grade 12 pre-calculus mathematics notebook chapter 3 ...synthetic division synthetic division is an...

Grade 12 Pre-Calculus Mathematics

Notebook

Chapter 3

Polynomial Functions

Outcomes: R11 & R12

3.1 Characteristics of Polynomial Functions R12 (p.106-113)

Polynomial Function = a function of the form where

1 2 2

1 2 2 1 0( ) ...n n n

n n nf x a x a x a x a x a x a

where n = a whole number

x = a variable

na and 0a are real numbers

Examples:

2

3 2

5 3

( ) 3 5

( ) 3 17

( ) 2 2

7 1

f x x

g x x x

h x x x x

y x x

Vocabulary

▪ end behaviour

▪ degree

▪ constant term ( 0a )

▪ leading coefficient

Ex1: Identify the functions that are not polynomials and state why.

For each, state the degree, leading coefficient, and the constant term of each

polynomial function.

a)

b)

c)

d)

Your turn:

a)

b)

c)

d) y =

Characteristics of Polynomial Functions

Compare the graphs of even and odd degree functions. How does the leading

term affect the general behaviour of the graph?

Look at the graphs on page 109 of your text and list any generalizations or

patterns you notice.

Characteristics of Polynomial Functions

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) b)

c) d)

Homework: Page 114 #1, 2, 3, 4,

x- 3 - 2 - 1 1 2 3 4 5 6

y

- 60

- 40

- 20

20

40

x- 4 - 3 - 2 - 1 1 2 3 4

y

- 6

- 5

- 4

- 3

- 2

- 1

1

2

3

4

5

6

x- 4 - 3 - 2 - 1 1 2 3 4

y

- 16

- 8

8

16

24

32

x- 4 - 3 - 2 - 1 1 2 3 4

y

- 6

- 5

- 4

- 3

- 2

- 1

1

2

3

4

5

6

3.2 The Remainder Theorem R11 (p.118-123)

Long Division

Divide the following expression: 3

1582

x

xx=

We can divide the expression above by using long division:

1583 2 xxx

After you divide, your answer can be written in two forms:

1) Dividend remainder

QuotientDivisor Divisor

OR

2) ( )Dividend Divisor Quotient remainder

Answer:

Note: The restriction on the variable is 3x since division by 0 is not

defined.

Note: To verify, multiply the divisor by the quotient and add the remainder.

Synthetic division

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. Use the divisor to solve for x. Example: (x + 2) → x = –2

3. Write only the coefficients of the equation. If any are missing, fill in

their spot with a zero. Make an “L” shape with the value of x outside.

4. Bring down the first coefficient.

5. Multiply by the divisor.

6. Add that number to the second coefficient.

7. Repeat steps 4-6 until there are no more coefficients to bring down.

8. Write the resulting numbers as the coefficients of a new equation.

The last number will be the remainder.

Divide: 3

1582

x

xx

Ex1: Divide: 3952 23 xxx

Answer: __________________________________________________

Ex2: Divide: x3 – 6x + 72 by x – 4.

Answer: __________________________________________________

Remainder Theorem

The remainder theorem allows us to obtain the value of the remainder

without actually dividing.

Ex1: Determine the remainder when the polynomial 21175)( 23 xxxxP

is divided by the following binomials:

a) x + 1 b) x – 3

Synthetic division:

Remainder Theorem:

Homework: Page 124 #1, 2, 3 (choose 2), 4 (choose 3), 5 (choose 2),

6 (choose 3), 8 (choose 2), 9, 11, C2

3.3 The Factor Theorem R11 (p.126-133)

The factor theorem tells us whether or not the divisor 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 ( ) 0P a .

Ex1: Determine whether or not 2x is a factor of 46)( 3 xxxf .

Ex2: Completely factor 3( ) 7 6P x x x .

To do this, we must first find factors of 3( ) 7 6P x x x .

Let’s use the Remainder Theorem.

There must be an easier way than randomly guessing infinitely many times…

Integral Zero Theorem

Expand the following expression:

1034521 23 xxxxxx

Note: The factors of the polynomial are x – 1, x + 2 and x – 5.

The zeroes 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.

Ex1: a) Find the possible factors of the following polynomial:

863 23 xxxxf → __________________________

b) Completely factor the polynomial above.

Ex2: Determine all of the possible zeroes of the following polynomial:

12832 23 xxxxf → _________________________

b) Factor the polynomial.

Ex3: Factor 72546)( 234 xxxxxP .

Homework: /

Page 133 #1, 2a, 2f, 3c, 3e, 4-6 (choose 3 each), 7, 11, C1, C2

3.3 Division of a Polynomial by a Binomial R11

1. Divide the polynomial P(x) x4 3x3 2x2 55x 11 by x 3.

Express your answer in the form .

2. Determine the quotient of the following polynomials divided by binomials.

a) (4w4 3w3 7w2 2w 1) (w 2) b)

c) (5y4 2y2 y 4) (y 1) d) (3x2 16x 5) (x 5)

e) (2x4 3x3 5x2 6x 1) (x 3) f) (4x3 5x2 7) (x 2)

3. Using the Remainder Theorem, determine the remainder when each of the following polynomials is

divided by (x 2). State whether or not (x 2) is a factor of each.

a) 4x4 3x3 2x2 x 5 b) 7x5 5x4 23x2 8 c) 8x3 1

4. Divide each of the following and state whether or not the binomial is a factor of the polynomial.

a) (3x3 4x2 6x 9) (x 1) b) (3x2 8x 4) (x 2) c) (x3 5x2 7x 9) (x 5)

5. Determine if (x 1) is a factor of each polynomial.

a) 4x4 3x3 2x2 x 5 b) 7x5 5x4 23x2 8 c) 2x4 3x3 5x2 7x 1

6. Determine is each polynomial has a factor of (x 2).

a) 3x3 2x2 10x 5 b) 2x4 3x3 5x2 + 36 c) 3x3 12x 2

d) Explain how we know that (x + 2) is not a factor of the polynomial in (a) without having to

calculate the remainder.

7. Factor each polynomial below.

a) x3 2x2 13x 10 b) x4 7x3 3x2 63x 108

c) x3 x2 26x 24 d) x4 26x2 25

e) x4 4x3 7x2 34x 24 f) x5 3x4 5x3 15x2 4x 12

8. Given (2 x3 5x2 k x 9) (x 3), determine the value of k if the remainder is 6.

9. When 4x2 - 8x - 20 is divided by x + k, the remainder is 12. Determine possible values

for k.

10. Each polynomial has x 3 as a factor. Determine the value of k in each case.

a) kx3 10x2 2x 3 b) 4x4 3x3 2x2 kx 9

Solutions

1.

2. a) 4w3 - 5w2 + 3w - 4 b) x3 + 4x2 - 5 c) 5y3 - 5y2 + 7y – 8

d) 3x – 1 e) 3

179602292 23

xxxx f)

2

4526134 2

xxx

3. a) -25 b) -44 c) -65

x + 2 is not a factor of any of the polynomials since the remainder is never equal to 0.

4. a) 1

231473 2

xxx , no b) 3x – 2, yes c)

5

4472

xx , no

5. a) no, r = -1 b) no, r = 43 c) yes, r = 0

6. a) no, r = 17 b) yes, r = 0 c) no, r = -2

d) The constant is 5 and 2 is not a factor of 5. Therefore, there will always be a remainder. (integral zero theorem)

7. a) (x 1)(x 2)(x 5) b) (x 3)2(x 3)(x 4) c) (x 1)(x 4)(x 6)

d) (x 1)(x 1)(x 5)(x 5) e) (x 4)(x 2)(x 1)(x 3) f) (x 3)(x 2)(x 1)(x 1)(x 2)

8. 2

9. 4 and -2

10. a) 3 b) -72

3.2 Long Division & Synthetic Division

1. Divide the following polynomials using long division.

a) 3 23 2 5 1x x x x c) 3 23 2 27 36x x x x

b) 3 22 5 6 2x x x x d)

34 9 12

2

x x

x

2. State any restrictions on the polynomials above.

3. Verify your answers from Question 1 by divide the polynomials using synthetic division.

4. The volume of a rectangular box, in 3cm , can be modeled by the polynomial function

3 2( ) 3 12 4V x x x x . Determine expressions for the width and the length of the box if the height

is 2x .

3.2 The Remainder Theorem

5. Use the remainder theorem to determine each remainder.

a) 26 15 1x x x b)

3 22 5 13 2

4

x x x

x

c)

4 22 3 5 2x x x x

6. For each dividend, determine the value of k if the remainder is –2.

a) 3 22 5 4 1x x x k x b) 3 24 10 3x x kx x

7. For what value of m will the polynomial 3 2( ) 6 4P x x x mx have the same remainder when it is

divided by 1x and 2x ?

3.3 The Factor Theorem

8. Determine which of the following binomials are factors of 3 2( ) 4 6P x x x x .

a) 1x b) 2x c) 3x d) 2x

9. Determine the possible integral zeros of each polynomial.

a) 3 2( ) 5 4P x x x x b)

4 2( ) 3 12P x x x x c)3 2( ) 2 3 17 30P x x x x

10. An artist creates a carving from a block of soapstone. The soapstone is in the shape of a rectangular

prism whose volume, in cubic feet, is represent by 3 2( ) 6 25 2 8V x x x x where x is a positive

real number. Determine the factors that represent possible dimensions of the block of soapstone, in terms

of x.

11. Factor:

a) 3 2( ) 4 6P x x x x c)

3 2( ) 2 3 3 2P x x x x

b) 3 2( ) 3 5 26 8P x x x x d)

4 3 2( ) 3 7 27 18P x x x x x

3.4 Equations and Graphs of Polynomial Functions R12 (p.136-147)

Ex1: a) Determine the zeroes of the following cubic function:

3 7 6 2 3 1f x x x x x x

b) Determine the y-intercept of the function.

c) Summarize what we know about this function.

d) Sketch the graph.

degree

leading coefficient

end behaviour

zeroes

y-intercept

intervals (sign diagram)

Ex2: a) Determine the roots of the following function:

44)( 23 xxxxP

b) Determine the y-intercept of the function.

c) Summarize what we know about this function.

d) Sketch the graph.

degree

leading coefficient

end behaviour

zeroes

y-intercept

intervals (sign diagram)

3.4 Sketching Polynomial Functions R12

1. State the zeroes and the y-intercept of each of the polynomial fucntions below.

a) 153 xxxy b) 1522 xxxxy c) 813 xxxy

2. Determine the intervals where the following functions are positive and negative

(i.e. sign diagrams).

a) 533 xxxy b) 541 xxxy c) 321 xxxxy

3. Sketch the following function. Be sure to include all intercepts.

a) 511 xxxy b) 43 xxxy

c) 3221 xxxxy

4. Explain what happens to the graph when the leading coefficient of a polynomial function is negative.

5. Sketch the graphs of the following functions. Be sure to include all intercepts.

a) 15239 23 xxxy b) 1644 23 xxxy

c) xxy 93 d) 152162 234 xxxxy

e) 3613)( 24 xxxf f) 1209489233 2345 xxxxxy

Solutions 1. a) x = -3, 5 and -1 y = -15 b) x = ±2, 5 and -1 y = -20 c) x = 3, -1 and 8 y = -24

2. a) positive on the interval (-5,-3) and (3, ∞) negative on the interval (-∞, -5) & (-3, 3)

b) positive on the interval (-∞, -4) and (1, 5) negative on the interval (-4,1) and (5, ∞)

c) positive on the interval (-∞, -2) and (-1, 0) and (3,∞) negative on the interval (-2,-1) and (0, 3)

3. a) b) c)

4. There is a reflection over the x-axis. The behaviour of the graph when x → ±∞ changes direction.

5. a) b)

c) d)

e) f)

Multiplicity of a Zero

If ( )P x has factor ( )x a n times, we say that x a is a zero of multiplicity n.

For example,

1x is a zero of multiplicity 2

2 3( 1) ( 2)( 4)y x x x 2x is a zero of multiplicity 1

4x is a zero of multiplicity 3

Multiplicity represents the number of times a factor is repeated.

Multiplicity 1 Multiplicity 2 Multiplicity 3 Multiplicity 4

Note: The effect of an even multiplicity is a bounce on the x-axis. . The effect of an odd multiplicity is a flattened area which is often hard

to see without the use of technology.

Ex1: Examples of multiplicities:

a) 21 xy b) 32 xy

Ex2: Sketch the following graphs:

a) 212

xxy b) 32 xxy

Ex.3 The zeroes of a polynomial function are –2, 3, and 5.

Write an equation to represent this function.

Ex. 4 The zeroes of a polynomial function are 0 and –4 (with a multiplicity

of 2). Write an equation to represent this function.

Homework: Page 147 #1, 3, 4, 7, 8, 10, C1, C2, C3

degree

leading coefficient

end behaviour

zeroes

y-intercept

intervals

(sign diagram)

degree

leading coefficient

end behaviour

zeroes

y-intercept

intervals

(sign diagram)

Sketching polynomial functions

Sketch the following functions. Be sure to clearly indicate the x and y-intercepts for each.

1. )2)(4)(1()( xxxxg 2. )5)(1( xxxy

3. )8)(3)(2)(6()( xxxxxf 4. 2)3)(1( xxy

5. )7()( 2 xxxf 22 )4()1( xxy 6. 22 )4()1( xxy

7. 64 23 xxxy 8. 33)( 23 xxxxf

3.4 Solving and Graphing Polynomial Equations R12

1. Solve each equation.

a) (x 5)(x 2)(x 3)(x 6) 0 b) 012354 xxx c) (3x 1)(4 x)(x 7) 0

d) –(x 4)3(x 2)2 0 e) 05523 xx f) 034

4 xx

. g) For questions (a) to (f), describe the degree of the polynomial as well as

the end behaviour.

2. Using the graph of the polynomial function to the right:

a) the smallest possible degree of the function

b) the sign of the leading coefficient

c) the x-intercepts of the graph and the factors of the function

d) the intervals over which the function is positive and the intervals over which

the function is negative.

3. Determine the equation of the smallest degree that corresponds to each polynomial function below.

a) a cubic function with zeroes at 3 (of multiplicity 2) and –1, and whose y-intercept is 18

b) a quintic function with zeroes at -2 (of multiplicity 3) and 4 (of multiplicity 2) and whose

y-intercept is –64.

c) a quartic function with zeroes at –1 (of multiplicity 2) and 5 (of multiplicity 2) and whose

y-intercept is 10.

d) Sketch the functions (a) to (c).

4. Sketch the following graph. State all intercepts.

a) 122

xxy b) 3232 xxxy c) 32123

xxy

d) 3241 xxy e) 323 xxy f) 22

11128 xxxy

5. Sketch the graph of each of the functions below. State all intercepts.

a) y x3 4x2 5x b) f (x) x4 19x2 6x 72 c) g (x) x5 14x4 69x3 140x2 100x

Solutions

1. a) x = -5, -2, 3 & 6 b) x = 4

5 , 3 & 2

1 c) x =

3

1 , 4 & 7 d) x = 0, -4 & -2 e) x =

2

5 & 5 f) x = 4 & -3

g) For (a), 4th degree polynomial with end behavior up in QI and QII.

For (b), 3rd degree polynomial with end behavior down in QIII and up in QI.

For (c), 3rd degree polynomial with end behavior up in QII and down in QIV.

For (d), 5th degree polynomial with end behavior up in QII and down in QIV.

For (e), 4th degree polynomial with end behavior down in QIII and QIV.

For (f), 5th degree polynomial with end behavior down in QIII and up in QI.

2. a) 3 b) negative c) 4, 2, 3 ; (x 4), (x 2), (x 3)

d) Positive intervals: (3, 2) and (, 4); negative intervals: (4, 3) and (2,

3. a) y 2(x 3)2(x 1) b) 2342

2

1 xxy

c)

d) « a » « b » « c »

4. a) b) c) d)

e) e) f)

5. a)

b) c)

3.4 Applications of Polynomial Functions R12

Ex1: The volume of air flowing into the lungs during one breath can be

represented by the polynomial function

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?

c) What percentage of the breath is spent inhaling?

x- 6 - 5 - 4 - 3 - 2 - 1 1 2 3 4 5 6

y

- 6

- 5

- 4

- 3

- 2

- 1

1

2

3

4

5

6

Ex2: A block of snow measures 3m by 4m by 5m. The block melts in such a

way that each dimension decreases in size at the same rate. At the end

of a warm, sunny day, the block has a volume of 24m3.

a) Draw the initial block of snow.

4 – x

3 – x

5 – x

b) Write a polynomial function to represent this situation.

c) Determine algebraically the new dimensions of the block.

)3611)(1(0

)3611)(1(0

3647120

)3)(4)(5(24

2

2

23

xxx

xxx

xxx

xxx

Ex3: The length of a swimming pool is 10m larger than the depth.

The width of the pool is 3m larger than the depth.

The City of Winnipeg charges $2 per m3 of water used to fill the pool.

The bill to fill the above pool is $240.

a) Represent this situation algebraically in terms of the pool’s depth, d.

b) Determine all possible values for the depth of this pool.

)6015)(2(0

12030130

)3)(10(120

2

23

ddd

ddd

ddd

c) Determine the real dimensions of the pool.

Ex4: 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 30cm and a width of 20cm.

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.

b) We would like a box with a volume of 1000cm3.

Determine the dimensions of the box that could be created with this piece of cardboard.

)5020)(5(0

250150250

100060010040

)220)(230(1000

2

23

23

xxx

xxx

xxx

xxx

Homework: Page 150 #12, 15, 16, 18

3.4 Applications of Polynomial Functions R12

1. Write an algebraic expression to calculate the product of 3 whole consecutive numbers if the smallest

number is x and the product is the function P(x). Next, determine the 3 numbers if the product equals –

504.

2. The product of 3 odd consecutive numbers is 315. Find the 3 numbers using a polynomial function.

3. The volume of a rectangular aquarium is modeled by the equation 20011019)( 23 xxxxV . The depth

of the aquarium can be represented by the expression 5x . Write a polynomial to represent the length and width

of the aquarium.

4. The Pan-Am pool contains 8000m3 of water. It has dimensions x for the depth, x + 6 for the length and 5x for the

width.

a) Determine all possible values for x.

b) Determine the dimensions of the pool.

5. A rectangular box has square ends. The measure of the box’s length is 12 cm longer than the width. The

volume is 135 cm3. Determine the dimensions of the box.

6. A rectangular prism measures 10 cm by 10 cm by 5 cm. When each dimension is increased by the same

quantity, the volume becomes 1008 cm3. Determine the dimensions of this new rectangular prism.

7. An open box with locking tabs is to be made from a square piece of cardboard with side length 28 cm.

This is done by cutting equal squares of side length x

cm from the corners and folding along the dotted

lines as shown. .

a) Write a polynomial equation to represent the

volume, V, of the box in terms of x.

b) Sketch the polynomial function that represents

this situation.

c) Determine the approximate value of x that

maximizes the volume of the box to the nearest

centimetre.

Note: use the graph to make your estimation.

8. A large ice cube with dimensions 6cm by 4cm by 4cm is placed on the counter where it uniformly melts.

Upon returning hours later, the ice cube has melted to a sixth of its original volume. Determine

algebraically the new dimensions of the ice cube.

9. The dimensions of a box are 9x cm, x2575 cm, and x313 cm. Determine the possible

dimensions of the box if each dimension is a whole number and if the volume of the box is 8000cm3.

Solutions

1. 21 xxxxP and –9, –8 and –7

2. 5, 7 and 9

3. width: x + 4, length: x + 10

4. a) x = -10, -6 and 10

b) x = 10 is the only real value of x. Thus, the dimensions are 10m by 16m by 50m

5. 3 cm by 3 cm by 15 cm


7. a) )7)(14(8)( xxxxV b)

c) x = 3cm


9. 10cm, 50cm and 16cm or 8cm, 100cm and 10cm

grade 12 pre-calculus mathematics notebook chapter 3 ...synthetic division synthetic division is an...

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