You Try It!
EXAMPLE 6. Simplify: (−3x2 )(4x3 ).
Solution: Use the commutative and associative properties to change the order
and regroup.
(−3x2 )(4x3 ) = [(−3)(4)](x2 x3 )
5
Reorder. Regroup.
Multiply: (−3)(4) = 12, x2 x3 = x5 .
= −12x
!
You Try It!
EXAMPLE 7. Simplify: (−2a2 b3 )(−5a3 b).
Solution: Use the commutative and associative properties to change the order
and regroup.
(−2a2 b3 )(−5a3 b) = [(−2)(−5)](a2 a3 )(b3 b)
5 4
= 10a b
Reorder. Regroup.
Multiply: (−2)(−5) = 10,
a2 a3 = a5 , and b3 b = b4 .
!
When multiplying monomials, it is much more efficient to make the required
calculations mentally. In the case of Example 5, multiply −5 and 7 mentally
to get
−5(7y) = −35y.
In the case of Example 6, multiply −3 and 4 to get −12, then repeat the base
x and add exponents to get
(−3x2 )(4x3 ) = −12x5 .
Finally, in the case of Example 7, multiply −2 and −5 to get 10, then repeat
the bases and add their exponents.
(−2a2 b3 )(−5a3 b) = 10a5 b4
Multiplying a Monomial and a Polynomial
Now we turn our attention to the product of a monomial and a polynomial.
You Try It!
EXAMPLE 8. Multiply: 5(3x2 − 4x − 8)
20
MODULE 4. INTEGER EXPONENTS AND POLYNOMIALS
Solution: We need to first distribute the 5 times each term of the polynomial.
Then we multiply the resulting monomials mentally.
5(3x2 − 4x − 8) = 5(3x2 ) − 5(4x) − 5(8)
= 15x2 − 20x − 40
!
You Try It!
EXAMPLE 9. Multiply: 2y(−3y + 5)
Solution: We need to first distribute the 2y times each term of the polynomial.
Then we multiply the resulting monomials mentally.
2y(−3y + 5) = 2y(−3y) + 2y(5)
= −6y 2 + 10y
!
You Try It!
EXAMPLE 10. Multiply: −3ab(a2 + 2ab − b2 )
Solution: We need to first distribute the −3ab times each term of the polynomial. Then we multiply the resulting monomials mentally.
−3ab(a2 + 2ab − b2 ) = −3ab(a2 ) + (−3ab)(2ab) − (−3ab)(b2 )
= −3a3 b + (−6a2 b2 ) − (−3ab3 )
= −3a3 b − 6a2 b2 + 3ab3
Alternate solution: It is far more efficient to perform most of the steps of
the product −3ab(a2 + 2ab − b2 ) mentally. We know we must multiply −3ab
times each of the terms of the polynomial a2 + 2ab − b2 . Here are our mental
calculations:
• −3ab times a2 equals −3a3 b.
• −3ab times 2ab equals −6a2 b2 .
• −3ab times −b2 equals 3ab3 .
Thinking in this manner allows us to write down the answer without any of
the steps shown in our first solution. That is, we immediately write:
−3ab(a2 + 2ab − b2 ) = −3a3 b − 6a2 b2 + 3ab3
4B. POLYNOMIAL ARITHMETIC
21
!
You Try It!
EXAMPLE 11. Multiply: −2z 2(z 3 + 4z 2 − 11)
Solution: We need to first distribute the −2z 2 times each term of the polynomial. Then we multiply the resulting monomials mentally.
−2z 2 (z 3 + 4z 2 − 11) = −2z 2 (z 3 ) + (−2z 2 )(4z 2 ) − (−2z 2 )(11)
= −2z 5 + (−8z 4 ) − (−22z 2)
= −2z 5 − 8z 4 + 22z 2
Alternate solution: It is far more efficient to perform most of the steps of
the product −2z 2(z 3 + 4z 2 − 11) mentally. We know we must multiply −2z 2
times each of the terms of the polynomial z 3 + 4z 2 − 11. Here are our mental
calculations:
• −2z 2 times z 3 equals −2z 5.
• −2z 2 times 4z 2 equals −8z 4 .
• −2z 2 times −11 equals 22z 2.
Thinking in this manner allows us to write down the answer without any of
the steps shown in our first solution. That is, we immediately write:
−2z 2 (z 3 + 4z 2 − 11) = −2z 5 − 8z 4 + 22z 2
!
Multiplying Polynomials
Now that we’ve learned how to take the product of two monomials or the
product of a monomial and a polynomial, we can apply what we’ve learned to
multiply two arbitrary polynomials.
Before we begin with the examples, let’s revisit the distributive property.
We know that
2 · (3 + 4) = 2 · 3 + 2 · 4.
Both sides are equal to 14. We’re used to having the monomial factor on the
left, but it can also appear on the right.
(3 + 4) · 2 = 3 · 2 + 4 · 2
Again, both sides equal 14. So, whether the monomial appears on the left or
right makes no difference; that is, a(b + c) and (b + c)a give the same result.
22
MODULE 4. INTEGER EXPONENTS AND POLYNOMIALS
You Try It!
EXAMPLE 12. Multiply: (2x − 5)(3x + 2)
Solution: Let’s imagine that (2x−5)(3x+2) has the form (b+c)a and multiply
3x + 2 times both terms in 2x − 5.
(2x − 5)(3x + 2) = 2x(3x + 2) − 5(3x + 2)
Now we distribute monomials times polynomials, then combine like terms.
= 6x2 + 4x − 15x − 10
= 6x2 − 11x − 10
Thus, (2x − 5)(3x + 2) = 6x2 − 11x − 10.
!
You Try It!
EXAMPLE 13. Multiply: (x + 5)(x2 + 2x + 7)
Solution: Let’s imagine that (x + 5)(x2 + 2x + 7) has the form (b + c)a and
multiply x2 + 2x + 7 times both terms in x + 5.
(x + 5)(x2 + 2x + 7) = x(x2 + 2x + 7) + 5(x2 + 2x + 7)
Now we distribute monomials times polynomials, then combine like terms.
= x3 + 2x2 + 7x + 5x2 + 10x + 35
= x3 + 7x2 + 17x + 35
Thus, (x + 5)(x2 + 2x + 7) = x3 + 7x2 + 17x + 35.
!
Speeding Things Up a Bit
Let’s rexamine Example 13 with the hope of unearthing a pattern that will
allow multiplication of polynomials to proceed more quickly with less work.
Note the first step of Example 13.
(x + 5)(x2 + 2x + 7) = x(x2 + 2x + 7) + 5(x2 + 2x + 7)
Note that the first product on the right is the result of taking the product
of the first term of x + 5 and x2 + 2x + 7. Similarly, the second product on
4B. POLYNOMIAL ARITHMETIC
23
the right is the result of taking the product of the second term of x + 5 and
x2 + 2x + 7. Next, let’s examine the result of the second step.
(x + 5)(x2 + 2x + 7) = x3 + 2x2 + 7x + 5x2 + 10x + 35
The first three terms on the right are the result of multiplying x times x2 +2x+7.
(x + 5)(x2 + 2x + 7) = x3 + 2x2 + 7x + · · ·
The second set of three terms on the right are the result of multiplying 5 times
x2 + 2x + 7.
(x + 5)(x2 + 2x + 7) = · · · + 5x2 + 10x + 35
These notes suggest an efficient shortcut. To multiply x + 5 times x2 + 2x + 7,
• multiply x times each term of x2 + 2x + 7, then
• multiply 5 times each term of x2 + 2x + 7.
• Combine like terms.
This process would have the following appearance.
(x + 5)(x2 + 2x + 7) = x3 + 2x2 + 7x + 5x2 + 10x + 35
= x3 + 7x2 + 17x + 35
You Try It!
EXAMPLE 14. Use the shortcut technique described above to simplify
(2y + 6)(3y 2 + 4y + 11)
Solution: Multiply 2y times each term of 3y 2 + 4y + 11, then multiply 6 times
each term of 3y 2 + 4y + 11. Finally, combine like terms.
(2y + 6)(3y 2 + 4y + 11) = 6y 3 + 8y 2 + 22y + 18y 2 + 24y + 66
= 6y 3 + 26y 2 + 46y + 66
!
You Try It!
EXAMPLE 15. Use the shortcut technique to simplify (a + b)2 .
Solution: To simplify (a + b)2 , we first must write a + b as a factor two times.
(a + b)2 = (a + b)(a + b)
24
MODULE 4. INTEGER EXPONENTS AND POLYNOMIALS
Next, multiply a times both terms of a + b, multiply b times both terms of
a + b, then combine like terms.
= a2 + ab + ba + b2
= a2 + 2ab + b2
Note that ab = ba because multiplication is commutative, so ab + ba = 2ab.
You Try It!
EXAMPLE 16. Use the shortcut technique to simplify
(x2 + x + 1)(x2 − x − 1).
Solution: This time the first factor contains three terms, so we multiply x2
times each term of x2 − x − 1, then x times each term of x2 − x − 1, and 1
times each term of x2 − x − 1. Then we combine like terms.
(x2 + x + 1)(x2 − x − 1) = x4 − x3 − x2 + x3 − x2 − x + x2 − x − 1
= x4 − x2 − 2x − 1
!
Special Products
This section is dedicated to explaining a number of important shortcuts for
multiplying binomials. These are extremely important patterns as they will
surface repeatedly in future work. It is essential that readers practice until
they become proficient using each of the patterns presented in this section.
The FOIL Method
Consider the product of two binomials (x + 3)(x + 6). We already know how
to find the product of these two binomials; we multiply x times both terms of
x + 6, then we multiply 3 times both terms of x + 6.
(x + 3)(x + 6) = x2 + 6x + 3x + 18
Normally combine like terms, but we halt the process at this point so as to
introduce the pattern called the FOIL method.
The letters in the word FOIL stand for “First,” “Outer,” “Inner,” and
“Last.” Let’s see how we can connect these terms to the product (x + 3)(x + 6).
25
4B. POLYNOMIAL ARITHMETIC
• The arrows indicate the terms in the “First” positions in each binomial.
If you multiply the terms in the “First” position, you get x2 .
(x + 3)(x + 6)
F
• The arrows indicate the terms in the “Outer” positions in each binomial.
If you multiply the terms in the “Outer” positions, you get 6x.
(x + 3)(x + 6)
O
• The arrows indicate the terms in the “Inner” positions in each binomial.
If you multiply the terms in the “Inner” positions, you get 3x.
(x + 3)(x + 6)
I
• The arrows indicate the terms in the “Last” positions in each binomial.
If you multiply the terms in the “Last” positions, you get 18.
(x + 3)(x + 6)
L
The following diagram shows the connection between “First,” “Outer,” “Inner,” “Last” and the answer.
F
(x + 3)(x + 6) = x2
+
O
I
6x + 3x +
L
18
You Try It!
EXAMPLE 17. Use the FOIL method to simplify: (x + 5)(x + 7)
Solution: Multiply the “First” positions: x2 . Multiply the “Outer” positions:
7x. Multiply the “Inner” positions: 5x. Multiply the “Last” positions: 35.
F
(x + 5)(x + 7) = x2
+
O
I
7x + 5x +
L
35
Combining like terms, (x + 5)(x + 7) = x2 + 12x + 35.
!
You Try It!
26
MODULE 4. INTEGER EXPONENTS AND POLYNOMIALS
EXAMPLE 18. Use the FOIL method to simplify: (2x − 7)(x − 4)
Solution: Multiply the “First” positions: 2x2 . Multiply the “Outer” positions: −8x. Multiply the “Inner” positions: −7x. Multiply the “Last” positions: 28.
(2x − 7)(x − 4) =
F
2x2
O
− 8x −
I
7x +
L
28
Combining like terms, (2x − 7)(x − 4) = 2x2 − 15x + 28.
!
At first glance, the FOIL method doesn’t look like much of a shortcut. After
all, if we simply use the distributive property on the product of Example 18,
we get the same quick result.
(2x − 7)(x − 4) = 2x(x − 4) − 7(x − 4)
= 2x2 − 8x − 7x + 28
= 2x2 − 15x + 28
The FOIL method becomes a true shortcut when we add the “Outer” and
“Inner” results in our head.
FOIL Shortcut. To multiply two binomials, follow these steps:
1. Multiply the terms in the “First” positions.
2. Multiply the terms in the “Outer” and “Inner” positions and combine
the results mentally (if they are like terms).
3. Multiply the terms in the “Last” positions.
You Try It!
EXAMPLE 19. Use the FOIL shortcut to simplify: (3x + 8)(2x − 1)
Solution: Each of the following steps is performed mentally.
1. Multiply the terms in the “First” positions: 6x2
2. Multiply the terms in the “Outer” and “Inner” positions and add the
results mentally: −3x + 16x = 13x
3. Multiply the terms in the “Last” positions: −8
4B. POLYNOMIAL ARITHMETIC
27
Write the answer with no intermediate steps: (3x + 8)(2x − 1) = 6x2 + 13x − 8.
!
You Try It!
EXAMPLE 20. Use the FOIL shortcut to simplify: (4y − 3)(5y + 2)
Solution: Each of the following steps is performed mentally.
1. Multiply the terms in the “First” positions: 20y 2
2. Multiply the terms in the “Outer” and “Inner” positions and add the
results mentally: 8y − 15y = −7y
3. Multiply the terms in the “Last” positions: −6
Write the answer with no intermediate steps: (4y − 3)(5y + 2) = 20y 2 − 7y − 6.
!
The Difference of Squares
We can use the FOIL shortcut to multiply (a + b)(a − b).
1. Multiply the terms in the “First” positions: a2
2. Multiply the terms in the “Outer” and “Inner” positions and add the
results mentally: ab − ab = 0
3. Multiply the terms in the “Last” positions: −b2
Thus, (a + b)(a − b) = a2 − b2 . Note how the right-hand side (a2 − b2 ) is the
difference of two squares.
The difference of squares. If you have identical terms in the “First” positions and identical terms in the “Last” positions, but one set is separated
with a plus sign while the other is separated by a minus sign, then proceed as
follows:
1. Square the “First” term.
2. Square the “Last” term.
3. Place a minus sign between the results
That is,
(a + b)(a − b) = a2 − b2
28
MODULE 4. INTEGER EXPONENTS AND POLYNOMIALS
Note: If you don’t have identical terms in the “First” and “Last” positions,
with one set separated with a plus sign and the other with a minus sign, then
you do not have the difference of squares pattern and you must find some other
way to multiply. For example, (x + 3)(x − 3) is an example of the difference of
squares pattern, but (2y + 3)(2y − 5) is not.
You Try It!
EXAMPLE 21. Use the difference of squares shortcut to simplify: (x+3)(x−
3)
Solution: Note how the terms in the “First” position are identical, as are the
terms in the “Last” position, with one set separated by a plus sign and the
other with a minus sign. Hence, this is the difference of squares pattern and
we proceed as follows:
1. Square the term in the “First” position: x2
2. Square the term in the “Last” position: (−3)2 = 9
3. Separate the squares with a minus sign.
That is:
(x + 3)(x − 3) = x2 − (3)2
= x2 − 9
Note: You should practice this pattern until you can go straight from the
problem statement to the answer without writing down any intermediate work,
as in (x + 3)(x − 3) = x2 − 9.
!
You Try It!
EXAMPLE 22. Use the difference of squares shortcut to simplify:
(8y + 7z)(8y − 7z)
Solution: Note how the terms in the “First” position are identical, as are the
terms in the “Last” position, with one set separated by a plus sign and the
other with a minus sign. Hence, this is the difference of squares pattern and
we proceed as follows:
1. Square the term in the “First” position: (8y)2 = 64y 2
2. Square the term in the “Last” position: (7z)2 = 49z 2
3. Separate the squares with a minus sign.
29
4B. POLYNOMIAL ARITHMETIC
That is:
(8y + 7z)(8y − 7z) = (8y)2 − (7z)2
= 64y 2 − 49z 2
Note: You should practice this pattern until you can go straight from the
problem statement to the answer without writing down any intermediate work,
as in (8y + 7z)(8y − 7z) = 64y 2 − 49z 2 .
!
You Try It!
EXAMPLE 23. Use the difference of squares shortcut to simplify:
(x3 − 5y 2 )(x3 + 5y 2 )
Solution: Note how the terms in the “First” position are identical, as are the
terms in the “Last” position, with one set separated by a plus sign and the
other with a minus sign. Hence, this is the difference of squares pattern and
we proceed as follows:
1. Square the term in the “First” position: (x3 )2 = x6
2. Square the term in the “Last” position: (5y 2 )2 = 25y 4
3. Separate the squares with a minus sign.
That is:
(x3 − 5y 2 )(x3 + 5y 2 ) = (x3 )2 − (5y 2 )2
= x6 − 25y 4
Note: You should practice this pattern until you can go straight from the
problem statement to the answer without writing down any intermediate work,
as in (x3 − 5y 2 )(x3 + 5y 2 ) = x6 − 25y 4 .
!
Squaring a Binomial
First, (a + b)2 = (a + b)(a + b), then use the FOIL shortcut.
1. Multiply the terms in the “First” positions: a2
2. Multiply the terms in the “Outer” and “Inner” positions and add the
results mentally: ab + ab = 2ab
3. Multiply the terms in the “Last” positions: b2
30
MODULE 4. INTEGER EXPONENTS AND POLYNOMIALS
Hence, the correct answer is (a + b)2 = a2 + 2ab + b2 . This gives us the correct
technique for squaring a binomial.
Squaring a binomial. To square a binomial, such as (a + b)2 , perform the
following steps:
1. Square the “First” term: a2
2. Multiply the “First” and “Last” terms and double the result: 2ab
3. Square the “Last” term: b2
That is:
(a + b)2 = a2 + 2ab + b2
You Try It!
EXAMPLE 24. Use the squaring a binomial shortcut to expand: (x + 5)2
Solution: Follow these steps:
1. Square the first term: x2
2. Multiply the “First” and “Last” terms and double the result: 2(x)(5) =
10x
3. Square the “Last” term: 52 = 25
Thus:
(x + 5)2 = x2 + 2(x)(5) + (5)2
= x2 + 10x + 25
Note: You should practice this pattern until you can go straight from the
problem statement to the answer without writing down any intermediate work,
as in (x + 5)2 = x2 + 10x + 25.
!
You Try It!
EXAMPLE 25. Use the squaring a binomial shortcut to expand: (3x + 7y)2
Solution: Follow these steps:
1. Square the first term: (3x)2 = 9x2
2. Multiply the “First” and “Last” terms and double the result: 2(3x)(7y) =
42xy
4B. POLYNOMIAL ARITHMETIC
31
3. Square the “Last” term: (7y)2 = 49y 2
Thus:
(3x + 7y)2 = (3x)2 + 2(3x)(7y) + (7y)2
= 9x2 + 42xy + 49y 2
Note: Practice this pattern until you can go straight from the problem statement to the answer without writing down any intermediate work, as in (3x +
7y)2 = 9x2 + 42xy + 49y 2 .
!
You Try It!
EXAMPLE 26. Use the squaring a binomial shortcut to expand: (4a2 −5b3 )2
Solution: We can take care of a minus sign by “adding the opposite”: (4a2 −
5b3 )2 = (4a2 + (−5b3 ))2 . Now follow these steps:
1. Square the first term: (4a2 )2 = 16a4
2. Multiply the “First” and “Last” terms and double the result: 2(4a2 )(−5b3 ) =
−40a2b3
3. Square the “Last” term: (−5b3 )2 = 25b6
Thus:
(4a2 − 5b3 )2 = (4a2 + (−5b3 ))2
= (4a2 )2 + 2(4a2 )(−5b3 ) + (−5b3 )2
= 16a4 − 40a2 b3 + 25b6
Note: Practice this pattern until you can go straight from the problem statement to the answer without writing down any intermediate work, as in (4a2 −
5b3 )2 = 16a4 − 40a2 b3 + 25b6 .
!
Example 26 shows us that if we are squaring a difference, the middle term
will be minus. That is, the only difference between (a + b)2 and (a − b)2 is the
sign of the middle term.
Squaring a binomial. Note that
(a + b)2 = a2 + 2ab + b2
and
(a − b)2 = a2 − 2ab + b2 .
32
MODULE 4. INTEGER EXPONENTS AND POLYNOMIALS
Dividing a Polynomial by a Monomial
We know that multiplication is distributive with respect to addition; that is,
a(b + c) = ab + ac. We use this property to perform multiplications such as:
x2 (2x2 − 3x − 8) = 2x4 − 3x3 − 8x2
However, it is also true that division is distributive with respect to addition.
Distributive property for division. If a, b, and c are any numbers, then:
a+b
a b
= +
c
c
c
For example, note that
4+6
4 6
= + .
2
2 2
This form of the distributive property can be used to divide a polynomial by
a monomial.
You Try It!
EXAMPLE 27. Divide x2 − 2x − 3 by x2 .
Solution: We use the distributive property, dividing each term by x2 .
x2
2x
3
x2 − 2x − 3
=
− 2 − 2
2
2
x
x
x
x
Now we reduce each term of the last result to lowest terms, canceling common
factors.
=1−
3
2
− 2
x x
!
You Try It!
EXAMPLE 28. Divide 2x3 − 3x + 12 by 6x3 .
Solution: We use the distributive property, dividing each term by 6x3 .
2x3 − 3x + 12
2x3
3x
12
=
− 3+ 3
6x3
6x3
6x
6x
33
4B. POLYNOMIAL ARITHMETIC
Now we reduce each term of the last result to lowest terms, canceling common
factors.
=
2
1
1
− 2+ 3
3 2x
x
!