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−1𝑥𝑛−1 + 𝑎𝑛−2𝑥𝑛−2 + ⋯ + 𝑎2𝑥2 + 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: Synthetic Division (Method #2)

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 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.