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Section 5.1
Polynomials Addition And Subtraction
OBJECTIVES
A Classify polynomials.
OBJECTIVES
B Find the degree of a polynomial and write descending order.
OBJECTIVES
C Evaluate a polynomial.
OBJECTIVES
D Add or subtract polynomials.
OBJECTIVES
E Solve applications involving sums or differences of polynomials.
DEFINITION
Degree of a Polynomial in One Variable
The degree of a polynomial in one variable is the greatest exponent of that variable.
DEFINITION
Degree of a Polynomial in Several Variables
The greatest sum of the exponents of the variables in any one term of the polynomial.
RULES
Properties for Adding Polynomials
P + Q = Q + P If P , Q, and R are polynomials,
Commutative Property
RULES
If P , Q, and R are polynomials,
P + Q + R = P + Q + R
Associative Property
Properties for Adding Polynomials
RULES
If P , Q, and R are polynomials,
P Q + R = PQ + PR
Q + R P = QP + RP
Properties for Adding Polynomials
RULES
Subtracting Polynomials
a – (b + c) = a – b – c
Section 5.1A,B
Exercise #1
Chapter 5
Classify as a monomial, binomial, or trinomial and give the degree.
xy3z4 – x7
Binomial.
Degree is determined by comparing
Degree 1st term: x1 y3 z4
1 + 3 + 4 = 8
Degree 2nd term: x7 7
Degree 8
Section 5.1D
Exercise #5
Chapter 5
Add 6x3 + 8x2 – 6x – 4 and 6 – 3x + x2 – 3x3 .
6x3 + 8x2 – 6x – 4
– 3x3 + x2 – 3x + 6
3x3
METHOD 1
+ 9x2 – 9x + 2
3 2 2 36 + 8 – 6 – 4 6 – 3 + – 3x x x + x x x
= 6x3 + 8x2 – 6x – 4 + 6 – 3x + x2 – 3x3
= 3x3 + 9x2 – 9x + 2
Add 6x3 + 8x2 – 6x – 4 and 6 – 3x + x2 – 3x3 .
= 6x3 – 3x3 + 8x2 + x2 – 6x – 3x – 4 + 6
METHOD 2
Section 5.1D
Exercise #6
Chapter 5
Subtract 8x3 – 6x2 + 5x – 3 from 5x3 + 3x2 + 3.
5x3 + 3x 2 + 3 –
METHOD 1
8x3 – 6x2 + 5x – 3
Subtract 8x3 – 6x2 + 5x – 3 from 5x3 + 3x2 + 3.
5x3 + 3x 2 + 3 –
5x3 + 3x 2 + 3
METHOD 1
8x3 – 6x2 + 5x – 3
+
Subtract 8x3 – 6x2 + 5x – 3 from 5x3 + 3x2 + 3.
5x3 + 3x 2 + 3 –
– 8x3 + 6x2 – 5x + 3
– 3x3
5x3 + 3x 2 + 3
METHOD 1
8x3 – 6x2 + 5x – 3
+ 9x2 – 5x + 6
+
3 2 3 25 + 3 + 3 – 8 – 6 + 5 – 3x x x x x
= – 3x3 + 9x2 – 5x + 6
= 5x3 + 3x 2 + 3 – 8x3 + 6x 2 – 5x + 3
Subtract 8x3 – 6x2 + 5x – 3 from 5x3 + 3x2 + 3.
= 5x3 – 8x3 + 3x2 + 6x2 – 5x + 3 + 3
METHOD 2
Section 5.2
Multiplication of Polynomials
OBJECTIVES
A Multiply a monomial by a polynomial.
OBJECTIVES
B Multiply two polynomials.
OBJECTIVES
C Use the FOIL method to multiply two binomials.
OBJECTIVES
D Square a binomial sum or difference.
OBJECTIVES
E Find the product of the sum and the difference of two terms.
OBJECTIVES
F Use the ideas discussed to solve applications.
RULES
Multiplication of Polynomials
If P , Q, and R are polynomials,
P Q = Q P P Q R = P Q R
To Multiply Two Binomials
USING FOIL
= x 2 + (b + a)x + ab
(x + a)(x + b)
= x 2 + bx + ax + ab First Outside Inside Last
RULETo Square a Binomial Sum
(x + a)2
= (x + a)(x + a)
= x 2 + 2ax + a2
RULETo Square a Binomial Difference
(x – a)2
= (x – a)(x – a)
= x 2 – 2ax + a2
Sum and Difference of Same Two Monomials
PROCEDURE
(x + a)(x – a) = x 2 – a2
(x – a)(x + a) = x 2 – a2
Section 5.2B,C
Exercise #8a
Chapter 5
2
M
ultip
– 2 – 4 – 5
ly.
x x x
x2 – 4x – 5
x3 – 4x2 – 5x
x3 – 6x2 + 3x + 10
x – 2
– 2x2 + 8x + 10
METHOD 1
2 2= – 4 – 5 + – 2 – 4 – 5x x x x x
= x3 – 4x2 – 5x – 2x2 + 8x + 10
= x3 – 4x2 – 2x2 – 5x + 8x + 10
= x3 – 6x2 + 3x + 10
2
M
ultip
– 2 – 4 – 5
ly.
x x x
METHOD 2
Section 5.2D
Exercise #9b
Chapter 5
Multiply.
= 9x2
23 – 4x y
2 2= 3 – 2(3 )(4 ) + 4x x y y
2 2 2 – = – 2 + x a x ax a
– 24xy + 16y2
Section 5.2E
Exercise #10
Chapter 5
Multiply.
2 2= 3 – 4x y
= 9x2 – 16y2
3 + 4 3 – 4x y x y
2 2 + – = – x a x a x a
Product of Sum and Difference ofSame Two Monomials
Section 5.3
The Greatest Common Factor and Factoring by Grouping
OBJECTIVES
A Factor out the greatest common factor in a polynomial.
OBJECTIVES
B Factor a polynomial with four terms by grouping.
GREATEST COMMON FACTOR
is the Greatest Common monomial Factor (GCF) of a polynomial in x if:
axn
1. a is the greatest integer that divides each coefficient.
GREATEST COMMON FACTOR
is the Greatest Common monomial Factor (GCF) of a polynomial in x if:
axn
2. n is the smallest exponent of x in all the terms.
PROCEDURE
Factoring by Grouping
1. Group terms with common factors using the associative property.
PROCEDURE
2. Factor each resulting binomial.
Factoring by Grouping
PROCEDURE
3. Factor out the binomial using the GCF, by the distributive property.
Factoring by Grouping
Section 5.3B
Exercise #12
Chapter 5
Factor completely.
2 5 3 2 = 3 2 + 2 + 5 + 5x x x x
6x7 + 6x5 + 15x4 + 15x2
2 3 2 2 = 3 2 + 1 + 5 + 1 x x x x
2 2 3 = 3 + 1 2 + 5 x x x
2 2 3 = 3 + 1 2 + 5x x x
Section 5.4
Factoring Trinomials
OBJECTIVES
A Factor a trinomial of the form . x 2 + bx + c
OBJECTIVES
B Factor a trinomial of the form using trial and error.
ax 2 + bx + c
OBJECTIVES
C Factor a trinomial of the form using the ac test.
ax 2 + bx + c
PROCEDURE
Factoring Trinomials
x2 + (b + a)x + ab
= (x + a)(x + b)
RULE
The ac Test
is factorable only if there are two integers whose product is ac and sum is b.
ax 2 + bx + c
Section 5.4A,B,C
Exercise #13b
Chapter 5
Factor completely.
= 2 + 5 – 2x y x y
2x2 + xy – 10y2The ac Method Find factors of ac (–20) whose sum is (1) and replace the middle term (xy).
= 2x2 – 4xy + 5xy – 10y2
= (2x 2 – 4xy) + (5xy – 10y 2)
= 2x(x – 2y) + 5y(x – 2y)
Section 5.5
Special Factoring
OBJECTIVES
A Factor a perfect square trinomial.
OBJECTIVES
B Factor the difference of two squares.
OBJECTIVES
C Factor the sum or difference of two cubes.
PROCEDURE
Factoring Perfect Square Trinomials
x 2 + 2 ax + a2 = (x + a)2
x 2 – 2 ax + a2 = (x – a)2
PROCEDURE
Factoring the Difference of Two Squares
x2 – a2 = (x + a)(x – a)
PROCEDURE
Factoring the Sum and Difference of Two Cubes
x3+ a3 = (x + a)(x 2 – ax + a2 )
x3 – a3 = (x – a)(x 2+ ax + a2 )
Section 5.5A
Exercise #15a
Chapter 5
Factor completely.
16x2 – 24xy + 9y2
Perfect Square Trinomial
2 = 4 – 3x y
22 2 – 2 + = – x ax a x a
2 2 = 16 – 2 12 + 9x xy y
Section 5.5
Exercise #16
Chapter 5
Factor completely.
22 2 2= + 4 – 4x y x y
x4 – 16y4
Difference of Two Squares
Factor x2 – 4y2
2 2= + 4 + 2 – 2x y x y x y
2 2 – = + – x a x a x a
2 22 2 = – 4x y
Section 5.5B
Exercise #17
Chapter 5
Factor completely.
x2 – 10x + 25 – y2
Perfect Square Trinomial
2 2 = – 5 – x y
= – 5 + – 5 – x y x y
= + – 5 – 5 x y x – y
x2 – 2ax + a2 = x – a 2
Difference of Two Squares
2 2 – = + – x a x a x a
Section 5.5c
Exercise #18a
Chapter 5
Factor completely.
2 = 3 + 2 9 x y x
27x3 + 8y3
3 3 = 3 + 2x y
Sum of Two Cubes
3 3 2 2 + = + – + x a x a x ax a
– 6xy + 4y2
Section 5.6
General Methods of Factoring
OBJECTIVES
A Factor a polynomial using the procedure given in the text.
PROCEDUREA General Factoring Strategy
1. Factor out the GCF, if there is one.
2. Look at the number of terms in the given polynomial.
PROCEDUREA General Factoring Strategy
If there are two terms, look for:
Difference of Two Squaresx 2 – a2 = (x + a)(x – a)
PROCEDUREA General Factoring Strategy
If there are two terms, look for:
Difference of Two Cubesx3 – a3 = (x – a)(x 2
+ ax + a2 )
PROCEDUREA General Factoring Strategy
If there are two terms, look for:
Sum of Two Cubesx3+ a3 = (x + a)(x 2 – ax + a2 )
PROCEDUREA General Factoring Strategy
If there are two terms, look for:
The sum of two squares, is not factorable. x 2 + a2
PROCEDUREA General Factoring Strategy
If there are three terms, look for:
Perfect square trinomial
x 2 + 2ab + b2 = (x + a)2
x 2 – 2ab + b2 = (x – a)2
PROCEDUREA General Factoring Strategy
If there are three terms, look for:
Trinomials of the form
ax 2 + bx + c (a > 0)
PROCEDUREA General Factoring Strategy
If a < 0, factor out – 1 first.
Use the ac method or trial and error.
PROCEDUREA General Factoring Strategy
If there are four terms:
Factor by grouping.
PROCEDUREA General Factoring Strategy
3. Check the result by multiplying the factors.
Section 5.6A
Exercise #20b
Chapter 5
Factor completely.
2 2 2 = 3 16 – 24 + 9y x xy y
48x2y2 – 72xy3 + 27y4
22 = 3 4 – 3y x y
Perfect Square Trinomial
22 2 – 2 + = – x ax a x a
2 22 = 3 4 – 2 12 + 3 y x xy y
Section 5.6A
Exercise #21
Chapter 5
Factor completely: 9x3y – 33x2y2 – 12xy3
= 3xy [3x2 – 12xy + xy – 4y2]
The ac Method Find factors of ac (–12) whose sum is (–11) and replace the middle term (–11xy).
2 2 = 3 3 – 12 + – 4xy x xy xy y [ ] = 3 3 – 4 + – 4xy x x y y x y [ ]
= 3 – 4 3 + xy x y x y [ ]
= 3xy [3x2 – 11xy – 4y2]
Factor completely: 9x3y – 33x2y2 – 12xy3
= 3xy [3x2 – 12xy + xy – 4y2]
= 3 3 + – 4xy x y x y
The ac Method Find factors of ac (–12) whose sum is (–11) and replace the middle term (–11xy).
2 2 = 3 3 – 12 + – 4xy x xy xy y [ ] = 3 3 – 4 + – 4xy x x y y x y [ ]
= 3xy [3x2 – 11xy – 4y2]
Section 5.6A
Exercise #22
Chapter 5
Factor completely.
16x3 – 12x2 – 4x + 3
2 = 4 – 3 4 – 1x x
= 4 – 3 2 + 1 2 – 1x x x
2 = 4 4 – 3x x
Difference of Two Squares
2 2 – = + – x a x a x a
– 1 4 – 3x
Section 5.7
Solving Equations by Factoring: Applications
OBJECTIVES
A Solve equations by factoring.
OBJECTIVES
B Use Pythagorean theorem to find the length of one side of a right triangle when the lengths of the other two sides are given.
OBJECTIVES
C Solve applications involving quadratic equations.
PROCEDURE
1. Set equation equal to 0.
2. Factor Completely.
3. Set each linear Factor equal to 0 and solve each.
O
F
F
DEFINITION
Pythagorean Theorem
In symbols,c2 = a2 + b2
a
b
c
Section 5.7A
Exercise #23b
Chapter 5
Solve. 6x2 + 7x = 3
3 – 1 2 + 3 = 0x x
3x = 1
6x2 + 7x – 3 = 0
x =
13
, –32
3x – 1 = 0 2x + 3 = 0or
2x = – 3
x =
13
x = – 32
or
O
F
F
