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Grade 12 Pre-Calculus Mathematics
[MPC40S]
Chapter 3
Polynomial Functions
Outcomes
R11, R12
12P.R.11. Demonstrate an understanding of factoring polynomials
of degree greater than 2 (limited to polynomials of degree ≤ 5with
integral coefficients). 11P.R.12 Graph and analyze polynomial
functions (limited to polynomial functions of degree ≤ 5).
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Chapter 3 – Homework
Section Page Questions
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Chapter 3: POLYNOMIAL FUNCTIONS
3.1 – Characteristics of Polynomial Functions Polynomial
Function: A function of the form
𝑓(𝑥) = 𝑎𝑛𝑥𝑛 + 𝑎𝑛−1𝑥
𝑛−1 + 𝑎𝑛−2𝑥𝑛−2 + ⋯ + 𝑎2𝑥
2 + 𝑎1𝑥 + 𝑎0 where 𝑛 = ____________________________ 𝑥 =
____________________________ 𝑎𝑛 and 𝑎0 = _____________________
Example
The following are examples of polynomial functions.
𝑓(𝑥) = 3𝑥 − 5 ℎ(𝑥) = 𝑥3 + 2𝑥2 + 𝑥 − 2 𝑔(𝑥) = 𝑥2 + 3𝑥 − 17 𝑦 = 𝑥5
+ 7𝑥3 − 1 Example #1
Identify the functions that are not polynomials and state
why.
a) 𝑔(𝑥) = √𝑥 + 5
______________________________________________________ b) 𝑦 = |𝑥|
____________________________________________________________
c) 𝑓(𝑥) = 3𝑥4
_________________________________________________________
d) 𝑦 = 𝑥1
8 − 7
__________________________________________________________
e) 𝑦 = 2𝑥3 + 3𝑥2 − 4𝑥 − 1
_______________________________________________
f) ℎ(𝑥) =1
𝑥
____________________________________________________________
R12
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End Behaviour:
________________________________________________________
________________________________________________________
Degree:
_______________________________________________________________
Constant Term:
________________________________________________________
________________________________________________________ Leading
Coefficient: ____________________________________________________
____________________________________________________ Example #2
For each polynomial function, state the degree, the leading
coefficient, and the constant term of each polynomial function.
a) 𝑦 = 3𝑥2 − 2𝑥5 + 4 - Degree _______ - Leading Coefficient
_______ - Constant Term _______
b) 𝑦 = −4𝑥3 − 4𝑥 + 3 - Degree _______ - Leading Coefficient
_______ - Constant Term _______ c) 𝑓(𝑥) = 3𝑥 − 5 - Degree _______ -
Leading Coefficient _______ - Constant Term _______
d) 𝑓(𝑥) = −6𝑥4 − 2𝑥 - Degree _______ - Leading Coefficient
_______ - Constant Term _______
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Compare the graphs of even degree and odd degree functions. How
does the leading term affect the general behaviour of the graph?
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
a) The equations and graphs of several even-degree polynomials are
shown below. Study these graphs and generalize the end behaviour of
even-degree polynomials. What do you notice about the end behaviour
of an even-degree polynomial when…
The leading coefficient is positive?
_________________________________
_________________________________
_________________________________
_________________________________
The leading coefficient is negative?
_________________________________
_________________________________
_________________________________
_________________________________
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Notes
The graph of a polynomial function must be smooth and
continuous
The graph has at most (n – 1) turning points
The function has at most n roots (x – intercepts)
All polynomial functions have y – intercept at 𝑎0, the constant
term of the function
b) The equations and graphs of several odd-degree polynomials
are shown below. Study these graphs and generalize the end
behaviour of odd-degree polynomials. What do you notice about the
end behaviour of an odd-degree polynomial when…
The leading coefficient is positive?
_________________________________
_________________________________
_________________________________
_________________________________
The leading coefficient is negative?
_________________________________
_________________________________
_________________________________
_________________________________
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Example #3
Match the following polynomials with its corresponding
graph.
1.) 𝑓(𝑥) = 2𝑥3 − 4𝑥2 + 𝑥 + 2
2.) 𝑔(𝑥) = −𝑥4 + 10𝑥2 + 5𝑥 − 6
3.) ℎ(𝑥) = −2𝑥5 + 5𝑥3 − 𝑥 + 1
4.) 𝑝(𝑥) = 𝑥4 − 5𝑥3 + 16 a) Answer: __________
x- 3 - 2 - 1 1 2 3 4 5 6
y
- 60
- 40
- 20
20
40
b) Answer: __________
x- 4 - 3 - 2 - 1 1 2 3 4
y
- 6
- 5
- 4
- 3
- 2
- 1
1
2
3
4
5
6
c) Answer: __________
x- 4 - 3 - 2 - 1 1 2 3 4
y
- 16
- 8
8
16
24
32
d) Answer: __________
x- 4 - 3 - 2 - 1 1 2 3 4
y
- 6
- 5
- 4
- 3
- 2
- 1
1
2
3
4
5
6
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Chapter 3: POLYNOMIAL FUNCTIONS
3.2 – The Remainder Theorem We are going to learn how to divide
a polynomial by a binomial. Long Division (Method #1) Example
#1
Divide the following expression 𝑥2+8𝑥+15
𝑥+3
After you divide, your answer can be written in two forms:
1) Dividend
Divisor= Quotient +
remainder
Divisor OR 2) Dividend = (Divisor)(Quotient) + remainder
Answer to the above example:
Dividend = Polynomial Divisor = Binomial (x – a) Quotient =
Answer
R11
Note: Since the remainder is 0, this tells us that (𝑥 + 3) is a
factor of the polynomial 𝑥2 + 8𝑥 + 15 Note: The restriction on the
variable is 𝑥 ≠ −3 since division by 0 is not defined.
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Synthetic Division (Method #2) Synthetic division is an
alternate form of long division that we can use to divide
polynomials. This type of division uses only the coefficients of
each equation. Steps: 1. Rearrange the equation in descending order
2. Write only the coefficients of the polynomial. If any are
missing, fill in their spot with a zero. 3. Bring down the first
coefficient. 4. Multiply by the divisor. 5. Add that number to the
second coefficient. 6. Repeat steps 4-6 until there are no more
coefficients to bring down. 7. Write the resulting numbers as the
coefficients of a new polynomial. The last number will be the
remainder. Example #2
Divide the following expression 𝑥2+8𝑥+15
𝑥+3
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Example #3
Divide 2𝑥4 + 2𝑥3 − 𝑥 + 4 by 𝑥 + 2 Example #4
Divide (2𝑥3 + 5𝑥2 + 9) ÷ (𝑥 + 3)
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Example #5
Divide 𝑥3 − 2𝑥2 + 6𝑥 − 12 by 𝑥 − 1 Example #6
Divide 𝑥3 − 7𝑥2 − 6𝑥 + 72 by 𝑥 − 4
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The Remainder Theorem The remainder theorem allows us to obtain
the value of the remainder without actually dividing. Example
#7
Use the remainder theorem to determine the remainder when the
polynomial
𝑃(𝑥) = 𝑥3 − 5𝑥2 − 17𝑥 + 21 is divided by the following
binomials. Verify your solution using either long division or
synthetic division. a) 𝑥 + 1 b) 𝑥 − 1
When 𝑃(𝑥) is divided by (𝑥 − 𝑎) the remainder is 𝑃(𝑎)
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Chapter 3: POLYNOMIAL FUNCTIONS
3.3 – The Factor Theorem The Factor Theorem tells us whether or
not the divisor (𝑥 − 𝑎) is a factor of the dividend. If there is no
remainder (i.e. the remainder = 0), then the divisor is a
factor.
The Factor Theorem states that (𝑥 − 𝑎) is a factor of 𝑃(𝑥) if
and only if 𝑃(𝑎) = 0
Example #1
a) Determine whether or not 𝑥 + 2 is a factor of 𝑃(𝑥) = 𝑥3 + 4𝑥2
+ 𝑥 − 6
b) If 𝑥 + 2 is a factor, completely factor 𝑃(𝑥) = 𝑥3 + 4𝑥2 + 𝑥 −
6
R11
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Example #2
Completely factor 𝑃(𝑥) = 𝑥3 − 7𝑥 + 6
To do this, we must find the factors of 𝑃(𝑥) = 𝑥3 − 7𝑥 + 6.
Let’s use the Remainder Theorem. Try (𝑥 + 1) 𝑃(−1) = (−1)3 − 7(−1)
+ 6 𝑃(−1) = ____________________
𝑃(−1) = ____________________________________________________ Try
(𝑥 + 2) 𝑃(−2) = (−2)3 − 7(−2) + 6 𝑃(−2) = ____________________
𝑃(−2) = ____________________________________________________ Try (𝑥
− 3) 𝑃(3) = (3)3 − 7(3) + 6 𝑃(3) = ____________________ 𝑃(3) =
____________________________________________________ Try (𝑥 − 1)
𝑃(1) = (1)3 − 7(1) + 6 𝑃(1) = ____________________ 𝑃(1) =
____________________________________________________ There must be
an easier way than randomly guessing infinitely many times…
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Integral Zero Theorem Expand the following expression: (𝑥 − 1)(𝑥
+ 2)(𝑥 − 5) = Note: The factors of the polynomial are (𝑥 − 1), (𝑥 +
2) and (𝑥 − 5) The zeros of the polynomial are 1, –2 and 5 Note:
When we multiply all of the factors, the constant is +10, which
means that only factors of 10 can be factors of the polynomial.
This is known as the Integral Zero Theorem The Integral Zero
Theorem states that if (𝑥 − 𝑎) is a factor of the polynomial
function
𝑃(𝑥) with integral coefficients, then 𝑎 is a factor of the
constant term of 𝑃(𝑥).
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Example #3
a) Find all of the possible zeros of the following
polynomial:
𝑓(𝑥) = 𝑥3 − 3𝑥2 − 6𝑥 + 8 →
__________________________________________ b) Completely factor the
polynomial above.
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Example #4
a) Find all of the possible zeros of the following
polynomial:
𝑓(𝑥) = 2𝑥3 − 3𝑥2 − 8𝑥 + 12 →
______________________________________ b) Completely factor the
polynomial above. c) Determine the zeros of 𝑓(𝑥).
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Example #5
a) Completely factor 𝑃(𝑥) = 𝑥4 − 5𝑥3 + 2𝑥2 + 20𝑥 − 24 b)
Determine the zeros of 𝑃(𝑥).
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Chapter 3: POLYNOMIAL FUNCTIONS
3.4 – Equations and Graphs of Polynomial Functions Example
#1
a) Determine the zeroes of the following cubic function
𝑓(𝑥) = 𝑥3 − 𝑥2 − 4𝑥 + 4 b) Determine the y – intercept of the
function c) Summarize what we know about this function
Degree
Leading Coefficient
End Behaviour
Zeroes
y – Intercept
Intervals
(sign Diagram)
d) Look at the sketch of the polynomial function.
R12
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Zero Zero Zero Multiplicity Multiplicity Multiplicity
Multiplicity of a Zero If 𝑃(𝑥) has a factor (𝑥 − 𝑎) that is
repeated 𝑛 times, we say that 𝑥 = 𝑎 is a zero of multiplicity 𝑛.
For example:
𝑦 = (𝑥 + 1)2(𝑥 − 2)(𝑥 − 4)3 {
𝑥 = −1 is a zero of multiplicity __________𝑥 = 2 is a zero of
multiplicity ___________𝑥 = 4 is a zero of multiplpcity
___________
Multiplicity is the number of times the zero of a polynomial
occurs. (The number of times a factor is repeated) The shape of the
graph of a function close to a zero (x – intercept) depends on its
multiplicity. Example #2
Determine the zeroes of each polynomial function and their
multiplicities from the given graphs. a) b) c)
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Example #3
Sketch the following graphs: a) 𝑦 = −(𝑥 − 1)2(𝑥 + 2) b) 𝑦 =
(𝑥)(𝑥 + 2)3
Degree
Degree
Leading Coefficient
Leading Coefficient
End Behaviour
End Behaviour
Zeroes
Zeroes
y – intercept
y – intercept
Intervals (Sign Diagram)
Intervals (Sign Diagram)
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Example #4
Sketch the graph of 𝑃(𝑥) = −𝑥3 + 4𝑥2 + 𝑥 − 4
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Example #5
Determine the equation for the following polynomial function
given the graph below. Example #6
The zeroes of a quartic function are at –2, –1, and 3, with
multiplicities of 1, 1, and 2 respectively. Determine the equation
of the function that satisfies this condition and passes through
the point (1, 24).
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Applications of Polynomial Functions Example #7
The volume of air flowing into the lungs during one breath can
be represented by the polynomial function 𝑉(𝑡) = −0.041𝑡3 + 0.18𝑡2
+ 0.202𝑡, where V is the volume in litres and t is the time in
seconds. This situation can be represented by the graph below.
What does the x-axis represent? What does the y-axis represent?
Determine any restrictions on the variables.
Using the graph above, answer the following questions: a)
Determine the maximum volume of air inhaled into the lungs. At what
time during the breath does this occur? b) How many seconds does it
take for one complete breath?
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Example #8
Bill is preparing to make an ice sculpture. He has a block of
snow that is 3 ft. wide, 4 ft. high and 5 ft. long. Bill wants to
reduce the size of the block of ice by removing the same amount
from each of the three dimensions. He wants to reduce the volume of
the ice block so that it is 24 ft.3 The drawing on the right
represents this situation. a) Write a polynomial function to
represent this situation. b) Determine algebraically the new
dimensions of the block.
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Example #9
A box is assembled by cutting the corners of a piece of
cardboard and then folding up the remaining sides. A piece of
cardboard has a length of 30 cm and a width of 20 cm. A square with
sides measuring x cm is cut from each of the corners of the
cardboard as shown in the diagram below.
a) Write an algebraic expression that represents the volume of
this box.
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b) We would like a box with a volume of 1000 cm3. Determine the
dimensions of the box that could be created with this piece of
cardboard.