#### Transcript Polynomial Functions algebra 2 notes chapter 6

```Properties of Exponents
• Product of Powers
Power of a Power
Power of a Product
Negative Exponents
Zero Exponent
Quotient of Powers
Power of a quotient
a m  a n  a m n
(a m ) n  a mn
(ab) m  a mb m
1
a m  m , a  0
a
a 0  1, a  0
am
mn

a
,a  0
n
a
m
am
a
   m ,b  0
b
b
(23 ) 5
3
 
5
Examples
2
(2) 3 (2) 5
 r 
 5 
s 
2
(7b 3 ) 2 b 5b
( xy 2 ) 2
x 3 y 1
Polynomial Functions
-exponents are whole numbers
-coefficients are real numbers
f ( x)  2 x  5 x  2 x  x  7
4
3
2
4 is the degree ( the highest exponent)
-7 is the constant term
The degree of a polynomial function is
the exponent of the leading term
when it is in standard form
•
•
•
•
•
•
Degree
0
1
2
3
4
type
Constant
Linear
Cubic
Quartic
You can evaluate polynomial functions
using
-direct substitution
-synthetic substitution
• Evaluate
f ( x)  2 x 4  8 x 2  5 x  7
EVALUATING A FUNCTION
given a value for x
DIRECT SUBSTITUTION:
- replace each x with the given value
- evaluate expression, following PEMDAS
Example: f(x) = 2x⁴ - 8x² + 5x – 7, for x = 3
2(3)⁴ - 8(3)² + 5(3) - 7
2(81) – 8(9) + 5(3) - 7
162 – 72 + 15 -7
98
SYNTHETIC SUBSTITUTION:
- write polynomial expression in standard form (include all degree terms)
- write only coefficients (including zeros)
-use the given value of x in the process below
Example: f(x) = 2x⁴ - 8x² + 5x – 7, for x = 3
2x⁴ + 0x³- 8x² + 5x – 7
2
0
-8
5
-7
2
6
6
18
10
30
35
105
98
X=3
The solution is the last number written.
End Behavior of Polynomial Functions
is determined by the degree (n) and leading coefficient (a)
Use your graphing calculator to investigate the end
behavior of several polynomial functions. Write a
paragraph to explain how the leading coefficient and
degree of the function affect the end behavior of these
graphs
For a>0 and n even
For a>0 and n odd
For a<0 and n even
For a <0 and n odd
END BEHAVIORS
WHAT THE GRAPH DOES AT THE ENDS?
DEGREE
0
1
2
3
4
POSITIVE
NEGATIVE
POLYNOMIAL GRAPHS
IT’S A MATTER OF DEGREES
DEGREE/TYPE
EXAMPLE
END BEHAVIORS
MAX # OF
MAX
ZEROS TURNING POINTS
0 /Constant
y=3
Horizontal
0 or infinity
0
1/linear
y = -2x + 4
Alternate
1
0
y = x2 + 2x – 1
Same
2
1
3/cubic
y = x3 – 3x2 + 2
Alternate
3
2
Same
4
3
n (odd)
Alternate
n
n-1
n (even)
Same
n
n-1
4/quartic
y = x4 – 4x3 – x2 + 12x – 2
Degre
even
Coefficie
nt
positive
End behavior of the
function
Graph of the function
Example: f (x) = x2
Degree
even
Negative
coefficient
Example: f
(x) = –x2
Degree
Odd
Positive
coefficient
Example: f
(x) = x3
Degree
Odd
Negative
coefficient
Example: f
(x) = –x3
• The time t ( in seconds)it takes a camera
battery to recharge after flashing n times can
be modeled by:
t  0.000015n  0.0034n  0.25n  5.3
3
2
• Find the recharge time after 100 flashes.
6.3 OPERATIONS ON POLYNOMIALS
Aka: combine like terms
EXAMPLE: (3 3 + 2 2 −  − 7) + ( 3 −10 2 + 8)
Horizontally:
SUBTRACTION: add the opposite of the second polynomial
EXAMPLE: (3 3 + 2 2 −  − 7) − ( 3 −10 2 + 8)
(3 3 + 2 2 −  − 7) + (− 3 +10 2 − 8)
Vertically:
MULTIPLY
EXAMPLE: ( − 3)(3 2 − 2 − 4)
Horizontally:
Vertically:
APPLICATIONS
OF POLYNOMIAL FUNCTIONS
From 1985 through 1995, the gross farm income G, and farm expenses, E (in billions of
dollars), in the United States can be modeled by
G(t) = −.246 2 + 7.88 + 159 and E(t) = .174 2 + 2.54 + 131
Where t is the number of years since 1985. Write a model for the net farm income, N,
for those years
N(t) = G(t) - E(t)
N(t) = (−.246 2 + 7.88 + 159) - (.174 2 + 2.54 + 131)
N(t) = −.42 2 + 5.34 + 28
From 1982 through 1995, the number of softbound books, N (in millions) sold in the
United States, and the average price per book, P (in dollars) can be modeled by
() = 1.36 2 + 2.53 + 1076  () = .314 + 3.42
Where t is the number of years since 1982. Write a model for the total revenue, R
received from the sales of softbound books.
R(t) = P(t) x N(t)
() = .42704 3 + 5.44562 2 + 346.5166 + 3679.92
What was the total revenue from softbound books in 1990?
Method #1: evaluate R with t = 8
Method #2: graph R and determine R(8)
\$7020 million (\$7.02 billion)
After vacation warm up
(32 ) 5
38
Simplify
 3 xy
9 x 3 y 4
Write in standard form
( x  x  7)  (2 x  4 x  3)
3
3
Graph
( x  3)( x  4 x  4)
y   x3  1
Use synthetic substitution to evaluate
y  2x  4x 1
3
for x=-2
2
SPECIAL PRODUCT PATTERNS
SUM x DIFFERENCE:
Example:
SQUARE OF A BINOMIAL:
(a + b)(a – b) = a² - b²
(x + 4)(x – 4) = x² - 16
(a + b)² = a² + 2ab + b²
Example:
(x + 4)² = x² + 8x + 16
NOTE: The square of a binomial is always a trinomial.
CUBE OF A BINOMIAL:
Example:
(a + b)³ = a³ + 3a²b + 3ab² + b³
(x + 5)³ = a³ + 3a²b + 3ab² + b³
x³ + 3x²b(5) + 3x(25)² + 125
x³ + 15x² + 75x + 125
FACTORING REVIEW
COMMON FACTOR: 6x² + 15x + 27 = 3(
TRINOMIAL: 2x² -5x – 12 = (
)(
)
)
PERFECT SQUARE TRINOMIAL: x² + 20x + 100 = (
DIFFERENCE OF TWO SQUARES: x² - 49 = (
)(
)(
)
)
MORE SPECIAL FACTORING PATTERNS
SUM OF 2 CUBES: a³ + b³ = (a + b)(a² - ab + b²)
Example: x³ + 27 = (x + 3)(x² - x(3) + 9
(x + 3)(x² - 3x + 9)
DIFFERENCE OF 2 CUBES: a³ - b³ = (a - b)(a² + ab + b²)
Try these:
x³ - 125
x³ + 64
27x³ - 8
343x³ + 1000
Warm-up
• Factor
y 3  27
8 x 2  10 x  3
x  125
3
5 x 3  40
ZERO PRODUCT RULE
(STILL GOOD!)
Solving polynomial equations:
1. Transform equation to make one side zero
2. Factor other side completely
3. Determine values to make each factor zero
Example: 2x⁵ + 24x = 14x³
2x⁵ - 14x³ + 24x = 0
2x(x⁴ - 7x² + 12) = 0
2x(x² - 3)(x² - 4) = 0
2x(x² - 3)(x - 2)(x + 2) = 0
Set each factor to zero:
2x = 0
x = 0
x² - 3 = 0
x = ±√3
x–2=0
x=2
x+2 = 0
x = -2
3 + 27 = 0
(X + 3)(X² – 3X + 9) = 0
X+3=0
OR
X² – 3X + 9 = 0
FACTOR BY GROUPING
Use for polynomials with 4 terms
3 − 3 2 + 6 − 18
Separate into 2 binomials:
Factor out GCF of each:
3 − 3 2 + 6 − 18
2 ( − 3) + 6( − 3)
Factor out new GCF:
TRY THESE:
3 + 6 2 + 7 + 42
3 − 2 2 − 16 + 32
253 − 252 −  + 1
93 + 182 − 4 − 8
( − 3)( 2 + 6)
CHECK:
(X² + 7)(X + 6)
(z² - 16)(z – 2)
(5p - 1)(5p + 1)(p – 1)
(3m - 2)(3m + 2)(m + 2)
• Suppose you have 250 cubic inches of clay
with which to make a rectangular prism for a
sculpture. If you want the height and width
each to be 5 inches less than the length, what
should the dimensions of the prism be? Solve
by factoring.
• You are building a bin to hold cedar mulch for
your garden. The bin will hold 162 cubic feet
of mulch. The dimensions of the bin are x feet
by 5x-9 feet by 5x-6 feet. How tall will the bin
be?
• In 1980 archeologists at the ruins of Caesara
discovered a huge hydraulic concrete block with a
volume of 330 cubic yards. The blocks dimensions
are x yards high by (13x – 11) yards long by
• (13x – 15) yards wide. What are the dimensions of
the block?
• You are building a bin to hold cedar mulch for your
garden. The bin will hold 162 cubic feet of mulch.
The dimensions of the bin are x ft. by (5x-6)ft. by
(5x-9) ft. How tall will the bin be?
LONG DIVISION REVIEW
32040 /15
213 6
LONG DIVISION - Remember 4th grade?
Write dividend “inside the house”
1st
Divide digit(s) in dividend by the divisor;
Multiply
Subtract
Bring down next digit
Repeat process as needed
15
3 2 0 4 0
3 0
2 0
1 5
54
4 5
9 0
9 0
0
POLYNOMIAL DIVISION
(2x⁴ + 3x³ + 5x – 1) /(x² - 2x + 2)
2x² +7x + 10
LONG DIVISION
Write dividend in standard form
(include all degrees)
Divide 1st term in dividend by
1st term in divisor
Multiply
Subtract
2x⁴ + 3x³ + 0x² + 5x – 1
2x⁴ - 4x³ + 4x²
7x³ - 4x² +5x
7x³ - 14x² + 14x
10x² -9x
-1
10x² - 20x + 20
Bring down next term
Repeat process as needed
Answer: 2x² + 7x + 10
X² - 2x + 2
11x – 21
X² - 2x + 2
11x - 21
• TRY THESE:
Divide x² + 6x + 8 by x + 4
SOLUTIONS:
X+2
Verify: (x + 4)(x + 2)
Divide x² + 3x – 12 by x – 3
6
X + 6 R 6 or x + 6 −3
Verify: (x – 3)(x + 6) + 6
SYNTHETIC DIVISION:
To divide polynomial f(x) by (x – k),
- write polynomial expression in standard form (include all degree terms)
- write only coefficients (including zeros)
- use the given value of k in the process below *
- The remainder is the last number written
- the other numbers in the answer are the coefficients/constant of the
quotient
Example: Divide x3 - 3x2 - 16x – 12 , by ( x – 6)
x3 - 3x2 - 16x – 12
1
-3
-16
-12
1
6
3
18
2
12
0
K=6
Quotient: 1x2 + 3x + 2
• TRY THESE:
Divide x² + 6x + 8 by x + 4
SOLUTIONS:
X+2
Verify: (x + 4)(x + 2)
Divide x² + 3x – 12 by x – 3
6
X + 6 R 6 or x + 6 −3
Verify: (x – 3)(x + 6) + 6
RELATED THEOREMS
REMAINDER THEOREM: If a polynomial f(x) is divided by x – k,
then the remainder, r, equals f(k).
Remember synthetic substitution?
Example: (x3 + 2x2 – 6x – 9) ⁄ (x – 2)
1
2
2
-6
8
-9
4
1
4
2
-5
K=2
f(2) = -5
Use the Remainder Theorem to evaluate P(-4)
for P(x) = 2x4 + 6x3 – 5x2 - 60
WORKOUT
-4
P(-4) = -12
2
6
-8
-5
8
0
-12
-60
48
2
-2
3
-12
-12
SPECIAL CASE
Use the Remainder Theorem to evaluate P(-3) for
P(x) = 2x3 + 11x2 + 18x + 9
2
11
18
9
2
-6
5
-15
3
-9
0
-3
Since P(-3) = 0:
1. 3 is a zero of P(x)
2. (x - ¯3) is a factor
(x + 3)
Quotient: 2x2 + 5x + 3
Note: the quotient is also factorable:
2x2 + 5x + 3 = (2x + 3) (x + 1)
Therefore, 2x3 + 11x2 + 18x + 9 = (x + 3) (2x + 3) (x + 1)
Try:
f ( x)  x 3  2 x 2  9 x  18
if one zero is 2
OBSERVATION
When dividing f(x) by (x-k), if the remainder is 0, then (x – k) is
__ ____________ of f(x).
Determine whether each divisor is a factor of each dividend:
a) (2x2 – 19x + 24) ÷ ( − 8)
yes
b) (x3 – 4x2 + 3x + 2) ÷ (x + 2)
no
FACTOR THEOREM:
A polynomial f(x) has a factor (x - k) if and only if f(k) = 0.
Factor f(x) = 2x3 + 7x2 - 33x – 18 given that f(-6) = 0
2
-6
2
7
-33
-18
-12
30
18
-5
-3
0
f(-6) = 0, so (x + 6) is a factor
Quotient: 2x2 – 5x -3
(which is the other factor, and can be factored into (2x + 1) (x – 3)
Therefore, 2x³ + 7x² - 33x – 18 = (x + 6)(2x + 1)(x – 3)
TRY THIS:
Given one zero of the polynomial function, find the other zeros.
F(x) = 15x3 – 119x2 – 10x + 16; 8
Since 8 is a zero, (x – 8) is a factor.
Since the quotient is 15x2 + x -2, it is also a factor.
Since 15x2 + x -2 can be factored into (5x + 2) (3x - 1).
The factors of 15x3 – 119x2 – 10x + 16 are
(x – 8) (5x + 2) (3x – 1)
Warm-up
• Divide using long division.
(3 x  2 x  5 x  1)  (3 x  1)
3
2
(4 x  7 x  8)  (2 x  1)
3
• The volume of a box is represented by the
function f ( x)  2 x  11x  10 x  8 The box is (x-4)
high and (2x+1) wide. Find the length.
• V=lwh
3
2
WRITING A FUNCTION
GIVEN THE ZEROS
Given: 2 and 4 are the zeros of the function f(x).
Write the function
f(x) = (x – 2) (x – 4)
f(x) = x2 – 6x + 8
Given: 3 and -4 and 1 are the zeros of the function f(x).
Write the function
f(x) = (x – 3) (x + 4) (x – 1)
f(x) = (x2 + x – 12) (x – 1)
f(x) = x3 – 13x + 12
Try these:
Given the zeros of a function, write
the function.
1.
2.
3.
4.
-1, 3, 4
-3, 1, 10
-2, 4, 5
1, 2
SOLUTIONS:
1. f (x) = x3 - 6x2 + 5x + 12
2. f (x) = x3 – 8x2 – 23x + 30
3. f (x) = x3 – 7x2 + 2x + 40
4. f (x) = x2 – 3x + 2
The Rational Zero Theorem
• If a polynomial function has integer
coefficients then every rational zero of the
function has the following form:
• P = factor of the constant term
• Q
• Find the rational zeros of
f ( x)  x 3  2 x 2  11x  12
• List the possible zeros
 1,2,3,4,6,12
• Test the zeros using synthetic division
• Divide out the factor and factor the remaining
trinomial to find the other zeros.
• (You may use your calculator to guide you)
•
• List all the possible rational zeros of the
function.
f ( x)  2 x 3  7 x 2  7 x  30
f ( x)  x 3  4 x 2  x  4
Find all zeros of the function.
f ( x)  x  4 x  11x  30
3
2
f ( x)  2 x  5 x  2 x  5
3
2
Molten Glass
At a factory, molten glass is poured into molds
to make paperweights. Each mold is a
rectangular prism whose height is 3 inches
greater than the length of each side of the
square base. A machine pours 20 cubic inches
of liquid glass into each mold. What are the
dimensions of the mold?
United States Exports
• For 1980 through 1996, the total exports E (in
billions of dollars) of the United States can be
modeled by
E  0.131t 3  5.033t 2  23.2t  233
• where t is the number of years since 1980. In
what year were the total exports about
• \$312.76 billion?
Fundamental Theorem of Algebra
• If f(x) is a polynomial of degree n and n is
greater than zero, then the equation f(x)=0
has at least one root in the set of complex
numbers.
• (written by Carl Friedrich Gauss)
• When all real and imaginary solutions are
counted (with all repeated solutions counted
individually), an nth degree polynomial
equation has exactly n solutions. Any nth
degree polynomial function has exactly n
zeros.
Turning points of a graph
• The graph of every polynomial function of degree
n has at most n-1 turning points. If the function
has n distinct real zeros then its graph has exactly
n-1 turning points.
• Polynomial functions have local maximum and
local minimum points, these are the turning
points.
• Quadratic functions have only one maximum or
minimum point.
Finding Turning Points
• Use your calculator to graph
f ( x)  x  3x  2
3
2
• Identify the x intercepts and the points where
the local maximums and minimums occur.
Maximizing a Polynomial Model
You are designing an open box to be made of a piece of
cardboard that is 10 inches by 15 inches. The box will be
formed by making the cuts at the corners and folding up the
sides so that the flaps are square. You want the box to have
the greatest volume possible. How long should you make
the cuts? What is the maximum volume? What will the
dimensions of the finished box be?
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