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Transcript
Algebra 1 Key Vocabulary Words
Chapter 9
Section 9.1: monomial
degree of monomial
polynomial
degree of polynomial
leading coefficient
binomial
trinomial
Section 9.2: algebra unit-tile (not a Key Vocabulary)
algebra x-tile (not a Key Vocabulary)
algebra x2-tile (not a Key Vocabulary)
distributive property of multiplication (see Ch 2.5)
FOIL pattern
area of rectangle (not a Key Vocabulary)
square of a binomial
difference of two squares (sum and difference pattern)
Section 9.3: perfect square trinomial
difference of two squares
Section 9.4: roots
solutions of quadratic equation
vertical motion
zero-product property
Section 9.5: zero of a function
Section 9.6: Trinomial (see Ch 9.1)
Section 9.7: perfect square trinomial (see Ch 9.3)
Section 9.8: factor completely
common factor
product of binomial factors
Permission for Use Granted by Dr. Kate Kinsella 10/08
Word
Monomial
9.1
Meaning
DEF:
A monomial is a
number, a variable, or the
product of a number and
one or more variables
with whole number
exponents.
Examples
Oral Practice
The following are monomials:
x
m2
The following are examples of polynomials: y3,
5mn, 4 and p.
____ and ____ are also examples of monomials.
-2xy
5
-2b2c
Writing Practice:
The following are examples of monomials: 3c, -4f2, 5xy, and 6.
The following are examples of monomials: ________________.
Degree of
Monomial
9.1
The Degree of a
Monomial is the sum of
the exponents of the
variables in the
monomial.
Monomial
Degree
x
1
m3
3
2xy
2
-3v12u2
14
5
0
The degree of the monomial 2x3 is 3 because the
exponent of the variable x is 3.
The degree of the monomial 2xy4 is 5 because the
exponent of the variable x is 1 and the exponent
of the variable y is 4 and 4 + 1 = 5
The degree of the monomial 12p3k4 is 7 because
the exponent of the variable p is 3 and the
exponent of the variable k is 4 and 3 + 4 = 7
Writing Practice:
The degree of the monomial -x7 is 7 because the exponent of the variable x is 7.
The degree of the monomial ______ is _____ because _______________________.
Permission for Use Granted by Dr. Kate Kinsella 10/08
Word
Polynomial
9.1
Meaning
A polynomial is a
monomial or a sum of
monomials.
Examples
Oral Practice
Below are examples of polynomials
2x3 – 3 is a polynomial.
2x (monomial)
-x + 3y – 5 is a polynomial
3x + y (sum of 2 monomials)
_______________ is a polynomial,
5m2 – 3m + 2 (sum of 3 monomials)
Writing Practice:
3f – 5 is a polynomial.
_______ is a polynomial.
Degree of
Polynomial
9.1
The Degree of a
Polynomial is the
greatest degree of its
terms.
Polynomial
Degree
x+5
1
m3 – 2m2 + 3
3
2xy + 5z5 – 1
5
-3v12 + 6u4v10 + w12
14
2x2 + 3xyz
3
The polynomial 3x4 – 5x3 + 2x3 – 1 is a 4th degree
polynomial.
The polynomial 2xy – y + 1 is a 2nd degree
polynomial.
The polynomial ______________ is a ___ degree
polynomial.
Writing Practice:
The polynomial 2x5 – 3x2 – 1 is a 5th degree polynomial.
The polynomial __________ is a ___ degree polynomial.
Permission for Use Granted by Dr. Kate Kinsella 10/08
Word
Leading Coefficient
9.1
Meaning
DEF:
When a polynomial is
written so that the
exponents of a variable
decrease from left to
right, the coefficient of
the first term is the
leading coefficient.
Examples
Oral Practice
The leading coefficient of -2x5 + 3x3 –
2x + 5 is –2.
The leading coefficient of 4v3 – 2v2 + v – 7 is 4.
The leading coefficient of 5m4 – 3m2 is
5.
The leading coefficient of 2x3 – 5x + 1 is ___.
Writing Practice:
The leading coefficient of 8m2 – 5m + 1 is 8.
The leading coefficient of ______________ is ____.
Binomial
9.1
DEF:
A polynomial with two
terms is called a
binomial.
The following are examples of
binomials.
Since the polynomial 2x + 8 has two terms then
it is a binomial.
2x + 3
-5mn + 6n
Since the polynomial 3m + 5n has _______ then
it is a _________.
x+y
-2x3 + 6xy
Writing Practice:
Since the polynomial –5v + 7 has two terms then it is a binomial.
Since the polynomial _______ has two terms then it is a ________.
Permission for Use Granted by Dr. Kate Kinsella 10/08
Word
Trinomial
9.1
Meaning
A polynomial with three
terms is called a
trinomial.
Examples
The following are examples of
trinomials
Oral Practice
Since the polynomial 2x – y + 3 has three terms
then it is a trinomial.
2x + 3y – 1
-5mn + 6n + 7
Since the polynomial ________ has ____ terms
then it is a _______.
x+y+z
-2x3 + 6xy – 7
Writing Practice:
Since the polynomial 5m + 6n – 1 has three terms then it is a trinomial.
Since the polynomial ________ has ____ terms then it is a _______.
FOIL Pattern
9.2
This pattern which means
F – first, O – outer, I –
inner, and L – last
helps you remember how
to use the distributive
property to multiply
binomials.
To get the product we use FOIL pattern
described as:
F
L
For the factors (x – 6)(x + 7), identify the FOIL
Products
First term products: (x)(x)
Outer term products: (x)(7)
Inner term products: (-6)(x)
Last term products: (-6)(7)
(2x + 3) (x + 4)
I
O
First term products: (2x)(x)
Outer term products: (2x)(4)
Inner term products: (3)(x)
Last term products: (3)(4)
For the factors (3x + 5)(x – 2), identify the FOIL
Products
First term products: (3x)(x)
Outer term products: (3x)(-2)
Inner term products: (5)(x)
Last term products: (5)(-2)
Writing Practice:
For the factors (x + 1)(x + 8), identify the FOIL Products. F – (x)(x) O – (x)(8) I - (1)(x) L – (1)(8)
For the factors (_______)(_______) identify the FOIL Products
Permission for Use Granted by Dr. Kate Kinsella 10/08
Word
Square of a
Binomial
9.2
Meaning
DEF:
Square of a binomial
Pattern:
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
Examples
To use the pattern, identify the 1st term
a and the 2nd term b. Plug into the
formula and simplify:
2
(2x + 3)
a = 2x and b = 3
(2x)2+2(2x)(3) + (3)2= 4x2 + 12x + 9
PRODUCT
Oral Practice
Use the pattern to get the product
(x – 5)2
a = x and b = -5
(x)2+2(x)(-5) + (-5)2= x2 – 10x + 25
PRODUCT
Use the pattern to get the product
(2x – 7)2
a = 2x and b = -7
( )2+2( )( ) + ( )2= ____________
PRODUCT
Writing Practice:
Find the product: (3x + 5)2 = 9x2 + 30x + 25
Find the product: (________)2 = _____________
Difference of 2
Squares
9.2
DEF:
(a + b)(a – b) = a2 – b2
To use the pattern, identify the 1st term
a and the 2nd term b and make sure
that they assume the (a + b)(a – b)
form.
(2x + 3)(2x – 3) = ?
a = 2x and b = 3
To use the pattern, identify the 1st term a and the
2nd term b and make sure that they assume the
(a + b)(a – b) form.
(x – 3)(x + 3) = ?
a = x and b = 3
Plug into a and b and simplify:
Plug into a and b and simplify:
2
2
2
(2x) – (3) = 4x – 9 (product)
(x)2 – (3)2 = x2 – 9 (product)
Find the product using the pattern:
(2x + 7)(2x – 7) = ( )2 – (
)2 = ________
Writing Practice:
Find the product using the pattern:
(x + 9)(x – 9) = (x)2 – (9)2 = x2 – 81
(_______)(_______) = ( )2 – (
)2 = ________
Permission for Use Granted by Dr. Kate Kinsella 10/08
Word
Perfect Square
Trinomial (PST)
9.3
Meaning
Examples
DEF:
a2 + 2ab + b2 = (a + b)2
2
2
a – 2ab + b = (a – b)
2
The following is a perfect square
trinomial:
4x2 + 4x + 1 because the middle term
is a product of the square root of the
first and last term and 2.
Square root of 4x2 is 2x and square
root of 1 is 1
Oral Practice
The trinomial x2 – 6x + 9 is a PST because the
middle term is 2(x)(3) and therefore can be
factored as (x – 3)2
The trinomial __________ is a PST because the
middle term is _____ and therefore can be
factored as _________.
2(2x)(1) = 4x, therefore can be
factored as
(2x + 1)2
Writing Practice:
The trinomial 9x2 – 12x + 4 is a PST because the middle term is 2(3x)(2) and therefore can be factored as (3x – 2)2
The trinomial ____________is a PST because the middle term is _______ and therefore can be factored as ________.
Roots
9.4
DEF:
The solutions of
equations. (Such as an
equation where one side
is zero and the other side
is a product of polynomial
factors)
For the quadratic equation
For the quadratic equation
(x + 3)(x – 2) = 0
(x + 5)(x + 7) = 0
Since x = -5 or x = -7
Since x = -3 or x = 2
The roots of the equation are –3
and 2
The roots of the equation are –5 and –7
For the quadratic equation
(x – 8)(x + 5) = 0
Since x = 8 or x = -5
The roots of the equation are _________.
Writing Practice:
(x)(x + 2) = 0, Since x = 0 or x = -2. The roots of the equation are 0 and –2.
______________ = 0, Since ______ or ______. The roots of the equation are __________.
Permission for Use Granted by Dr. Kate Kinsella 10/08
Word
Solutions of
Quadratic Equation
9.4
Meaning
DEF:
The solutions of a
quadratic equation the
values of the x that
satisfies the equation. It
varies from none, one or
two solutions.
Examples
Here are examples of Quadratic
Equations and their Solutions:
x2 – 1 = 0
The solutions of this quadratic equation
are x = 1. When these values are
plugged into the equation, it makes a
true statement.
Oral Practice
x2 – 2x + 1 = 0
The solution of this quadratic equation is x = 1.
When 1 is plugged in the equation it makes the
equation true.
x2 – 4x + 3 = 0
The solution of this quadratic equation is _____
When ____ is plugged in the equation it makes
the equation true.
Writing Practice:
The solution to x2 – 9 = 0 is 3. When these values are plugged in the equation, it makes the equation true.
The solution to ________ is ___. When these values are plugged in the equation, it makes the equation true.
The Vertical Motion Model
Vertical Motion
DEF:
An equation for an object thrown upwards with an
Model
The height h (in feet) of a projectile can initial velocity of 11 ft/sec is given as:
9.4
The Vertical Motion
be modeled by:
Let v = 10 ft/sec and s = 0
Model can describe the
path of a projectile. A
h = -16t2 + vt + s
h = -16t2 + (11)t + (0)
projectile is an object
h = -16t2 + 11t
thrown into the air but
Where t is time in seconds the object
has no power to keep
has been in the air, v is the initial
itself in the air.
velocity, and s is the initial height in
An equation for an object thrown upwards with an
feet.
initial velocity of ________ is given as:
Let v = _______ and s = 0
The model can be used to find an
equation that gives the height of an
h = -16t2 + (___)t + (__)
object as a function of time (in
h = ______________
seconds) given the initial velocity.
(s = 0 at this point)
Writing Practice:
An equation for an object thrown upwards with an initial velocity of 12 ft/sec is given as: h = -16t2 + 12t
An equation for an object thrown upwards with an initial velocity of _______ is given as: _____________
Permission for Use Granted by Dr. Kate Kinsella 10/08
Word
Meaning
Zero-Product
Property
9.4
DEF:
Let a and b be real
numbers. If ab = 0, then
a = 0 or b = 0.
Examples
The zero product property states that:
If (x +3)(x – 2) = 0,
Oral Practice
Fill in the correct conclusion:
If (2x – 5)(x + 8) = 0,
Then (2x – 5)= 0 or (x + 8) = 0
Then (x +3) = 0 or (x – 2) = 0
Fill in the correct conclusion:
If (x – 9)(3x – 7) = 0,
Then ______________________
Writing Practice:
If (x + 5)(x + 7) = 0, then (x + 5) = 0 or (x + 7) = 0
If (
)(
) = 0, then _____________________
Zero of a Function
9.5
DEF:
The zero of a function
y = f(x) is a value of x for
which f(x) = 0. (or y = 0)
Given:
Given:
F(x) = x – 5
F(x) = 2x – 1
Setting F(x) = 0 and solving for x, the value of
x=5
Therefore, 5 is a zero of the function because
F(5) = 0
Setting F(x) = 0 and solving for x, the
value of x = 1/2
Therefore, 1/2 is a zero of the
function because F(1/2) = 0
F(x) = 2x + 6
Setting F(x) = 0 and solving for x, the value of
x = -3
Therefore, ____________________ because
________
Writing Practice:
If F(-2) = 0, then –2 is a zero of the function.
If _______, then ___ is a zero of the function.
Permission for Use Granted by Dr. Kate Kinsella 10/08
Word
Factor Completely
9.8
Meaning
DEF:
A factorable polynomial
with integer coefficients is
said to be factored
completely if it is
written as a product of
unfactorable polynomials
with integer coefficients.
Examples
Given 2x2 – 5x – 12 = (2x + 3)(x – 4)
Since it is written as (2x + 3)(x – 4)
with integer coefficients, then the
polynomial 2x2 – 5x – 12 is factored
completely.
Oral Practice
Given 15x2 + 11x + 2 = (3x + 1)(5x + 2)
Since it is written as (3x + 1)(5x + 2) with
integer coefficients, then the polynomial
15x2 + 11x + 2 is factored completely.
Given 5x2 – 14x – 3 = (x – 3)(5x + 1)
Since it is written as (x – 3)(5x + 1) with
integer coefficients, then the polynomial
5x2 – 14x – 3 is factored completely.
Writing Practice:
Given 2x2 + 5x – 3 = (x + 3)(2x – 1). Since the factors have integer coefficients, then it is factored completely.
Given __________ = ____________. Since the factors have _____________, then ___________________.
Common Factor
9.8
DEF:
A common factor is a
whole number that is a
factor of each number.
Given the numbers and their factors:
25: 1, 5, 25
10: 1, 2, 5, 10
The number 5 is a common factor.
Given the numbers and their factors:
12: 1, 2, 3, 4, 6, 12
20: 1, 2, 4, 5, 10, 20
The numbers 3 and 4 are common factors.
Given the numbers and their factors:
24: 1, 2, 3, 4, 6, 8, 12, 24
36: 1, 2, 3, 4, 6, 9, 12, 36
The numbers _____________ are common
factors.
Writing Practice:
The common factor(s)of 12 and 36 are 2, 3, 4, 6, and 12
The common factor(s)of ____ and ____ are _____________
Permission for Use Granted by Dr. Kate Kinsella 10/08
Word
Product of
Binomial Factors
9.8
Meaning
DEF:
If a polynomial is factored
completely and the
factors are written as a
binomial, the factors are
said to be written as
product of binomial
factors.
Examples
The factors of 15x2 + 11x + 2 are
written as product of binomial
factors as (3x + 1)(5x + 2) because
the factors are binomials.
Oral Practice
The factors of 5x2 – 14x – 3 are written as
product of binomial factors as (x – 3)(5x + 1)
because the factors are binomials.
The factors of ___________are written as
product of binomial factors as ____________
because the factors are binomials.
Writing Practice:
The factors of 2x2 + 5x – 3 are written as product of binomial factors as (x + 3)(2x – 1) because the factors are binomials.
The factors of __________ are written as product of binomial factors as ____________ because the factors are binomials.
Permission for Use Granted by Dr. Kate Kinsella 10/08