Adding and Subtracting
Polynomials
Section 9-1
Vocabulary
• Monomial
• Degree of a Monomial
• Polynomial
• Standard Form of a Polynomial
• Degree of a Polynomial
• Binomial
• Trinomial
A monomial is a number, a variable, or a product of numbers and
variables with whole-number exponents.
The degree of a monomial is the sum of the exponents of the
variables. A constant has degree 0.
Definitions
Find the degree of each monomial.
A. 4p4q3
The degree is 7. Add the exponents of the variables:
4 + 3 = 7.
B. 7ed
The degree is 2. Add the exponents of the variables:
1+ 1 = 2. C. 3
The degree is 0. Add the exponents of the variables:
0 = 0.
Example: Degree of a
Monomial
Find the degree of each monomial.
a. 1.5k2m
The degree is 3. Add the exponents of the variables:
2 + 1 = 3.
b. 4x
The degree is 1. Add the exponents of the variables:
1 = 1.
b. 2c3
The degree is 3. Add the exponents of the variables:
3 = 3.
Your Turn:
You can add or subtract monomials by adding or subtracting like terms.
4a3b2 + 3a2b3 – 2a3b2
Like terms
Not like terms
The variables have the same powers.
The variables have different powers.
Add or subtract like terms by adding or subtracting the coefficients of the like terms.
Like Terms
Identify the like terms in each polynomial.
A. 5x3 + y2 + 2 – 6y2 + 4x3
B. 3a3b2 + 3a2b3 + 2a3b2 – a3b2
Identify like terms. 5x + y + 2 – 6y + 4x 3 2 2 3
Like terms: 5x3 and 4x3, y2 and –6y2
3a b + 3a b + 2a b – a b 3 2 2 3 3 3 2 2 Identify like terms.
Like terms: 3a3b2, 2a3b2, and –a3b2
Example: Identify Like Terms
Identify the like terms in the polynomial.
C. 7p3q2 + 7p2q3 + 7pq2
Identify like terms.
There are no like terms.
7p3q2 + 7p2q3 + 7pq2
Example: Identify Like Terms
Identify the like terms in each polynomial.
A. 4y4 + y2 + 2 – 8y2 + 2y4
B. 7n4r2 + 3n2r3 + 5n4r2 + n4r2
Identify like terms. 4y + y + 2 – 8y + 2y 4 2 2 4
Like terms: 4y4 and 2y4, y2 and –8y2
7n4r2 + 3n2r3 + 5n4r2 + n4r2 Identify like terms.
Like terms: 7n4r2, 5n4r2, and n4r2
Your Turn:
Identify the like terms in the polynomial.
C. 9m3n2 + 7m2n3 + pq2
Identify like terms.
There are no like terms.
9m3n2 + 7m2n3 + pq2
Your Turn:
Simplify.
A. 4x2 + 2x2
2 6x Combine coefficients:
4 + 2 = 6
Identify like terms. 4x2 + 2x2
Example: Add or Subtract
Monomials
Simplify.
B. 3n5m4 - n5m4
Identify like terms. 3n5m4 - n5m4
Combine coefficients: 3 - 1 = 2. 2n5m4
Example: Add or Subtract
Monomials
Simplify.
A. 2x3 - 5x3
Identify like terms.
Combine coefficients: 2 - 5 = -3
2x3 - 5x3
-3x3
Your Turn:
Simplify.
B. 2n5p4 + n5p4
Identify like terms.
Combine coefficients: 2 + 1 = 3
2n5p4 + n5p4
3n5p4
Your Turn:
A polynomial is a monomial or a sum or difference of
monomials.
Example: 3x4 + 5x2 – 7x + 1
This polynomial is the sum of the
monomials 3x4, 5x2, -7x, and 1.
The degree of a polynomial is the degree of the term
with the greatest degree.
Example: The degree of 3x4 + 5x2 – 7x + 1 is 4.
Definitions
Find the degree of each polynomial.
A. 11x7 + 3x3
11x7: degree 7 3x3: degree 3
The degree of the polynomial is the
greatest degree, 7.
Find the degree of
each term.
B.
Find the degree of
each term.
The degree of the polynomial is the greatest degree, 4.
:degree 3 :degree 4
–5: degree 0
Example: Degree of a
Polynomial
Find the degree of each polynomial.
a. 5x – 6
5x: degree 1 Find the degree of
each term. The degree of the polynomial is the
greatest degree, 1.
b. x3y2 + x2y3 – x4 + 2
x3y2: degree 5
The degree of the polynomial is the
greatest degree, 5.
Find the degree of
each term.
–6: degree 0
x2y3: degree 5
–x4: degree 4 2: degree 0
Your Turn:
The terms of a polynomial may be written in any order.
However, polynomials that contain only one variable
are usually written in standard form.
The standard form of a polynomial that contains one
variable is written with the terms in order from greatest
degree to least degree. When written in standard form,
the coefficient of the first term is called the leading
coefficient.
Example: 3x4 + 5x2 – 7x + 1 and 3 is the leading
coefficient.
Definitions
Write the polynomial in standard form. Then give
the leading coefficient.
6x – 7x5 + 4x2 + 9
Find the degree of each term. Then arrange them in descending order:
6x – 7x5 + 4x2 + 9 –7x5 + 4x2 + 6x + 9
Degree 1 5 2 0 5 2 1 0
–7x5 + 4x2 + 6x + 9. The standard form is The leading
coefficient is –7.
Example: Standard Form
Write the polynomial in standard form. Then give the
leading coefficient.
16 – 4x2 + x5 + 9x3
Find the degree of each term. Then arrange them in descending order:
16 – 4x2 + x5 + 9x3 x5 + 9x3 – 4x2 + 16
Degree 0 2 5 3 0 2 3 5
The standard form is The leading
coefficient is 1.
x5 + 9x3 – 4x2 + 16.
Your Turn:
Write the polynomial in standard form. Then give the
leading coefficient.
Find the degree of each term. Then arrange them in descending order:
18y5 – 3y8 + 14y
18y5 – 3y8 + 14y –3y8 + 18y5 + 14y
Degree 5 8 1 8 5 1
The standard form is The leading
coefficient is –3.
–3y8 + 18y5 + 14y.
Your Turn:
Some polynomials have special names based on
their degree and the number of terms they have.
Degree Name
0
1
2
Constant
Linear
Quadratic
3
4
5
6 or more 6th,7th,degree and so on
Cubic
Quartic
Quintic
Name Terms
Monomial
Binomial
Trinomial
Polynomial 4 or more
1
2
3
By Degree
By # of Terms
Classify each polynomial according to its degree and
number of terms.
A. 5n3 + 4n
Degree 3 Terms 2
5n3 + 4n is a cubic binomial.
B. 4y6 – 5y3 + 2y – 9
Degree 6 Terms 4
4y6 – 5y3 + 2y – 9 is a
6th-degree polynomial.
C. –2x
Degree 1 Terms 1
–2x is a linear monomial.
Example: Classifying
Polynomials
Classify each polynomial according to its degree and number of
terms.
a. x3 + x2 – x + 2
Degree 3 Terms 4
x3 + x2 – x + 2 is a cubic
polynomial.
b. 6
Degree 0 Terms 1 6 is a constant monomial.
c. –3y8 + 18y5 + 14y
Degree 8 Terms 3
–3y8 + 18y5 + 14y is an 8th-
degree trinomial.
Your Turn:
Just as you can perform operations on numbers, you can perform operations on polynomials. To add or subtract polynomials, combine like terms.
Adding and Subtracting
Polynomials
Combine like terms.
A. 12p3 + 11p2 + 8p3
12p3 + 11p2 + 8p3
12p3 + 8p3 + 11p2
20p3 + 11p2
Identify like terms.
Rearrange terms so that like
terms are together. Combine like terms.
B. 5x2 – 6 – 3x + 8
5x2 – 6 – 3x + 8
5x2 – 3x + 8 – 6
5x2 – 3x + 2
Identify like terms.
Rearrange terms so that like
terms are together. Combine like terms.
Example: Simplifying
Polynomials
Combine like terms.
C. t2 + 2s2 – 4t2 – s2
t2 – 4t2 + 2s2 – s2
t2 + 2s2 – 4t2 – s2
–3t2 + s2
Identify like terms.
Rearrange terms so that like
terms are together. Combine like terms.
D. 10m2n + 4m2n – 8m2n
10m2n + 4m2n – 8m2n
6m2n
Identify like terms.
Combine like terms.
Example: Simplifying
Polynomials
Like terms are constants or terms with the same variable(s) raised to the same power(s).
Remember!
a. 2s2 + 3s2 + s
Combine like terms.
2s2 + 3s2 + s
5s2 + s
b. 4z4 – 8 + 16z4 + 2
4z4 – 8 + 16z4 + 2
4z4 + 16z4 – 8 + 2
20z4 – 6
Identify like terms.
Combine like terms.
Identify like terms.
Rearrange terms so that like
terms are together. Combine like terms.
Your Turn:
c. 2x8 + 7y8 – x8 – y8
Combine like terms.
2x8 + 7y8 – x8 – y8
2x8 – x8 + 7y8 – y8
x8 + 6y8
d. 9b3c2 + 5b3c2 – 13b3c2
9b3c2 + 5b3c2 – 13b3c2
b3c2
Identify like terms.
Combine like terms.
Identify like terms.
Rearrange terms so that like
terms are together. Combine like terms.
Your Turn:
Polynomials can be added in either vertical or horizontal form.
In vertical form, align the like terms and add:
In horizontal form, use the Associative and Commutative Properties to regroup and combine like terms.
(5x2 + 4x + 1) + (2x2 + 5x + 2)
= (5x2 + 2x2 + 1) + (4x + 5x) + (1 + 2)
= 7x2 + 9x + 3
5x2 + 4x + 1
+ 2x2 + 5x + 2
7x2 + 9x + 3
Adding Polynomials
Add.
A. (4m2 + 5) + (m2 – m + 6)
(4m2 + 5) + (m2 – m + 6)
(4m2 + m2) + (–m) +(5 + 6)
5m2 – m + 11
Identify like terms.
Group like terms
together. Combine like terms.
B. (10xy + x) + (–3xy + y)
(10xy + x) + (–3xy + y)
(10xy – 3xy) + x + y
7xy + x + y
Identify like terms.
Group like terms
together. Combine like terms.
Example: Adding
Polynomials
Add.
(6x2 – 4y) + (3x2 + 3y – 8x2 – 2y)
Identify like terms.
Group like terms together
within each polynomial.
Combine like terms.
(6x2 – 4y) + (3x2 + 3y – 8x2 – 2y)
(6x2 + 3x2 – 8x2) + (3y – 4y – 2y)
Use the vertical method. 6x2 – 4y
+ –5x2 + y
x2 – 3y Simplify.
Example: Adding
Polynomials
Add.
Identify like terms.
Group like terms
together.
Combine like terms.
Your Turn:
Add (5a3 + 3a2 – 6a + 12a2) + (7a3 – 10a).
(5a3 + 3a2 – 6a + 12a2) + (7a3 – 10a)
(5a3 + 7a3) + (3a2 + 12a2) + (–10a – 6a)
12a3 + 15a2 – 16a
Identify like terms.
Group like terms
together.
Combine like terms.
Your Turn:
To subtract polynomials, remember that subtracting is the same as adding the opposite (distributing the negative). To find the opposite of a polynomial, you must write the opposite of each term in the polynomial:
–(2x3 – 3x + 7)= –2x3 + 3x – 7
Subtracting Polynomials
Subtract.
(x3 + 4y) – (2x3)
(x3 + 4y) + (–2x3)
(x3 + 4y) + (–2x3)
(x3 – 2x3) + 4y
–x3 + 4y
Rewrite subtraction as addition of
the opposite.
Identify like terms.
Group like terms together.
Combine like terms.
Example: Subtracting
Polynomials
Subtract.
(7m4 – 2m2) – (5m4 – 5m2 + 8)
(7m4 – 2m2) + (–5m4 + 5m2 – 8)
(7m4 – 5m4) + (–2m2 + 5m2) – 8
(7m4 – 2m2) + (–5m4 + 5m2 – 8)
2m4 + 3m2 – 8
Rewrite subtraction as
addition of the opposite.
Identify like terms.
Group like terms together.
Combine like terms.
Example: Subtracting
Polynomials
Subtract.
(–10x2 – 3x + 7) – (x2 – 9)
(–10x2 – 3x + 7) + (–x2 + 9)
(–10x2 – 3x + 7) + (–x2 + 9)
–10x2 – 3x + 7
–x2 + 0x + 9
–11x2 – 3x + 16
Rewrite subtraction as
addition of the opposite.
Identify like terms.
Use the vertical method.
Write 0x as a placeholder.
Combine like terms.
Example: Subtracting
Polynomials
Subtract.
(9q2 – 3q) – (q2 – 5)
(9q2 – 3q) + (–q2 + 5)
(9q2 – 3q) + (–q2 + 5)
9q2 – 3q + 0 + − q2 – 0q + 5
8q2 – 3q + 5
Rewrite subtraction as
addition of the opposite.
Identify like terms.
Use the vertical method.
Write 0 and 0q as
placeholders.
Combine like terms.
Your Turn:
Subtract.
(2x2 – 3x2 + 1) – (x2 + x + 1)
(2x2 – 3x2 + 1) + (–x2 – x – 1)
(2x2 – 3x2 + 1) + (–x2 – x – 1)
–x2 + 0x + 1 + –x2 – x – 1
–2x2 – x
Rewrite subtraction as
addition of the opposite.
Identify like terms.
Use the vertical method. Write 0x as a placeholder.
Combine like terms.
Your Turn:
Practice
• 9-1 Exercises Pg. 459-460 #2-50 even
Multiplying and Factoring
Section 9-2
To multiply monomials and polynomials, you will use some of the properties of exponents that you learned earlier in this chapter.
Multiplying Polynomials
Multiply.
A. (6y3)(3y5)
(6y3)(3y5)
18y8
Group factors with like bases together.
B. (3mn2) (9m2n)
(3mn2)(9m2n)
27m3n3
Multiply.
Group factors with like bases together.
Multiply.
(6 • 3)(y3 • y5)
(3 • 9)(m • m2)(n2 • n)
Multiplying Monomials
Multiply.
Group factors with like bases
together.
Multiply.
Multiplying Monomials
(4s2t2)(st)(-12st2)
(4s2t2)(st)(-12st2)
(4 • -12)(s2 • s • s)(t2 • t • t2)
-48s4t5
When multiplying powers with the same base, keep the base and add the exponents.
x2 x3 = x2+3 = x5
Remember!
Multiply.
a. (3x3)(6x2)
(3x3)(6x2)
(3 • 6)(x3 • x2)
18x5
Group factors with like bases together.
Multiply.
Group factors with like bases together.
Multiply.
b. (2r2t)(5t3)
(2r2t)(5t3)
(2 • 5)(r2)(t3 • t)
10r2t4
Your Turn:
Multiply.
Group factors with like
bases together.
Multiply.
c.
Your Turn:
(3x2y)(2x3z2)(y4z5)
(3x2y)(2x3z2)(y4z5)
(3 • 2)(x2 • x3)(y • y4)(z2 • z5)
6x5y5z7
To multiply a polynomial by a monomial, use the Distributive Property.
Multiplying Monomials
and Polynomials
Multiply.
4(3x2 + 4x – 8)
4(3x2 + 4x – 8)
(4)3x2 +(4)4x – (4)8
12x2 + 16x – 32
Distribute 4.
Multiply.
Example: Multiplying a
Polynomial by a Monomial
6pq(2p – q)
(6pq)(2p – q)
Multiply.
(6pq)2p + (6pq)(–q)
(6 2)(p p)(q) + (–1)(6)(p)(q q)
12p2q – 6pq2
Distribute 6pq.
Group like bases together.
Multiply.
Example: Multiplying a
Polynomial by a Monomial
Multiply.
a. 2(4x2 + x + 3)
2(4x2 + x + 3)
2(4x2) + 2(x) + 2(3)
8x2 + 2x + 6
Distribute 2.
Multiply.
Your Turn:
Multiply.
b. 3ab(5a2 + b)
3ab(5a2 + b)
(3ab)(5a2) + (3ab)(b)
(3 5)(a a2)(b) + (3)(a)(b b)
15a3b + 3ab2
Distribute 3ab.
Group like bases
together.
Multiply.
Your Turn:
Multiply.
c. 5r2s2(r – 3s)
5r2s2(r – 3s)
(5r2s2)(r) – (5r2s2)(3s)
(5)(r2 r)(s2) – (5 3)(r2)(s2 s)
5r3s2 – 15r2s3
Distribute 5r2s2.
Group like bases
together.
Multiply.
Your Turn:
When multiplying a polynomial by a negative
monomial, be sure to distribute the negative sign.
Helpful Hint
Multiply.
d. –5y3(y2 + 6y – 8)
–5y3(y2 + 6y – 8)
–5y5 – 30y4 + 40y3
Multiply each term in
parentheses by –5y3.
Your Turn:
Factoring
• Factoring a polynomial reverses the
multiplication process (factoring is
unmultiplying).
• When factoring a monomial from a
polynomial, the first step is to find the
greatest common factor (GCF) of the
polynomial’s terms.
Factors that are shared by two or more whole numbers are
called common factors. The greatest of these common factors is
called the greatest common factor, or GCF.
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 32: 1, 2, 4, 8, 16, 32
Common factors: 1, 2, 4
The greatest of the common factors is 4.
Greatest Common Factor
Find the GCF of each pair of numbers.
100 and 60
factors of 100: 1, 2, 4,
5, 10, 20, 25, 50, 100
factors of 60: 1, 2, 3, 4, 5,
6, 10, 12, 15, 20, 30, 60
The GCF of 100 and 60 is 20.
List all the factors.
Circle the GCF.
List the factors.
Example: GCF of Two Numbers
Find the GCF of each pair of numbers.
12 and 16
factors of 12: 1, 2, 3, 4, 6, 12
factors of 16: 1, 2, 4, 8, 16
The GCF of 12 and 16 is 4.
List all the factors.
Circle the GCF.
List the factors.
Your Turn:
You can also find the GCF of monomials that
include variables. To find the GCF of monomials,
write the prime factorization of each coefficient and
write all powers of variables as products. Then find
the product of the common factors.
GCF of Monomials
Find the GCF of each pair of monomials.
15x3 and 9x2
15x3 = 3 5 x x x
9x2 = 3 3 x x
3 x x = 3x2
Write the factorization of each
coefficient and write powers as
products.
Align the common factors.
Find the product of the common
factors.
The GCF of 3x3 and 6x2 is 3x2.
Example: GCF of a
Monomial
Find the GCF of each pair of monomials.
8x2 and 7y3
8x2 = 2 2 2 x x
7y3 = 7 y y y
Write the factorization of
each coefficient and write
powers as products.
Align the common factors.
There are no common
factors other than 1. The GCF 8x2 and 7y is 1.
Example: GCF of a
Monomial
If two terms contain the same variable raised to
different powers, the GCF will contain that variable
raised to the lower power.
Helpful Hint
Find the GCF of each pair of monomials.
18g2 and 27g3
18g2 = 2 3 3 g g
27g3 = 3 3 3 g g g
3 3 g g
The GCF of 18g2 and 27g3 is 9g2.
Write the factorization of each
coefficient and write powers as
products.
Align the common factors.
Find the product of the common
factors.
Your Turn:
Find the GCF of each pair of monomials.
16a6 and 9b
9b = 3 3 b
16a6 = 2 2 2 2 a a a a a a
Write the
factorization of
each coefficient
and write powers
as products.
Align the common
factors.
There are no common factors
other than 1.
The GCF of 16a6 and 7b is 1.
Your Turn:
Find the GCF of each pair of monomials.
8x and 7x2
8x = 2 2 2 x
7v2 = 7 x x
Write the prime factorization of each
coefficient and write powers as
products.
Align the common factors.
The GCF of 8x and 7x2 is x.
Your Turn:
Recall that the Distributive Property states that ab + ac =a(b + c).
The Distributive Property allows you to “factor” out the GCF of
the terms in a polynomial to write a factored form of the
polynomial.
A polynomial is in its factored form when it is written as a
product of monomials and polynomials that cannot be factored
further. The polynomial 2(3x – 4x) is not fully factored because
the terms in the parentheses have a common factor of x.
Factoring out a Monomial
Factor each polynomial. Check your answer.
2x2 – 4
2x2 = 2 x x
4 = 2 2
2
Find the GCF.
The GCF of 2x2 and 4 is 2.
Write terms as products using the GCF as a
factor. 2x2 – (2 2)
2(x2 – 2)
Check 2(x2 – 2)
2x2 – 4
Multiply to check your answer.
The product is the original polynomial.
Use the Distributive Property to factor out the
GCF.
Example: Factoring the GCF
Aligning common factors can help you find the greatest
common factor of two or more terms.
Writing Math
Factor each polynomial. Check your answer.
8x3 – 4x2 – 16x
2x2(4x) – x(4x) – 4(4x)
4x(2x2 – x – 4)
4x(2x2 – x – 4)
8x3 – 4x2 – 16x
8x3 = 2 2 2 x x x
4x2 = 2 2 x x
16x = 2 2 2 2 x
2 2 x = 4x
Find the GCF.
The GCF of 8x3, 4x2, and 16x is 4x.
Write terms as products using the
GCF as a factor.
Use the Distributive Property to factor
out the GCF.
Multiply to check your answer.
The product is the original polynomials.
Check
Example: Factoring the GCF
Factor each polynomial.
–14x – 12x2
– 1(14x + 12x2) Both coefficients are negative.
Factor out –1.
Find the GCF.
The GCF of 14x and 12x2 is 2x.
–1[7(2x) + 6x(2x)]
–1[2x(7 + 6x)]
–2x(7 + 6x)
Write each term as a product using
the GCF.
Use the Distributive Property to
factor out the GCF.
14x = 2 7 x
12x2 = 2 2 3 x x
2 x = 2x
Example: Factoring the GCF
When you factor out –1 as the first step, be sure to include it
in all the other steps as well.
Caution!
Factor each polynomial.
3x3 + 2x2 – 10
10 = 2 5
Find the GCF.
There are no common factors
other than 1.
The polynomial cannot be factored further.
3x3 + 2x2 – 10
3x3 = 3 x x x
2x2 = 2 x x
Example: Factoring the GCF
Factor each polynomial. Check your answer.
5b + 9b3
5b = 5 b
9b = 3 3 b b b
b
5(b) + 9b2(b)
b(5 + 9b2)
b(5 + 9b2) Check
5b + 9b3
Find the GCF.
The GCF of 5b and 9b3 is b.
Multiply to check your answer.
The product is the original
polynomial.
Write terms as products using the
GCF as a factor.
Use the Distributive Property to factor
out the GCF.
Your Turn:
Factor each polynomial.
9d2 – 82
Find the GCF.
There are no common factors
other than 1.
The polynomial cannot be factored further.
9d2 – 82
9d2 = 3 3 d d
82 = 2 2 2 2 2 2
Your Turn:
Factor each polynomial.
–18y3 – 7y2
– 1(18y3 + 7y2) Both coefficients are negative.
Factor out –1.
Find the GCF.
The GCF of 18y3 and 7y2 is y2.
18y3 = 2 3 3 y y y
7y2 = 7 y y
y y = y2
Write each term as a product using
the GCF.
Use the Distributive Property to
factor out the GCF..
–1[18y(y2) + 7(y2)]
–1[y2(18y + 7)]
–y2(18y + 7)
Example: Factoring the GCF
Factor each polynomial.
8x4 + 4x3 – 2x2
8x4 = 2 2 2 x x x x
4x3 = 2 2 x x x
2x2 = 2 x x
2 x x = 2x2
4x2(2x2) + 2x(2x2) –1(2x2)
2x2(4x2 + 2x – 1)
Check 2x2(4x2 + 2x – 1)
8x4 + 4x3 – 2x2
The GCF of 8x4, 4x3 and –2x2 is 2x2.
Multiply to check your answer.
The product is the original polynomial.
Write terms as products using the
GCF as a factor.
Use the Distributive Property to factor
out the GCF.
Find the GCF.
Your Turn:
To write expressions for the length and width of a rectangle
with area expressed by a polynomial, you need to write the
polynomial as a product. You can write a polynomial as a
product by factoring it.
The area of a court for the game squash is 9x2 + 6x m2.
Factor this polynomial to find possible expressions for
the dimensions of the squash court.
A = 9x2 + 6x
= 3x(3x) + 2(3x)
= 3x(3x + 2)
Possible expressions for the dimensions of the squash court are
3x m and (3x + 2) m.
The GCF of 9x2 and 6x is 3x.
Write each term as a product using the
GCF as a factor.
Use the Distributive Property to factor
out the GCF.
Example: Application
What if…? The area of the solar panel on another
calculator is (2x2 + 4x) cm2. Factor this polynomial
to find possible expressions for the dimensions of the
solar panel.
A = 2x2 + 4x
= x(2x) + 2(2x)
= 2x(x + 2)
The GCF of 2x2 and 4x is 2x.
Write each term as a product using the
GCF as a factor.
Use the Distributive Property to factor
out the GCF.
Possible expressions for the dimensions of the solar panel are 2x cm,
and (x + 2) cm.
Your Turn:
Assignment
• 9-2 Exercises Pg. 463-464 #1-12 eoo, 13-41
odd
Multiplying Binomials
Section 9-3
Multiplying Polynomials
3 Methods for multiplying polynomials
1. Using the Distributive Property
• Can be used to multiply any two polynomials
2. Using a Table or The Box Method
• Can be used to multiply any two polynomials
3. Using FOIL
• Can only be used to multiply two binomials
To multiply a binomial by a binomial, you can apply the Distributive Property more than once:
(x + 3)(x + 2) = x(x + 2) + 3(x + 2) Distribute x and 3.
Distribute x and 3 again.
Multiply.
Combine like terms.
= x(x + 2) + 3(x + 2)
= x(x) + x(2) + 3(x) + 3(2)
= x2 + 2x + 3x + 6
= x2 + 5x + 6
Method 1: Distributive Property
Multiply.
(s + 4)(s – 2)
(s + 4)(s – 2)
s(s – 2) + 4(s – 2)
s(s) + s(–2) + 4(s) + 4(–2)
s2 – 2s + 4s – 8
s2 + 2s – 8
Distribute s and 4.
Distribute s and 4 again.
Multiply.
Combine like terms.
Example: Multiply Using
Distributive Property
Multiply.
(a + 3)(a – 4)
(a + 3)(a – 4)
a(a – 4)+3(a – 4)
a(a) + a(–4) + 3(a) + 3(–4)
a2 – a – 12
a2 – 4a + 3a – 12
Distribute a and 3.
Distribute a and 3 again.
Multiply.
Combine like terms.
Your Turn:
+ 8 (y – 4)
= (y2 – 4y)
= y2 – 4y + 8y – 32
= y (y – 4)
+ (8y – 32)
= y2 + 4y – 32
Your Turn:
Multiply.
(y + 8)(y – 4)
The product can be simplified using the FOIL
method: multiply the First terms, the Outer terms, the
Inner terms, and the Last terms of the binomials.
2 First Last
Inner
Outer
Method 2: FOIL
4. Multiply the Last terms. (x + 3)(x + 2) 3 2 = 6
3. Multiply the Inner terms. (x + 3)(x + 2) 3 x = 3x
2. Multiply the Outer terms. (x + 3)(x + 2) x 2 = 2x
F
O
I
L
(x + 3)(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6
F O I L
1. Multiply the First terms. (x + 3)(x + 2) x x = x2
Example: Multiply Using FOIL
“First Outer Inner Last”, shortcut for distributing, only
works with binomial-binomial products. Multiply
(x + 3)(x + 2)
= z (z) + z (-12) -6 (z) -6 (-12) F O I L
= z2 - 12z – 6z + 72
= z2 - 18z + 72
= 5x (2x) + 5x (8) -4 (2x) -4 (8) F O I L
= 10x2 + 40x – 8x – 32
= 10x2 + 32x – 32
Example: FOIL
Multiply.
A. (m – 2)(m – 8) B. (x + 3)(x + 4)
(m – 2)(m – 8) (x + 3)(x + 4)
m2 – 8m – 2m + 16 x2 + 4x + 3x + 12
FOIL
m2 – 10m +16 x2 + 7x +12
Your Turn:
Multiply.
(x – 3)(x – 1)
(x – 3)(x – 1)
(x x) + (x(–1)) + (–3 x)+ (–3)(–1) ●
x2 – x – 3x + 3
x2 – 4x + 3
Use the FOIL method.
Multiply.
Combine like terms.
Your Turn:
Multiply.
(2a – b2)(a + 4b2)
(2a – b2)(a + 4b2)
2a(a) + 2a(4b2) – b2(a) + (–b2)(4b2)
2a2 + 8ab2 – ab2 – 4b4
2a2 + 7ab2 – 4b4
Use the FOIL method.
Multiply.
Combine like terms.
Your Turn:
To multiply polynomials with more than two terms, you can use the Distributive Property several times. Multiply (5x + 3) by (2x2 + 10x – 6):
(5x + 3)(2x2 + 10x – 6) = 5x(2x2 + 10x – 6) + 3(2x2 + 10x – 6)
= 5x(2x2 + 10x – 6) + 3(2x2 + 10x – 6)
= 5x(2x2) + 5x(10x) + 5x(–6) + 3(2x2) + 3(10x) + 3(–6)
= 10x3 + 50x2 – 30x + 6x2 + 30x – 18
= 10x3 + 56x2 – 18
Multiply.
(x – 5)(x2 + 4x – 6)
(x – 5 )(x2 + 4x – 6)
x(x2 + 4x – 6) – 5(x2 + 4x – 6)
x(x2) + x(4x) + x(–6) – 5(x2) – 5(4x) – 5(–6)
x3 + 4x2 – 5x2 – 6x – 20x + 30
x3 – x2 – 26x + 30
Distribute x and –5.
Distribute x and −5
again.
Simplify.
Combine like terms.
Example:
Multiply.
(x + 3)(x2 – 4x + 6)
(x + 3 )(x2 – 4x + 6)
x(x2 – 4x + 6) + 3(x2 – 4x + 6)
Distribute x and 3.
Distribute x and 3 again.
x(x2) + x(–4x) + x(6) +3(x2) +3(–4x) +3(6)
x3 – 4x2 + 3x2 +6x – 12x + 18
x3 – x2 – 6x + 18
Simplify.
Combine like terms.
Your Turn:
The width of a rectangular prism is 3 feet less than the height, and the length of the prism is 4 feet more than the height.
Write a polynomial that represents the area of the base of the
prism.
Write the formula for the area of a
rectangle.
Substitute h – 3 for w and h + 4
for l.
A = l w
A = l w
A = (h + 4)(h – 3)
Multiply. A = h2 + 4h – 3h – 12
Combine like terms. A = h2 + h – 12
The area is represented by h2 + h – 12.
Example: Application
The width of a rectangular prism is 3 feet less than the height, and the length of the prism is 4 feet more than the height.
Find the area of the base when the height is 5 ft.
A = h2 + h – 12
A = h2 + h – 12
A = 52 + 5 – 12
A = 25 + 5 – 12
A = 18
Write the formula for the area the base of
the prism.
Substitute 5 for h.
Simplify.
Combine terms.
The area is 18 square feet.
Your Turn:
Assignment
• 9-3 pg. #1-19 all #20, 30-38 odd, #39
9-4
Multiplying Special
Cases
There are formulas (shortcuts) that
work for certain polynomial
multiplication problems.
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 – 2ab + b2
(a - b)(a + b) = a2 - b2
Being able to use these formulas will help you in the future when you have to factor. If you do not remember the formulas, you can always multiply
using distributive, FOIL, or the box method.
Let’s try one!
1) Multiply: (x + 4)2
You can multiply this by rewriting this as (x + 4)(x + 4)
OR
You can use the following rule as a shortcut:
(a + b)2 = a2 + 2ab + b2
For comparison, I’ll show you both ways.
1) Multiply (x + 4)(x + 4)
First terms:
Outer terms:
Inner terms:
Last terms:
Combine like terms.
x2 +8x + 16
x +4
x
+4
x2
+4x
+4x
+16
Now let’s do it with the shortcut!
x2
+4x
+4x
+16
Notice you
have two of
the same
answer?
1) Multiply: (x + 4)2
using (a + b)2 = a2 + 2ab + b2
a is the first term, b is the second term
(x + 4)2
a = x and b = 4
Plug into the formula a2 + 2ab + b2
(x)2 + 2(x)(4) + (4)2
Simplify.
x2 + 8x+ 16
This is the
same answer!
That’s why
the 2 is in
the formula!
2) Multiply: (3x + 2y)2
using (a + b)2 = a2 + 2ab + b2
(3x + 2y)2
a = 3x and b = 2y
Plug into the formula
a2 + 2ab + b2
(3x)2 + 2(3x)(2y) + (2y)2
Simplify
9x2 + 12xy +4y2
Multiply (2a + 3)2
1. 4a2 – 9
2. 4a2 + 9
3. 4a2 + 36a + 9
4. 4a2 + 12a + 9
Multiply: (x – 5)2
using (a – b)2 = a2 – 2ab + b2
Everything is the same except the signs!
(x)2 – 2(x)(5) + (5)2
x2 – 10x + 25
4) Multiply: (4x – y)2
(4x)2 – 2(4x)(y) + (y)2
16x2 – 8xy + y2
Multiply (x – y)2
1. x2 + 2xy + y2
2. x2 – 2xy + y2
3. x2 + y2
4. x2 – y2
5) Multiply (x – 3)(x + 3)
First terms:
Outer terms:
Inner terms:
Last terms:
Combine like terms.
x2 – 9
x -3
x
+3
x2
+3x
-3x
-9
This is called the difference of squares.
x2
+3x
-3x
-9
Notice the
middle terms
eliminate
each other!
5) Multiply (x – 3)(x + 3) using
(a – b)(a + b) = a2 – b2
You can only use this rule when the binomials
are exactly the same except for the sign.
(x – 3)(x + 3)
a = x and b = 3
(x)2 – (3)2
x2 – 9
6) Multiply: (y – 2)(y + 2)
(y)2 – (2)2
y2 – 4
7) Multiply: (5a + 6b)(5a – 6b)
(5a)2 – (6b)2
25a2 – 36b2
Multiply (4m – 3n)(4m + 3n)
1. 16m2 – 9n2
2. 16m2 + 9n2
3. 16m2 – 24mn - 9n2
4. 16m2 + 24mn + 9n2
Practice!!
• Pg. 477- 478 #2-8 even, #15-20 all, 28-39
even, #44-52 even
• # 43 answer in complete sentences and turn
in for 5 points.
Factoring x2 + bx + c
Section 9-5
Earlier you learned how to multiply two binomials using the
Distributive Property or the FOIL method. In this lesson, you
will learn how to factor a trinomial into two binominals.
Factoring a Trinomial
x2 + bx + c
Notice that when you multiply (x + 2)(x + 5), the constant term
in the trinomial is the product of the constants in the binomials.
(x + 2)(x + 5) = x2 + 7x + 10
Factoring these trinomials is based on
reversing the FOIL process.
To factor a simple trinomial of the form
x 2 + bx + c (leading coefficient is 1),
express the trinomial as the product of
two binomials. For example,
x 2 + 10x + 24 = (x + 4)(x + 6).
Factoring a Trinomial
x2 + bx + c
Look at the product of (x + a) and (x + b).
(x + a)(x + b) = x2 + ax + bx + ab
x2 ab
ax
bx
The coefficient of the middle term is the sum of a and b. The
constant term is the product of a and b.
= x2 + (a + b)x + ab
Factoring a Trinomial
x2 + bx + c
Recall by using the FOIL method that
F O I L
(x + 2)(x + 4) = x 2 + 4x + 2x + 8
= x 2 + 6x + 8
To factor x 2 + bx + c into (x + one #)(x + another #), note that b is the sum of the two numbers and c is the product of the two numbers.
So we’ll be looking for 2 numbers whose product is c and whose sum is b.
Note: there are fewer choices for the product, so that’s why we start there first.
Factoring a Trinomial
x2 + bx + c
Procedure for Factoring Trinomials
of the Form x 2 + bx + c
Factor:
1. List the factors of 20 (two numbers that
multiply to equal the constant):
2. Select the pairs those sum is 12 (the
middle term).
3. Write the two
binomial factors:
4. Check using FOIL:
2 12 20x x
2
2
10 2
2 10 20
12 20
x x
x x x
x x
20 1×20 2×10 4×5
Factoring a Trinomial
x2 + bx + c
Factoring Trinomials x2 + bx + c TIP
When c is positive, its factors have the same
sign. The sign of b tells you whether the
factors are positive or negative. When b is
positive, the factors are positive and when b
is negative, the factors are negative.
2 12 20
10 2
x x
x x
2 9 20
5 4
x x
x x
x2 + 6x + 5
Factor each trinomial. Check your answer.
(x + )(x + ) b = 6 and c = 5; look for factors of 5
whose sum is 6.
Factors of 5 Sum
1 and 5 6 The factors needed are 1 and 5.
(x + 1)(x + 5)
Check (x + 1)(x + 5) = x2 + 5x + x + 5 Use the FOIL method.
The product is the
original trinomial.
= x2 + 6x + 5
Example: c is Positive
Factor each trinomial. Check your answer.
x2 + 6x + 9
(x + )(x + ) b = 6 and c = 9; look for factors of 9
whose sum is 6.
The factors needed are 3 and 3.
(x + 3)(x + 3)
Check (x + 3)(x + 3 ) = x2 + 3x + 3x + 9 Use the FOIL method. The product is the
original trinomial. = x2 + 6x + 9
Factors of 9 Sum
1 and 9 10
3 and 3 6
Example: c is Positive
Factor each trinomial. Check your answer.
x2 – 8x + 15
b = –8 and c = 15; look for
factors of 15 whose sum is –8.
The factors needed are –3 and –5 .
Factors of –15 Sum
–1 and –15 –16
–3 and –5 –8
(x – 3)(x – 5)
Check (x – 3)(x – 5 ) = x2 – 5x – 3x + 15 Use the FOIL method. The product is the
original trinomial. = x2 – 8x + 15
(x + )(x + )
Example: c is Positive
Factor each trinomial. Check your answer.
x2 + 8x + 12
b = 8 and c = 12; look for factors of
12 whose sum is 8.
The factors needed are 2 and 6 .
Factors of 12 Sum
1 and 12 13 2 and 6 8
(x + 2)(x + 6)
Check (x + 2)(x + 6 ) = x2 + 6x + 2x + 12 Use the FOIL method.
The product is the
original trinomial.
= x2 + 8x + 12
(x + )(x + )
Your Turn:
Factor each trinomial. Check your answer.
x2 – 5x + 6
(x + )(x+ ) b = –5 and c = 6; look for
factors of 6 whose sum is –5.
The factors needed are –2 and –3.
Factors of 6 Sum
–1 and –6 –7 –2 and –3 –5
(x – 2)(x – 3)
Check (x – 2)(x – 3) = x2 – 3x – 2x + 6 Use the FOIL method.
The product is the
original trinomial.
= x2 – 5x + 6
Your Turn:
Factor each trinomial. Check your answer.
x2 + 13x + 42
b = 13 and c = 42; look for
factors of 42 whose sum is
13.
The factors needed are 6 and 7.
(x + 6)(x + 7)
Check (x + 6)(x + 7) = x2 + 7x + 6x + 42 Use the FOIL method.
The product is the
original trinomial. = x2 + 13x + 42
(x + )(x + )
Factors of 42 Sum
1 and 42 43
6 and 7 13
2 and 21 23
Your Turn:
Factor each trinomial. Check your answer.
x2 – 13x + 40
(x + )(x+ )
b = –13 and c = 40; look for factors
of 40 whose sum is –13.
The factors needed are –5 and –8.
(x – 5)(x – 8)
Check (x – 5)(x – 8) = x2 – 8x – 5x + 40 Use the FOIL method.
The product is the
original polynomial.
= x2 – 13x + 40
Factors of 40 Sum
–2 and –20 –22
–4 and –10 –14
–5 and –8 –13
Your Turn:
When c is negative, its factors have opposite
signs. The sign of b tells you which factor is
positive and which is negative. The factor
with the greater absolute value has the same
sign as b.
2 9 22
11 2
x x
x x
Factoring Trinomials x2 + bx + c TIP
Factor each trinomial.
x2 + x – 20
(x + )(x + ) b = 1 and c = –20; look for
factors of –20 whose sum is
1. The factor with the greater
absolute value is positive.
The factors needed are 5 and
–4.
Factors of –20 Sum
–1 and 20 19
–2 and 10 8
–4 and 5 1
(x – 4)(x + 5)
Example: c is Negative
Factor each trinomial.
x2 – 3x – 18
b = –3 and c = –18; look for
factors of –18 whose sum is
–3. The factor with the
greater absolute value is
negative. Factors of –18 Sum
1 and –18 –17
2 and – 9 – 7
3 and – 6 – 3
The factors needed are 3 and
–6. (x – 6)(x + 3)
(x + )(x + )
Example: c is Negative
If you have trouble remembering the rules for which
factor is positive and which is negative, you can try
all the factor pairs and check their sums.
Helpful Hint
Factor each trinomial. Check your answer.
x2 + 2x – 15
(x + )(x + )
Factors of –15 Sum
–1 and 15 14
–3 and 5 2
(x – 3)(x + 5)
b = 2 and c = –15; look for
factors of –15 whose sum is 2.
The factor with the greater
absolute value is positive.
The factors needed are –3 and
5.
Check (x – 3)(x + 5) = x2 + 5x – 3x – 15 Use the FOIL method.
The product is the
original polynomial.
= x2 + 2x – 15
Your Turn:
Factor each trinomial. Check your answer.
x2 – 6x + 8
(x + )(x + ) b = –6 and c = 8; look for
factors of 8 whose sum is –6.
The factors needed are –4
and –2.
Factors of 8 Sum
–1 and –6 –7
–2 and –4 –6
(x – 2)(x – 4)
Check (x – 2)(x – 4) = x2 – 4x – 2x + 8 Use the FOIL method.
The product is the
original polynomial.
= x2 – 6x + 8
Your Turn:
x2 – 8x – 20
Factor each trinomial. Check your answer.
(x – 10)(x + 2)
Factors of –20 Sum
1 and –20 –19
2 and –10 –8
b = –8 and c = –20; look for
factors of –20 whose sum is
–8. The factor with the
greater absolute value is
negative.
The factors needed are –10
and 2.
(x + )(x + )
Check (x – 10)(x + 2) = x2 + 2x – 10x – 20 Use the FOIL method.
The product is the
original polynomial.
= x2 – 8x – 20
Your Turn:
Practice
More Practice
Assignment
• Pg 483-484 # 1-29 odd, 43-54 all
Factoring Trinomials of the
type:
𝑎𝑥2 + 𝑏𝑥 + 𝑐
9-6
Factoring
• When the quadratic term (first one) has a
coefficient other than 1, factoring becomes
more difficult. These polynomials are in the
form .
• We follow four steps to factor these
polynomials:
2ax bx c
2ax bx c
Steps for “Bottoms Up”
1. Multiply the first coefficient and the last term
together.
2. Find the pair of factors that will combine to
produce the middle term.
3. Put each factor of the pair over a [which is
the leading coefficient] and reduce the
fractions.
4. Use “Bottoms Up” to write the factors of the
trinomial.
What is “Bottoms Up”?
• Once you have your two reduced fractions, how do you
write the factors of the trinomial?
• Let’s suppose these are your fractions:
• Do the following:
becomes _____ and becomes ______
1 2
and2 3
2
31
2 2x 1 3x 2
1
should be written as2
1
2
FHS Polynomials 145
2 x 20 = 40 We need to find a pair of factors of 24 that combine to give us +5
Multiply the first and last coefficients
22x 3x 20
5 8and
2 2
2x 5 x 4
Example 1
1 and -40 2 and -20 4 and -10 5 and -8
-1 and 40 -2 and 20 -4 and 10 -5 and 8
5 4and
2 1
4 x 6 = 24 We need to find a pair of factors of 24 that combine to give us +5
Multiply the first and last coefficients
26x 5x 4
3 8and
6 6
2x 1 3x 4
Example 1
1 and -24 2 and -12 3 and -8 4 and -6
-1 and 24 -2 and 12 -3 and 8 -4 and 6
1 4and
2 3
Example 2
Multiply 8 x 15 = 120. Find all pairs of factors of 120.
We need a pair that combines to 26.
4x 3 2x 5
28x 26x 15 1 and 120
2 and 60
3 and 40
4 and 30
5 and 24
6 and 20
8 and 15
10 and 12
6 20and
8 8
3 5and
4 2
Practice!!
• Pg. 487-488 #1-12 all, #13-27 eoo,
• #44-47 all
Factoring Special Cases
Section 9-7
Perfect Square Trinomial
Factor the polynomial 25x 2 +
20x + 4.
The result is (5x + 2)2, an example of a
binomial squared.
Any trinomial that factors into a single
binomial squared is called a perfect square
trinomial.
Perfect Square Trinomials
A perfect square trinomial results after
squaring a binomial
• Example: (2x – 5)2
= 4x2
Multiply it out
using FOIL
+ 25 – 10x – 10x
= 4x2 – 20x + 25
The first and last terms are perfect squares.
The middle term is double the product of the
square roots of the first and last terms.
(2x – 5)2 = (2x – 5) (2x – 5)
Perfect Square Trinomials
(a + b)2 = a 2 + 2ab + b 2
(a – b)2 = a 2 – 2ab + b 2
So if the first and last terms of our polynomial to be
factored can be written as expressions squared, and the
middle term of our polynomial is twice the product of
those two expressions, then we can use these two previous
equations to easily factor the polynomial.
a 2 + 2ab + b 2 = (a + b)2
a 2 – 2ab + b 2 = (a – b)2
Perfect Square Trinomials
So how do we factor a Prefect Square
Trinomial
When you have to factor a perfect square
trinomial, the patterns make it easier
Example: Factor 36x2 + 60x + 25
First you have to recognize that it’s a
perfect square trinomial
Perfect
Square
Perfect
Square
Square
Root
Square
Root
Product
Doubled
Product
6x 5 30x
And so, the trinomial factors as: (6x + 5)2
Example – First verify it is a Perfect Square
Trinomial
1. m2 – 4m + 4
2. 9w2 + 24w + 16
3. 16x2 – 72x + 81
4. 25h2 – 100h + 64
5. a2 + 6a + 9
= (m – 2)2
= (3w + 4)2
= (4x – 9)2
= (5h – 16)(5h – 4)
= (a + 3)2
Not a perfect square trinomial! 5h 8 40h
m 2 2m
3w 4 12w
4x 9 36x
a 3 3a
Your Turn:
1. m2 – 6m + 9
2. 4w2 + 28w + 49
3. 81x2 – 18x + 1
4. 9a2 + 12a + 4
= (m – 3)2
= (2w + 7)2
= (9x – 1)2
= (3a + 2)2
m 3 3m
2w 7 14w
9x 1 9x
3a 2 6a
Review: Perfect-Square Trinomial
Difference of Two Squares
D.O.T.S.
Conjugate Pairs
The following pairs of binomials are called conjugates.
Notice that they all have the same terms, only the sign
between them is different.
(3x + 6) and (3x - 6)
(r - 5) and (r + 5)
(2b - 1) and (2b + 1)
(x2 + 5) and (x2 - 5)
Multiplying Conjugates
When we multiply any conjugate pairs, the middle terms always
cancel and we end up with a binomial.
(3x + 6)(3x - 6)
(r - 5)(r + 5)
(2b - 1)(2b + 1)
= 9x2 - 36
= r2 - 25
= 4b2 - 1
Difference of Two Squares
Binomials that look like this are called a Difference of Squares:
9x2 - 36
The first term
is a Perfect
Square!
The second term
is a Perfect
Square!
Only TWO terms (a binomial)
A MINUS
between!
Difference of Two Squares
A binomial is the difference of two square if
1.both terms are squares and
2.the signs of the terms are different.
9x 2 – 25y 2
– c 4 + d 4
a b a b a b2 2 ( )( )
Factoring the Difference of Two Squares
A Difference of
Squares!
A Conjugate Pair!
Difference of Two Squares
Factor the polynomial x 2 – 9.
The first term is a square and the last term, 9, can be
written as 32. The signs of each term are different, so we
have the difference of two squares
Therefore x 2 – 9 = (x – 3)(x + 3).
Note: You can use FOIL method to verify that the
factorization for the polynomial is accurate.
Example:
a b a b a b2 2 ( )( )
Difference of Two Squares
Example: Factor x2 - 64
x2 = x • x 64 = 8 • 8
= (x + 8)(x - 8)
Example: Factor 9t2 - 25
9t2 = 3t • 3t 25 = 5 • 5
= (3t + 5)(3t - 5)
Factor x 2 – 16.
Since this polynomial can be written as x 2 – 42,
x 2 – 16 = (x – 4)(x + 4).
Factor 9x2 – 4.
Since this polynomial can be written as (3x)2 – 22,
9x 2 – 4 = (3x – 2)(3x + 2).
Factor 16x 2 – 9y 2.
Since this polynomial can be written as (4x)2 – (3y)2,
16x 2 – 9y 2 = (4x – 3y)(4x + 3y).
Difference of Two Squares
Example:
A Sum of Squares?
A Sum of Squares, like x2 + 64,
can NOT be factored!
It is a PRIME polynomial.
Factor x 8 – y 6.
Since this polynomial can be written as
(x 4)2 – (y 3)2,
x 8 – y 6 = (x 4 – y 3)(x 4 + y 3).
Factor x2 + 4.
Oops, this is the sum of squares, not the difference
of squares, so it can’t be factored. This polynomial is a
prime polynomial.
Difference of Two Squares
Example:
Factor 36x 2 – 64.
Remember that you should always factor out any
common factors, if they exist, before you start any other
technique.
Factor out the GCF.
36x 2 – 64 = 4(9x 2 – 16)
Since the polynomial can be written as (3x)2 – (4)2,
(9x 2 – 16) = (3x – 4)(3x + 4).
So our final result is 36x 2 – 64 = 4(3x – 4)(3x + 4).
Difference of Two Squares
Example:
Recognize a difference of two squares: the coefficients of
variable terms are perfect squares, powers on variable terms
are even, and constants are perfect squares.
Reading Math
Recognizing D.O.T.S.
Determine whether each binomial is a difference of two squares. If
so, factor. If not, explain.
3p 2 – 9q 4
3p 2 – 9q 4
3q 2 3q 2 3p2 is not a perfect square.
3p 2 – 9q 4 is not the difference of two squares because 3p 2 is
not a perfect square.
Your Turn:
Determine whether each binomial is a difference of two squares. If
so, factor. If not, explain.
100x 2 – 4y 2
Write the polynomial as
(a + b)(a – b).
a = 10x, b = 2y
The polynomial is a difference
of two squares.
100x 2 – 4y 2
2y 2y 10x 10x
(10x)2 – (2y)2
(10x + 2y)(10x – 2y)
100x2 – 4y2 = (10x + 2y)(10x – 2y)
Your Turn:
Determine whether each binomial is a difference of two squares. If
so, factor. If not, explain.
x 4 – 25y 6
Write the polynomial as
(a + b)(a – b).
a = x2, b = 5y3
The polynomial is a difference
of two squares.
(x 2)2 – (5y 3)2
(x 2 + 5y 3)(x 2 – 5y 3)
x 4 – 25y 6 = (x 2 + 5y 3)(x 2 – 5y 3)
5y 3 5y 3 x 2 x 2
x 4 – 25y 6
Your Turn:
Determine whether each binomial is a difference of two squares.
If so, factor. If not, explain.
1 – 4x 2
Write the polynomial as
(a + b)(a – b).
a = 1, b = 2x
The polynomial is a
difference of two squares.
(1) – (2x)2
(1 + 2x)(1 – 2x)
1 – 4x 2 = (1 + 2x)(1 – 2x)
2x 2x 1 1
1 – 4x 2
Your Turn:
Determine whether each binomial is a difference of two squares.
If so, factor. If not, explain.
p 8 – 49q 6
Write the polynomial as
(a + b)(a – b).
a = p4, b = 7q3
The polynomial is a
difference of two squares.
(p 4)2 – (7q 3)2
(p 4 + 7q 3)(p 4 – 7q 3)
p 8 – 49q 6 = (p 4 + 7q 3)(p 4 – 7q 3)
7q 3 7q 3 ● p 4 p 4 ●
p 8 – 49q 6
Your Turn:
Determine whether each binomial is a difference of two squares. If
so, factor. If not, explain.
16x 2 – 4y 5
4x 4x 4y5 is not a perfect square.
16x 2 – 4y 5 is not the difference of two squares because 4y 5 is
not a perfect square.
16x 2 – 4y 5
Your Turn:
Review: D.O.T.S.
Practice
• Pg. 493-474 # 1-43 odd, #55
Factoring by Grouping
Section 9-8
Definition
Factoring by Grouping
– Using the distributive property to factor polynomials
with four or more terms.
– terms can be put into groups and then factored---- each
group will have a “like” factor used in regrouping.
Factoring by Grouping
• Polynomials with four or more terms like 3xy –
21y + 5x – 35, can sometimes be factored by
grouping terms of the polynomials.
• One method is to group the terms into binomials
that can be factored using the distributive
property.
• Then use the distributive property again with a
binomial as the common factor.
A polynomial can be factored by grouping if all of the
following conditions exist.
1. There are four or more terms.
2. Terms have common factors that can be grouped
together, and
3. There are two common factors that are identical.
Symbols: ax + bx + ay + by = (ax + bx) + (ay + by)
= x(a + b) + y(a + b)
= (x + y)(a + b)
Group, factor
Regroup
Factor by Grouping
Factor each polynomial by grouping. Check your
answer.
6h4 – 4h3 + 12h – 8
(6h4 – 4h3) + (12h – 8)
2h3(3h – 2) + 4(3h – 2)
2h3(3h – 2) + 4(3h – 2)
(3h – 2)(2h3 + 4)
Group terms that have a common number
or variable as a factor.
Factor out the GCF of each group.
(3h – 2) is another common factor.
Factor out (3h – 2).
Example: Factor by
Grouping
Check your answer.
Check (3h – 2)(2h3 + 4) Multiply to check your solution.
3h(2h3) + 3h(4) – 2(2h3) – 2(4)
6h4 + 12h – 4h3 – 8
The product is the original
polynomial. 6h4 – 4h3 + 12h – 8
Example: Continued
Factor each polynomial by grouping.
5y4 – 15y3 + y2 – 3y
(5y4 – 15y3) + (y2 – 3y)
5y3(y – 3) + y(y – 3)
5y3(y – 3) + y(y – 3)
(y – 3)(5y3 + y)
Group terms.
Factor out the GCF of each
group.
(y – 3) is a common factor.
Factor out (y – 3).
Example: Factor by
Grouping
Factor each polynomial by grouping.
6b3 + 8b2 + 9b + 12
(6b3 + 8b2) + (9b + 12)
2b2(3b + 4) + 3(3b + 4)
2b2(3b + 4) + 3(3b + 4)
(3b + 4)(2b2 + 3)
Group terms.
Factor out the GCF of each
group.
(3b + 4) is a common factor.
Factor out (3b + 4).
Your Turn:
Factor each polynomial by grouping.
4r3 + 24r + r2 + 6
(4r3 + 24r) + (r2 + 6)
4r(r2 + 6) + 1(r2 + 6)
4r(r2 + 6) + 1(r2 + 6)
(r2 + 6)(4r + 1)
Group terms.
Factor out the GCF of each
group.
(r2 + 6) is a common factor.
Factor out (r2 + 6).
Your Turn:
If two quantities are opposites, their sum is 0
(Additive Inverses).
(5 – x) + (x – 5)
5 – x + x – 5
– x + x + 5 – 5
0 + 0
0
Helpful Hint
Recognizing opposite binomials can help you factor
polynomials. The binomials (5 – x) and (x – 5) are
opposites. Notice (5 – x) can be written as –1(x – 5).
–1(x – 5) = (–1)(x) + (–1)(–5)
= –x + 5
= 5 – x
So, (5 – x) = –1(x – 5)
Distributive Property.
Simplify.
Commutative Property of
Addition.
Additive Inverse
Factor 2x3 – 12x2 + 18 – 3x
2x3 – 12x2 + 18 – 3x
(2x3 – 12x2) + (18 – 3x)
2x2(x – 6) + 3(6 – x)
2x2(x – 6) + 3(–1)(x – 6)
2x2(x – 6) – 3(x – 6)
(x – 6)(2x2 – 3)
Group terms.
Factor out the GCF of each
group.
Simplify. (x – 6) is a common
factor.
Factor out (x – 6).
Write (6 – x) as –1(x – 6).
Example: Factoring with
Opposite Groups
Factor each polynomial. Check your answer.
15x2 – 10x3 + 8x – 12
(15x2 – 10x3) + (8x – 12)
5x2(3 – 2x) + 4(2x – 3)
5x2(3 – 2x) + 4(–1)(3 – 2x)
5x2(3 – 2x) – 4(3 – 2x)
(3 – 2x)(5x2 – 4)
Group terms.
Factor out the GCF of each
group.
Simplify. (3 – 2x) is a
common factor.
Factor out (3 – 2x).
Write (2x – 3) as –1(3 – 2x).
Your Turn:
Factor each polynomial. Check your answer.
8y – 8 – x + xy
(8y – 8) + (–x + xy)
8(y – 1)+ (x)(–1 + y)
(y – 1)(8 + x)
8(y – 1)+ (x)(y – 1)
Group terms.
Factor out the GCF of each
group.
(y – 1) is a common factor.
Factor out (y – 1) .
Your Turn:
Factor each polynomial by grouping.
1. 2x3 + x2 – 6x – 3
2. 7p4 – 2p3 + 63p – 18
(7p – 2)(p3 + 9)
(2x + 1)(x2 – 3)
Your Turn:
Factoring Procedure
Methods of Factoring
Assignment
• Pg. 499-500 # 2-38 even, #46