mhf 4u lesson 2.0 review of factoring ex. factor each of ... · mhf 4u lesson 2.2 the factor and...

MHF 4U Lesson 2.0 Review of Factoring

Ex. Factor each of the following completely.

a) 243 352842 ppp b) 1032 xx

c) 2850 b d) 22 187 baba

e) yzxzxyx 2 f) 1572 2 xx

g) 3108 2 xx h) 22 49284 baba

i) 3613 24 xx j) 22)(4 cba

k) 251016 24 yyx

WS 2.0

MHF 4U Lesson 2.1 Dividing Polynomials

Dividing polynomials can be done in more than one way. It is important to use the most efficient way

in order to solve problems in the simplest manner.

I. Dividing by Factoring.

Whenever it is possible to divide polynomials by factoring, it is the

simplest and most efficient way to solve the problem.

Failing to recognize this will cause you to waste time and effort

on a more inefficient method of solving the problem.

II. Dividing using Long Division

Ex. 1 a) Divide 12823 32 xxx by 1x and express your answer in quotient form.

b) State any restrictions on the variable.

c) Write a corresponding statement that can be used to check the division.

d) Verify your answer.

6

61122

x

xx

Ex. 2 a) Divide 1294 3 xx by 12 x and express your answer in quotient form.

b) Write a corresponding statement that can be used to check the division.

Ex. 3 The volume, V, in cubic centimeters, of a rectangular box is given by 8147)( 23 xxxxV .

Determine expressions for possible dimensions of the box if the height, h, in centimeters is given

by 2x .

Ex. Divide each of the following using synthetic division.

a) )2()32( 2 xxx b) )3()953( 2 xxx

c) )12()3648( 23 xxxx d) )1()1( 3 xx

Pg. 168 # 2, 3, 4, (5 – 10)doso, 11, 12

MHF 4U Lesson 2.2 The Factor and Remainder Theorems

The Remainder Theorem

The remainder theorem states: When a polynomial f(x), is divided by x – a, the remainder is equal to f(a).

Ex. 1 a) Given: f(x) = 2332 23 xxx evaluate each of the following.

(i) f(2) (ii) f(-3) (iii) f(1)

b) Divide by each of the following.

(i) 2x (ii) 3x (iii) 1x

The Factor Theorem

From the remainder theorem, we have seen that the remainder can be found by determining the value

of f(a). By extrapolating, we can determine that if the remainder is zero, the function is evenly divisible

by the divisor.

The factor theorem states: x – a is a factor of f(x) if and only if (iff) f(a) = 0.

Ex. 2 Determine whether or not 2x is a factor of 43)( 23 xxxxf .

Ex. 3 Factor completely.

a) 652 23 xxx b) 182773 234 xxxx

c) 2723 23 xxx

Ex. 4 When 22 23 nxmxx is divided by 1x the remainder is -12 and 2x is a factor. Determine the

values of m and n.

Ex. 5 If when 54 23 kxxx is divided by 2x the remainder is 7, what is the value of

k ?

Pg. 176 # (1 – 7)doso, 9, 10, 12, 14

MHF 4U Lesson 2.3 Sum and Difference of Cubes

A sum or difference of cubes is in the form 33 ba or 33 ba .

Ex. 1 Factor 33 ba using the factor theorem.

If we use the factor theorem on 33 ba , we can see that 33 ba = ))(( 22 bababa .

Ex. 2 Factor each of the following completely.

a) 643 x b) 813 x

c) 18 3 x d) 2764 12 x

e) 63 64125

8yx f) 3615 216125 yx

g) 16)1(2 3 x h) 33 )2()3( yx

Pg. 182 # 1 - 5

MHF 4U INV 2.4 Exploring Polynomial Functions – Investigation

Polynomial functions are functions in the form ......)( 21 nnn cxbxaxxf , where a, b, c, … are real

numbers and each exponent is a WHOLE number.

1 2 3 4 5–1–2–3–4–5 x

2

4

6

8

10

12

14

16

18

20

–2

–4

–6

–8

–10

–12

–14

–16

–18

–20

y

Equation in expanded form:

_____________________________________

Degree: ________

Sign of Leading Coefficient (SOLC): _________

Number of real roots: __________

Starts in Q ______, Ends in Q ________.

Number of Turning Points: __________

1 2 3 4 5–1–2–3–4–5 x

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y


_____________________________________

Degree: ________





1 2 3 4 5–1–2–3–4–5 x

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y


_____________________________________

Degree: ________





1 2 3 4 5–1–2–3–4–5 x

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y


_____________________________________

Degree: ________





1 2 3 4 5–1–2–3–4–5 x

6

12

18

24

30

36

42

48

54

60

–6

–12

–18

–24

–30

–36

–42

–48

–54

–60

y

1 2 3 4 5–1–2–3–4–5 x

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y


_____________________________________

Degree: ________






_____________________________________

Degree: ________





1 2 3 4 5–1–2–3–4–5 x

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y


_____________________________________

Degree: ________





1 2 3 4 5–1–2–3–4–5 x

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y


_____________________________________

Degree: ________





1 2 3 4 5–1–2–3–4–5 x

2

4

6

8

10

12

14

16

18

20

–2

–4

–6

–8

–10

–12

–14

–16

–18

–20

y


___________________________________

Degree: ________

Sign of Leading Coefficient (SOLC): _______




1 2 3 4 5–1–2–3–4–5 x

15

30

45

60

75

90

105

120

135

150

–15

–30

–45

–60

–75

–90

–105

–120

–135

–150

y


___________________________________

Degree: ________





1 2 3 4 5–1–2–3–4–5 x

5

10

15

20

25

30

35

40

45

50

–5

–10

–15

–20

–25

–30

–35

–40

–45

–50

y

1 2 3 4 5–1–2–3–4–5 x

5

10

15

20

25

30

35

40

45

50

–5

–10

–15

–20

–25

–30

–35

–40

–45

–50

y


___________________________________

Degree: ________






___________________________________

Degree: ________





1 2 3 4 5–1–2–3–4–5 x

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y


_____________________________________

Degree: ________





HW: FINISH INVESTIGATION

Pg. 127 # 1 – 5, 7

MHF 4U INV 2.5 (Part I) Graphs of Polynomial Functions

1. Sketch each of the following.

2. What do you notice about all graphs that have:

a) Odd degree and negative leading coefficient

b) Odd degree and positive leading coefficient

c) Even degree and negative leading coefficient

d) Even degree and positive leading coefficient

3. Predict that general characteristics of the graph of a function that has:

a) a degree of 5 and a negative leading coefficient

b) a degree of 6 and a positive leading coefficient

c) check your predictions by sketching one possible graph from parts a) and b).

MULTIPLICITY

Pg. 136 # 1 - 6

MHF 4U Inv. 2.5 (Part II) Graphs of Polynomial Functions

1. This time try to complete the table before sketching the curves.

Polynomial

Odd or

Even

Degree

Sign of

Leading

Coefficient

Number of

Turning Points

Number

of real

zeros

End Behaviour

x x

e) )4)(1(3 xxy

f) 3)1( xy

g) )3()2(2 2 xxy

h) )4)(2( xxxy

i) )2)(1( 2 xxy

j) )2()1( 22 xxy

k) 22 )2)(3( xxy

l) 2)3)(1( xxxy

m) 3)4)(3(2 xxy

n) 22 )4()3( xxy

o) 23 )4()3(2 xxy

p) )1( 2 xxy

2. Now use the information above to sketch the above functions. Pay close attention to the end behaviours

and to the zeros (x-intercepts). Use the graphing calculators to check your graphs only.

Now we will examine the multiplicity and behavior of each zero in the previous functions.

Polynomial x–int Multiplicity &

Behaviour x–int

Multiplicity &

Behaviour x–int

Multiplicity &

Behaviour

a) 23 xy

b) 43 xy

c) 2)1( xy

d) )3)(2(2 xxy

e) )4)(1(3 xxy

f) 3)1( xy

g) )3()2(2 2 xxy

h) )4)(2( xxxy

i) )2)(1( 2 xxy

j) )2()1( 22 xxy

k) 22 )2)(3( xxy

l) 2)3)(1( xxxy

m) 3)4)(3(2 xxy

n) 22 )4()3( xxy

o) 23 )4()3(2 xxy

p) )1( 2 xxy

5. Given the polynomial y = –2(x + 1)3(x – 2)(x – 3)(x

2 – 9), determine with the help of the tables

above, but without the use of graphing technology:

a) the quadrants in which the graph originates/terminates.

b) the zeros of the function.

c) the x-intercepts of the function.

d) the y -intercept of the function.

e) the behaviour of the graph at each of the x -intercepts.

f) Sketch the graph of the function. Pg 137 # 7 - 15

SUMMARY – Graphs of Polynomial Functions

Functions with an odd degree

When the leading coefficient is positive, the graph When the leading coefficient is negative, the graph

extends from the 3rd quadrant to the 1st quadrant. extends from the 2nd quadrant to the 4th quadrant.

– Opens up to the right – Opens down to the right

* All cubic functions are symmetrical about a point.

Functions with an even degree

When the leading coefficient is positive, the graph When the leading coefficient is negative, the graph

extends from the 2nd quadrant to the 1st quadrant. extends from the 3rd quadrant to the 4th quadrant.

– Opens up – Opens down

* All quadratic functions are symmetrical about a line.

MHF 4U Investigation/Lesson 2.6 More Polynomial Functions in Factored Form

1. Draw a sketch of each graph using the properties of polynomial functions, clearly identifying all the

intercepts. Check your sketches with a graphing calculator (TI-83 or Desmos).

a) f (x)= (x - 4)(x + 3)

x

y

b) f (x) = -(x – 1)(x + 4)(x + 1)

x

y

c) f (x) = (2x - 1)(x + 1)2

x

y

c) f (x) = 2x(x -2)3

x

y

d) f (x) = - (2x - 3)2(x + 2)2

x

y

f) f (x) = x(x - 2)(x + 1)(2x+3)

x

y

g) f (x) = x3(x-4)

x

y

h) f (x) = (x +3)2(x - 3)3

x

y

i) f (x) = x(x +2)(x -1)(x-3)(x+ 4)

x

y

Ex. Sketch a possible graph of the function f(x) = –(x + 2)(x – 4)3(x – 1).

Finding Equations of Polynomial Functions

Ex. A quadratic function passes through the points (1, 0), (-2, 0), and (2, -12). Algebraically determine

the equations of this function.

Ex. Each member of a family of cubic functions has zeros of -2, 3, and 5

2.

a) Write the equation of the family of curves.

1 2 3 4 5 6 7 8 9 10 11 12 13–1–2–3–4–5–6–7–8–9–10–11–12–13 x

25

50

75

100

125

150

175

200

225

250

–25

–50

–75

–100

–125

–150

–175

–200

–225

–250

y

b) Determine the equation of the member of the family that has a y-intercept of 6.

Ex. Determine the equation of the following functions.

a) A quartic function has zeros at –1, 0, 3, and 3 and passes through the point (2, 9).

b)

Pg. 146 # 1 – 6, 8, 9, 12, 13

MHF4U INV 2.7 Transformations of Cubic and Quartic Functions HW: p. 155 # 1 - 5, 6doso, 7, 10

Parent Function: 3xy . Point: (2, 8)

Transformation 1 New

Pt.


Pt.


Pt.


Pt.

12 3 xy

33 xy

43

1 3 xy

86233 xy

Parent Function: 4xy . Point: (2, 16)

Transformation

1

New

Pt.

Transformation

2

New

Pt.

Transformation

3

New

Pt.

Transformation

4

New

Pt.

32 4 xy

41 xy

1434 xy

1023

1

2

14

xy

Review: Pg. 184 # 1 – 3, (4, 5)a, 6, 7, 8be,

9bd, 10bd, 12d, 13, 14bc, 15ac, 17, 18

mhf 4u lesson 2.0 review of factoring ex. factor each of ... · mhf 4u lesson 2.2 the factor and...

Documents