mcgraw-hill ryerson, mathematics 11 - lagout science/3_theory/mathematics... · methods used in...

CONTENTS Using McGraw-Hill Ryerson Calculus & Advanced Functions, Solutions v CHAPTER 1 Functions and Models 1 CHAPTER 2 Polynomials 23 CHAPTER 3 Limits 99 CHAPTER 4 Derivatives 168 CHAPTER 5 The Chain Rule and Its Applications 263 CHAPTER 6 Extreme Values: Curve Sketching and Optimization Problems 311 CHAPTER 7 Exponential and Logarithmic Functions 437 CHAPTER 8 Trigonometric Functions and Their Derivatives 513

iv

Using McGraw-Hill Ryerson Calculus & Advanced Functions, Solutions McGraw-Hill Ryerson Calculus & Advanced Functions, Solutions provides complete model solutions to the following: for each numbered section of McGraw-Hill Ryerson Calculus & Advanced Functions, - every odd numbered question in the Practise - all questions in the Apply, Solve, Communicate Solutions are also included for all questions in these sections: - Review - Chapter Check - Problem Solving: Using the Strategies Note that solutions to the Achievement Check questions are provided in McGraw-Hill Ryerson Calculus & Advanced Functions, Teacher’s Resource. Teachers will find the completeness of the McGraw-Hill Ryerson Calculus & Advanced Functions, Solutions helpful in planning students’ assignments. Depending on their level of ability, the time available, and local curriculum constraints, students will probably only be able to work on a selection of the questions presented in the McGraw-Hill Ryerson Calculus & Advanced Functions student text. A review of the solutions provides a valuable tool in deciding which problems might be better choices in any particular situation. The solutions will also be helpful in deciding which questions might be suitable for extra practice of a particular skill. In mathematics, for all but the most routine of practice questions, multiple solutions exist. The methods used in McGraw-Hill Ryerson Calculus & Advanced Functions, Solutions are generally modelled after the examples presented in the student text. Although only one solution is presented per question, teachers and students are encouraged to develop as many different solutions as possible. An example of such a question is Page 30, Question 7, parts b) and c). The approximate values can be found by substitution as shown or by using the Value operation on the graphing calculator. Discussion and comparison of different methods can be highly rewarding. It leads to a deeper understanding of the concepts involved in solving the problem and to a greater appreciation of the many connections among topics. Occasionally different approaches are used. This is done deliberately to enrich and extend the reader’s insight or to emphasize a particular concept. In such cases, the foundations of the approach are supplied. Also, in a few situations, a symbol that might be new to the students is introduced. For example in Chapter 3 the dot symbol is used for multiplication. When a graphing calculator is used, there are often multiple ways of obtaining the required solution. The solutions provided here sometimes use different operations than the one shown in the book. This will help to broaden students’ skills with the calculator. There are numerous complex numerical expressions that are evaluated in a single step. The solutions are developed with the understanding that the reader has access to a scientific calculator, and one has been used to achieve the result. Despite access to calculators, numerous problems offer irresistible challenges to develop their solution in a manner that avoids the need for one, through the order in which algebraic simplifications are performed. Such challenges should be encouraged.

There are a number of situations, particularly in the solutions to Practise questions, where the reader may sense a repetition in the style of presentation. The solutions were developed with an understanding that a solution may, from time to time, be viewed in isolation and as such might require the full treatment. The entire body of McGraw-Hill Ryerson Calculus & Advanced Functions, Solutions was created on a home computer in Textures. Graphics for the solutions were created with the help of a variety of graphing software, spreadsheets, and graphing calculator output captured to the computer. Some of the traditional elements of the accompanying graphic support are missing in favour of the rapid capabilities provided by the electronic tools. Since many students will be working with such tools in their future careers, discussion of the features and interpretation of these various graphs and tables is encouraged and will provide a very worthwhile learning experience. Some solutions include a reference to a web site from which data was obtained. Due to the dynamic nature of the Internet, it cannot be guaranteed that these sites are still operational.

CHAPTER 1 Functions and Models

1.1 Functions and Their Use in Modelling

Practise

Section 1.1 Page 18 Question 1

a) x ∈ [−2, 2]0 1 2-2 -1

b) x ∈ [4, 13]0 4 12 168-8 -4 13

c) x ∈ (−4,−1)0-5 -4 -3 -2 -1

d) x ∈ (0, 4)40 1 2 3

e) x ∈ (−∞, 2)0 1 2-2 -1

f) x ∈ (−1,∞)0 1 2-1 3

g) x ∈ (−∞,−1]-5 -4 -3 -2 -1

h) x ∈ [0,∞)40 1 2 3


f (−x) = 2(−x)3a)= −2x3

= −f (x)

f (x) = 2x3 is odd.

–8

–6

–4

–2

0

2

4

6

8

y

–4 –3 –2 1 2 3 4x

g(−x) = (−x)3 − 4b)= −x3 − 46= g(x) or − g(x)

g(x) = −x3 − 4 is neither even nor odd.

–8

–6

–4

–2

0

2

4

6

8

y

–4 –3 –2 1 2 3 4x

h(−x) = 1 − (−x)2c)= 1 − x2

= h(x)

h(x) = 1 − x2 is even.

–8

–6

–4

–2

0

2

4

6

8

y

–4 –3 –2 1 2 3 4x

1.1 Functions and Their Use in Modelling MHR 1


f (−x) = − |−x|a)= − |x|= f (x)

The function f (x) = − |x| is even.

–4

–3

–2

–1

0

1

2

3

4

y

–4 –3 –2 –1 1 2 3 4x

g(−x) = |−x| + 1b)= |x| + 1= g(x)

The function g(x) = |x| + 1 is even.

–2

–1

0

1

2

3

4

5

6

y

–4 –3 –2 –1 1 2 3 4x

h(−x) =∣

∣2(−x)3∣

∣c)

=∣

∣2x3∣

∣

= h(x)

The function h(x) =∣

∣2x3∣

∣ is even.

–2

–1

0

1

2

3

4

5

6

y

–4 –3 –2 –1 1 2 3 4x


a) Use f (x) =1

x2.

f (3) =1

32i)

=19

f (−3) =1

(−3)2ii)

=19

f

(

13

)

=1

(

13

)2iii)

=119

= 9

f

(

14

)

=1

(

14

)2iv)

=11

16

= 16

f

(

1k

)

=1

(

1k

)2v)

=11k2

= k2

f

(

k

1 + k

)

=1

(

k1+k

)2vi)

=1k2

(1+k)2

=(1 + k)2

k2

b) Use f (x) =x

1 − x.

f (3) =3

1 − 3i)

= −32

f (−3) =−3

1 − (−3)ii)

= −34

f

(

13

)

=13

1 − 13iii)

=1323

=12

2 MHR Chapter 1

f

(

14

)

=14

1 − 14iv)

=1434

=13

f

(

1k

)

=1k

1 − 1kv)

=1k

k−1k

=1

k − 1

f

(

k

1 + k

)

=k

1+k

1 − k1+kvi)

=k

1+k(1+k)−k

1+k

= k


Verbally: Answers will vary.

Visual representation with a scatter plot:

GRAPHING CALCULATOR

Algebraic representation using linear regression:

GRAPHING CALCULATOR

The graphing calculator suggests the functionf (x)

.= 134.9690x − 258 290.


a) The calculator suggests y = −0.75x+ 11.46 as theline of best fit.

GRAPHING CALCULATOR

b) y(12).= 2.375

c) When y = 0, x.= 15.136.

GRAPHING CALCULATOR

Apply, Solve, Communicate


Consider the function h(x), where

h(x) = f (x)g(x)

If f (x) and g(x) are odd functions, then

h(−x) = f (−x)g(−x)= −f (x)[−g(x)]= f (x)g(x)

= h(x)

The product of two odd functions is an even function.


Consider the function h(x), where

h(x) =f (x)g(x)

If f (x) and g(x) are even functions, then

h(−x) =f (−x)g(−x)

=f (x)g(x)

= h(x)

The quotient of two even functions is an even function.



The product of an odd function and an even function isan odd function. Consider h(x), where

h(x) = f (x)g(x)

If f (x) is odd and g(x) is even, then

h(−x) = f (−x)g(−x)= −f (x)g(x)= −h(x)


a) Since y = f (x) is odd, f (−x) = −f (x) for all x inthe domain of f . Given that 0 is in the domain off , we have f (−0) = −f (0) ⇒ f (0) = −f (0) ⇒2f (0) = 0 ⇒ f (0) = 0.

b) An odd function for which f (0) 6= 0 is f (x) =|x|x


a) Let V be the value, in dollars, of the computer equipment after t years. From the given information, the points

(t, V ) = (0, 18 000) and (t, V ) = (4, 9000) are on the linear function. The slope of the line is m =9000 − 18 000

4 − 0or −2250. The linear model can be defined by the function V (t) = 18 000 − 2250t.

b) V (6) = 18 000 − 2250(6) or $4500c) The domain of the function in the model is t ∈ [0, 8]. (After

8 years the equipment is worthless.)d) The slope represents the annual depreciation of the com-

puter equipment.f) Since the function is given to be linear, its slope does not

change.

e)GRAPHING CALCULATOR


a)GRAPHING CALCULATOR

b) The line of best fit is y = 0.72x − 1301.26.GRAPHING CALCULATOR

c) The model suggests the population reaches 145 940 inthe year 2010.

d) The model suggests the population was 0 in 1807. No.e) The model suggests the population will reach 1 000 000

in the year 3196. Answers will vary.f) No. Explanations will vary.

GRAPHING CALCULATOR

4 MHR Chapter 1




c) The model suggests the population will reach 348 870in the year 2010.

d) The model suggests the population was 0 in 1941.e) The model suggests the population will reach 1 000 000

in the year 2137. No.f) Explanations will vary.

GRAPHING CALCULATOR




c) Estimates will vary. The model suggests annual petexpenses of approximately $319.85.

d) Estimates will vary. The model suggests an annualincome of approximately $48 905.56 results in an-nual pet expenses of $400.

e) The model predicts pet expenses of approximately$62 959.85. Explanations will vary.

f) The model suggests an annual income of $ − 40.15.Explanations will vary.

g) No. Explanations will vary.

GRAPHING CALCULATOR




c) As speed increases the slope of the curve increases.d) Answers will vary. This curve is steeper, and its

slope increases more quickly.

b) Answers will vary. Using the quadratic regressionfeature of the calculator yields an approximation ofd = 0.008s2 + 0.002s + 0.059.

GRAPHING CALCULATOR

Section 1.1 Page 20 Question 22a) A scatter plot of the data appears below.

GRAPHING CALCULATOR

b) The ExpReg feature of the calculator approximatesthe data with V = 954.19(1.12)t.

GRAPHING CALCULATOR

c) The slope increases with time.d) The model predicts the value of the investment after

10 years to be approximately $2978.09.

GRAPHING CALCULATOR


All constant functions are even functions. The constant function f (x) = 0 is both even and odd.


The sum of an odd function and an even function can be neither odd nor even, unless one of the functions is y = 0.Let f (x) be even and g(x) be odd, and let h(x) = f (x) + g(x).If h(x) is even, then

h(−x) = h(x)f (−x) + g(−x) = f (x) + g(x)

f (x) − g(x) = f (x) + g(x)2g(x) = 0

g(x) = 0

If h(x) is odd, then

h(−x) = −h(x)f (−x) + g(−x) = −f (x) − g(x)

f (x) − g(x) = −f (x) − g(x)2f (x) = 0

f (x) = 0

6 MHR Chapter 1


Yes; only the function f (x) = 0. If a function f (x) is both even and odd, then

f (−x) = f (x)f (−x) = −f (x)

Thus,

f (x) = −f (x)2f (x) = 0

f (x) = 0


a) A possible domain is x ∈ [0, 100). Explanationsmay vary.

c) The calculator confirms that an estimate of 80% ofpollutant can be removed for $50 000.

GRAPHING CALCULATOR

b) Define y1 =25√x

√100 − x

. The costs of removal for the

percent given appear in the table below as $14 434,$25 000, $43 301 and $248 750.

GRAPHING CALCULATOR

d) Since an attempt to evaluate C(100) results in divisionby zero, the model suggests that no amount of moneywill remove all the pollutant.

GRAPHING CALCULATOR


1.2 Lies My Graphing Calculator Tells Me



a) The equations of the two vertical asymptotes arex = −3 and x = 2. There are no x-intercepts. The

y-intercept is −16

.

GRAPHING CALCULATOR

b) Factoring gives y =1

(x + 4)(x − 2). The equations

of the two vertical asymptotes are x = −4 and x = 2.There are no x-intercepts. The y-intercept is −0.125

or −18

.

GRAPHING CALCULATOR

c) Factoring gives y =x + 1

(2x + 5)(3x − 2). The equa-

tions of the two vertical asymptotes are x = −52

and x =23

. Setting y = 0 yields an x-intercept of

−1. Setting x = 0 yields a y-intercept of −1

10.

GRAPHING CALCULATOR

d) Factoring the numerator and denominator gives

y =(x − 5)(x + 4)(x − 6)(x + 5)

. The equations of the two

vertical asymptotes are x = −5 and x = 6. The

x-intercepts are −4 and 5. The y-intercept is23

.

GRAPHING CALCULATOR

Section 1.2 Page 28 Question 2a) y = −2x2 − 2x − 0.6 has no x-intercepts.

GRAPHING CALCULATOR

b) The Zero operation of the graphing calculator re-veals x-intercepts of approximately 0.731 and 1.149.

GRAPHING CALCULATOR

8 MHR Chapter 1

c) y = x3 + 2x2 − 5x + 2.3 has an x-intercept ofapproximately −3.578.

GRAPHING CALCULATOR

d) The Zero operation of the graphing calculator re-veals an x-intercept of approximately 3.582. Zoom-ing in on the interval around x = 0.459 several timesreveals no point of intersection of the function withthe x-axis.

GRAPHING CALCULATOR

Section 1.2 Page 28 Question 3a) The Intersect operation suggests a point of inter-

section at (1, 2). Substitution confirms this result.

GRAPHING CALCULATOR

b) Zooming in on the interval around x = 0.5 revealsthat the curves do not intersect.

GRAPHING CALCULATOR

c) Zooming in on the interval around x = −2 reveals that the curves intersect twice in this neighbourhood. Thecoordinates are (−2,−2) and approximately (−2.1401,−2.5406). The calculator identifies the third point of inter-section with the approximate coordinates (5.1401, 77.541).

GRAPHING CALCULATOR GRAPHING CALCULATOR

d) Zooming in on the interval around x = −1 reveals that the curves do not intersect.

GRAPHING CALCULATOR

1.2 Lies My Graphing Calculator Tells Me MHR 9


a) To avoid division by zero, x − 2 6= 0, so the domain is x ∈

Section 1.2 Page 28 Question 5a)

GRAPHING CALCULATORb) Answers may vary.c) It will not work in any window. In the window

given in part a), it will work for y = cos(2x) andy = cos(96x).


For each of these solutions, use the Zoom menu until the segment of the graph appears linear.

a) Answers may vary.

GRAPHING CALCULATOR

b) Answers may vary.

GRAPHING CALCULATOR

c) Answers may vary.

GRAPHING CALCULATOR

d) Answers may vary.

GRAPHING CALCULATOR

Section 1.2 Page 28 Question 7In the window x ∈ [−94, 94], y ∈ [−62, 62], the graphappears linear as |x| → ∞.

GRAPHING CALCULATOR

In the window x ∈ [−9.4, 9.4], y ∈ [−6.2, 6.2], thevertical asymptotes of x = 2 and x = 3 are revealed.

GRAPHING CALCULATOR

1.2 Lies My Graphing Calculator Tells Me MHR 11

In the window x ∈ [−1.7, 7.7], y ∈ [−40, 140], thepart of the function between the vertical asymptotes ishighlighted.

GRAPHING CALCULATOR


a) The graphing of a function is due in part to the cal-culator sampling elements within the domain from

Xmin to Xmax. The function y =1

x − 1.73is un-

defined at x = 1.73. To graph the function prop-erly, one sample must fall exactly at 1.73 so thediscontinuity in the graph can be detected. Trans-lating the ZDecimal window to the right 1.73 unitsresults in a correct graph. Thus, use the windowx ∈ [−2.97, 6.43], y ∈ [−3.1, 3.1]

GRAPHING CALCULATOR

b) The graphing of a function is due in part to the cal-culator sampling elements within the domain from

Xmin to Xmax. The function y =1

x − 2.684is un-

defined at x = 2.684. To graph the function prop-erly, one sample must fall exactly at 2.684 so thediscontinuity in the graph can be detected. Translat-ing the ZDecimal window to the right 2.684 unitsresults in a correct graph. Thus, use the windowx ∈ [−2.016, 7.384], y ∈ [−3.1, 3.1]

GRAPHING CALCULATOR


The domain is restricted to 2.68 − x2 > 0 or x2 < 2.68. Thus, |x| <√

2.68, or |x| < 1.6371.

The Minimum operation of the calculator helps approximate the range: y ∈ [0.6108,∞).


12 MHR Chapter 1

Review of Key Concepts

1.1 Functions and Their Use in Modelling

Section Review Page 30 Question 1

a)x y

−2 0

−1 0.5

1 3

3 1.5

b) Domain: x ∈ [−5,∞)

Section Review Page 30 Question 2a)

x y

−6 5 + |−6 + 3| = 8

−5 5 + |−5 + 3| = 7

−4 5 + |−4 + 3| = 6

−3 5 + |−3 + 3| = 5

−2 5 + |−2 + 3| = 6

−1 5 + |−1 + 3| = 7

0 5 + |0 + 3| = 8

b)

f(x)=5+|x+3|

–4

–2

0

2

4

6

8

10

12

14

y

–12 –10 –8 –6 –4 –2 2 4 6x

c) The function can be described as the sum of 5 andthe distance from x to −3 on a number line.


a) x ≤ 0 is written as x ∈ (−∞, 0].b) −4 < x is written as x ∈ (−4,+∞).c) −5 ≤ x ≤ 5 is written as x ∈ [−5, 5].


a) For all but x = ±2, f (x) = f (−x). f (x) is neither even nor odd.b) For each x-value, g(x) = −g(−x). g(x) is an odd function.c) For each x-value, h(x) = h(−x). h(x) is an even function.


f (−x) = (−x)2 + (−x)a)= x2 − x6= f (x) or −f (x)

f (x) is neither even nor odd.

g(−x) =∣

∣(−x)2 − 3∣

∣b)

=∣

∣x2 − 3∣

∣

= g(x)

g(x) is an even function.

h(−x) = 5(−x)c)= −5x= −h(x)

h(x) is an odd function.

r(−x) = (−x)2 − |−x|d)= x2 − |x|= r(x)

r(x) is an even function.

Review of Key Concepts MHR 13

s(−x) =1

(−x)3e)

= −1

x3

= −s(x)

s(x) is an odd function.

t(−x) =(

(−x)3)3

f)

= −(

x3)3

= −t(x)

t(x) is an odd function.

Section Review Page 30 Question 6Consider the function h(x), where

h(x) = f (x)g(x)

If f (x) and g(x) are even functions, then (1) becomes,

h(−x) = f (−x)g(−x)= f (x)g(x)

= h(x)

The product of two even functions is an even function.

Section Review Page 30 Question 7a) The calculator suggests d

.= −1.14t+ 26.57 as the line of best fit.

GRAPHING CALCULATOR

d(14) = −1.14(14) + 26.57b)= 10.61

d(23) = −1.14(23) + 26.57c)= 0.35


The QuadReg feature suggests the model y.= 0.09x2 + 0.08x + 0.03.

GRAPHING CALCULATOR

14 MHR Review of Key Concepts


a) The calculator suggests P.= 1231.4t−2 417 986.2

as the line of best fit, where P is the number ofpassengers, in thousands, and t is the year.

GRAPHING CALCULATOR

b) P (1995).= 38 656 800. This result is higher than

the actual value.c) The model suggests that the number of passengers

will reach 50 000 000 in 2004. No. Answers willvary.

GRAPHING CALCULATOR

Section Review Page 31 Question 10a)

GRAPHING CALCULATOR

b) The data suggest the best fuel economy of 10.1 L/100 kmis achieved at 60 km/h.

c) The QuadReg feature suggests the modelF

.= 0.001 56s2 − 0.192 68s + 16.043 16.

GRAPHING CALCULATOR

d) The curve would be higher, and the slope would be steeper. The lowest point might change.e) Translation of the curve downward would yield reduced fuel consumption for the same speeds.


a) To avoid division by zero, x + 1 6= 0. The domain is x ∈ 0 for all real numbers. The domain is x ∈

Chapter Test

Section Chapter Test Page 32 Question 1a) For f (x) = x2,

i) f (1) = 12 or 1

ii) f (−1) = (−1)2 or 1

iii) f (2) = (2)2 or 4

iv) f(

12

)

=(

12

)2

or14

b) For f (x) = 1 − x3,

i) f (1) = 1 − 13 or 0

ii) f (−1) = 1 − (−1)3 or 2

iii) f (2) = 1 − (2)3 or −7

iv) f(

12

)

= 1 −(

12

)3

or78

Section Chapter Test Page 32 Question 2

a) −4 < x < 10 is written x ∈ (−4, 10). b) x ≤ 5 is written x ∈ (−∞, 5]. c) 0 ≤ x is written x ∈ [0,+∞).


a) Since the graph of the function is rotationally symmetric with respect to the origin, the function is odd.b) Since the graph of the function is symmetric with respect to neither the origin nor the y-axis, the function is neither

odd nor even.c) Since the graph of the function is symmetric with respect to the y-axis, the function is even.


For each x-value, since f (x) 6= f (−x) and f (x) 6= −f (−x), f (x) is neither even nor odd.

For each x-value, g(x) = g(−x). g(x) is an even function.

For each x-value, h(x) = −h(−x). h(x) is an odd function.

Section Chapter Test Page 32 Question 5a)

y=x^3

–6

–4

–2

0

2

4

6

y

–6 –4 –2 2 4 6x

b) The slope has large positive values, decreases to 0 at x = 0,and then increases to large positive values.

c) The function is rotationally symmetric with respect to theorigin (odd).

d) y = x5 has a sharper turn on (−1, 1), and is steeper outsidethis interval.

Section Chapter Test Page 32 Question 6a) The calculator suggests the model V

.= 0.674r + 2.904.

GRAPHING CALCULATOR

b) V (7).= 7.624

c) V (14).= 12.343

GRAPHING CALCULATOR

16 MHR Chapter 1

Section Chapter Test Page 32 Question 7a)


b) Answers may vary. The PwrReg feature of the calculator suggests the model P.= 0.193l0.507, where P is the

period in seconds and l is the length of the pendulum in centimetres.c) P (6)

.= 0.479

d) The Solve feature of the calculator suggests a l.= 25.586 cm would yield a period of 1 s.

e) Shortening the pendulum reduces the period; the clock takes less time for each “tick” .

Section Chapter Test Page 33 Question 8a) To avoid division by 0, x2 − 1 6= 0. The domain of f (x) is x ∈ 0. The domain of f (x) is x ∈ (−3,+∞).c) The denominator can be written as (x− 3)(x+ 1). To avoid division by 0, (x− 3)(x+ 1) 6= 0. The domain of f (x)

is x ∈ 0 for all real numbers, the domain of f (x) is

Challenge Problems

Section Challenge Problems Page 34 Question 1

Let x be the number of hits the player has made to this point in the season.

x

322= 0.289

x = 0.289(322) (1)

Let y be the number of hits remaining to achieve a batting average of .300.x + y

322 + 53= 0.300 (2)

Substitute (1) into (2).0.289(322) + y

375= 0.3

y = 0.3(375) − 0.289(322).= 19.4

The batter must have 20 hits to achieve a batting average of .300. This equates to a batting average of2053

or 0.377 for

the rest of the season.

Section Challenge Problems Page 34 Question 2Let x be the date in the top left corner of the 3 by 3 square. Expressions for the remainingdates appear in the respective cells of the diagram. Let T be the total of the dates.

T = x + (x + 1) + (x + 2) + (x + 7) + (x + 8) + (x + 9) + (x + 14) + (x + 15) + (x + 16)

= 9x + 72

= 9(x + 8)

x x +1 x +2

x +7 x +8 x +9

x +14 x +15 x +16


Let T be the equivalent temperature on both scales.

T − 32212 − 32

=T − 0

100 − 0100T − 3200 = 180T

−80T = 3200T = −40

−40◦ is an equivalent temperature on both scales.


The amount of tape remaining on the reel is proportional to the area, A, of tape showing. Let r be the distance fromthe centre to the outer edge of the reel of tape.

Aremaining

Aoriginal=

34

a)

πr2 − π(2)2

π(6)2 − π(2)2=

34

πr2 − 4π36π − 4π

=34

πr2 − 4π32π

=34

πr2 − 4π = 24ππr2 = 28π

r = ±2√

7

Since r ≥ 0, r = 2√

7.

Aremaining

Aoriginal=

12

b)

πr2 − π(2)2

π(6)2 − π(2)2=

12

πr2 − 4π36π − 4π

=12

πr2 − 4π32π

=12

πr2 − 4π = 16ππr2 = 20π

r = ±2√

5

Since r ≥ 0, r = 2√

5.

Aremaining

Aoriginal=

14

c)

πr2 − π(2)2

π(6)2 − π(2)2=

14

πr2 − 4π36π − 4π

=14

πr2 − 4π32π

=14

πr2 − 4π = 8ππr2 = 12π

r = ±2√

3

Since r ≥ 0, r = 2√

3.

2 4

r

18 MHR Chapter 1


Double the first number, triple the second number, and add the two resulting numbers together.


(√a + x +

√a − x)2 = (a + x) + (a − x) + 2

√a − x

√a + x

= 2a + 2√a − x

√a + x

= 2(a +√

a2 − x2)

Since x ∈ (0, a],√

a2 − x2 <√

a2, thus,

(√a + x +

√a − x)2 < 2

(

a +√

a2)

< 2(a + a)

< 4a

< (2√a)2 (1)

Take the square root of both sides of (1).√a + x +

√a − x < 2

√a

2*sqrt(a)

sqrt(a+x)+sqrt(a-x)

–10

–8

–6

–4

–20

2

4

6

8

10

y

–10 –8 –6 –4 –2 2 4 6 8 10x


The graph can be developed by applying a sequence of transformations.

i) Graph the function f (x) = |x|.

f(x)=|x|

–20

–10

0

10

20

y

–20 –10 10 20x

ii) Reflect the graph in the x-axis: g(x) = − |x|.

g(x)=-|x|

–20

–10

0

10

20

y

–20 –10 10 20x

iii) Translate upward 10 units: h(x) = 10 − |x|.

h(x)=10-|x|

–20

–10

0

10

20

y

–20 –10 10 20x

iv) Apply the absolute value transformation: y = |10 − |x||.

y=|10-|x||

–20

–10

0

10

20

y

–20 –10 10 20x

Challenge Problems MHR 19


Rewrite the relation as |y| = 15− |x − 3| − |x − 7|. Use the definitionof absolute value to reconstruct the relation as a piecewise set.

|y| ={

15 − |x − 3| − |x − 7| ; y ≥ 0−(15 − |x − 3| − |x − 7|) ; y < 0 (1)

|x − 3| ={

x − 3 ; x ≥ 33 − x ; x < 3 (2)

|x − 7| ={

x − 7 ; x ≥ 77 − x ; x < 7 (3)

Assemble the pieces within the respective regions defined by the in-tersections of the intervals. The results are summarized in the tablebelow.

–10–8–6–4–2

0

2468

10

y

–2 2 4 6 8 10 12x

Interval y < 0 y ≥ 0

(−∞, 3)y = (3 − x) + (7 − x) − 15= −2x − 5

y = 15 − (3 − x) − (7 − x)= 2x + 5

[3, 7)y = (x − 3) + (7 − x) − 15= −11

y = 15 − (x − 3) − (7 − x)= 11

[7,∞)y = (x − 3) + (x − 7) − 15= 2x − 25

y = 15 − (x − 3) − (x − 7)= −2x + 25


The calculator assists in determining the roots of the numerator and denominator. The roots of x3 + 1.11x2 − 1.4872are −1.32, −1 and 1.21. The roots of x2 + 0.11x − 1.5972 are −1.32 and 1.21. These results define holes in

y =x3 + 1.11x2 − 1.4872x2 + 0.11x − 1.5972

at x = −1.32 and x − 1.21. For the holes to appear visually, window settings must en-sure that these domain values are sampled. Since both values are multiples of 0.11, suitable window settings forthe domain would be [−47(0.11), 47(0.11)] or [−5.17, 5.17]. To reflect proportionality, the range should be set to[−31(0.11), 31(0.11)] or [−3.41, 3.41]. The holes are depicted in the thick graphs below.


20 MHR Chapter 1

Using the Strategies

Section Problem Solving Page 37 Question 1

Once the female captain is identified, there are 3 remaining females from which 2 must be chosen. There are 3 waysthis can satisfied. From the four males, 3 must be chosen. There are 4 possible combinations for the male members.

As a result there are 4 × 3 or 12 possible combinations of members that could represent the school.


Fill the 5-L container and empty it into the 9-L container; then fill the 5-L container again and pour water into the 9-Lcontainer to fill it. There is now 1 L of water in the 5-L container. Empty the 9-L container, pour the 1 L of water fromthe 5-L container into the 9-L container, refill the 5-L container and pour it into the 9-L container. There are now 6 Lof water in the 9-L container.


Juan and Sue cross in 2 min; Sue returns in 2 min; Alicia and Larry cross in 8 min; Juan returns in 1 min; Juan andSue cross in 2 min. The total time to cross is 15 min. Hint: Alicia and Larry must cross together, and someone mustbe on the opposite side to return the flashlight.


From each vertex of convex n-gon, n − 3 diagonals can be drawn to non-adjacent vertices. Since each diagonal must

be counted only once, a convex n-gon hasn(n − 3)

2diagonals. Let the number of sides in the polygons be x and y

respectively. Solve the following system of equations.

x + y = 11

y = 11 − x (1)x(x − 3)

2+

y(y − 3)2

= 14

x2 − 3x + y2 − 3y = 28 (2)Substitute (1) into (2).

x2 − 3x + (11 − x)2 − 3(11 − x) = 28x2 − 3x + 121 − 22x + x2 − 33 + 3x = 28

2x2 − 22x + 60 = 0x2 − 11x + 30 = 0(x − 5)(x − 6) = 0

x = 5 or 6

the convex polygons that satisfy the requirements are a pentagon and a hexagon.


a) The top vertex of each pyramid meets at the centreof the cube with each face of the cube being a baseof the pyramid.

b) The dimensions are 20 cm by 20 cm by 10 cm.

A

D

B

C

E

20 cm

10 cm

20 cm

Problem Solving-Using the Strategies MHR 21


Yes. The following table gives the first month in the year offering a Friday the 13th.

January 1st First month (not a leap year) First month (leap year)

Sunday Friday January 13 Friday January 13

Monday Friday April 13 Friday September 13

Tuesday Friday September 13 Friday June 13

Wednesday Friday June 13 Friday March 13

Thursday Friday February 13 Friday February 13

Friday Friday August 13 Friday May 13

Saturday Friday May 13 Friday October 13

Section Problem Solving Page 37 Question 7Let r be the length of the diagonal of the inner square. Theradius of the circle is then r. Determine the side length ofthe square.

x2 + x2 = (2r)2

2x2 = 4r2

x2 = 2r2

Construct OP such that OP ⊥ XY. In 4XOP,

sin 6 OXP =OPOX

OX =OP

sin 6 OXP

=r

sin 30◦

= 2r

tan 6 OXP =OPPX

PX =OP

tan 6 OXP

=r

tan 30◦

=√

3r

O

X

Z

x

P

Y

r

r r

Q

Since XQ = XO + OQ, the height of 4XYZ is 2r + r or 3r. Since QY = PX =√

3r, the ratio of the areas can bedetermined.

AtriangleAsquare

=(√

3r)(3r)

2r2

=3√

32

The ratio of the area of the triangle to the area of the square is3√

32

.

22 MHR Chapter 1

CHAPTER 2 Polynomials

2.2 Dividing a Polynomial by a Polynomial

Practise


a) x + 5

x + 3)

x2 + 8x + 15

x2 + 3x

5x + 15

5x + 15

0

Restriction: x 6= −3

b) a − 2a − 5

)

a2 − 7a + 10a2 − 5a

−2a + 10−2a + 10

0

Restriction: a 6= 5

c) y + 3

y − 4)

y2 − y − 12y2 − 4y

3y − 123y − 12

0

Restriction: y 6= 4

d) t − 2t + 2

)

t2 + 0t − 4t2 + 2t

−2t − 4−2t − 4

0

Restriction: t 6= −2


a) x + 3

2x + 5)

2x2 + 11x + 15

2x2 + 5x

6x + 15

6x + 15

0


b) y + 3

3y − 1)

3y2 + 8y − 33y2 − y

9y − 39y − 3

0

Restriction: y 6=13

c) r − 65r − 1

)

5r2 − 31r + 65r2 − r

−30r + 6−30r + 6

0

Restriction: r 6=15

d) 2t − 32t + 1

)

4t2 − 4t − 34t2 + 2t

−6t − 3−6t − 3

0

Restriction: t 6= −12

e) 2r − 73r − 2

)

6r2 − 25r + 146r2 − 4r

−21r + 14−21r + 14

0

Restriction: r 6=23

f) 2x + 3

5x + 3)

10x2 + 21x + 9

10x2 + 6x

15x + 9

15x + 9

0


2.2 Dividing a Polynomial by a Polynomial MHR 23

g) 2x + 5

4x − 3)

8x2 + 14x + 15

8x2 − 6x20x + 15

20x − 1530

Restriction: x 6=34


a) 2x2 + 3

x − 1)

2x3 − 2x2 + 3x − 32x3 − 2x2

3x − 33x − 3

0

Restriction: x 6= 1

b) 3z2 + 5

z + 2)

3z3 + 6z2 + 5z + 10

3z3 + 6z2

5z + 10

5z + 10

0

Restriction: z 6= −2

c) 2m2 − 32m + 1

)

4m3 + 2m2 − 6m − 34m3 + 2m2

−6m − 3−6m − 3

0

Restriction: m 6= −12

d) 3n2 − 42n − 3

)

6n3 − 9n2 − 8n + 126n3 − 9n2

−8n + 12−8n + 12

0

Restriction: n 6=32

e) 2d + 3

5d2 + 0d + 2)

10d3 + 15d2 + 4d + 6

10d3 + 0d2 + 4d

15d2 + 6

15d2 + 6

0

Restriction: none

f) 4x − 32x2 + 0x + 1

)

8x3 − 6x2 + 4x − 38x3 + 0x2 + 4x

−6x2 − 3−6x2 − 3

0

Restriction: none

g) 4s + 1

3s2 + 0s − 5)

12s3 + 3s2 − 20s − 512s3 + 0s2 − 20s

3s2 − 53s2 − 5

0

Restriction: s 6= ±√

153

h) 3t − 47t2 + 0t − 2

)

21t3 − 28t2 − 6t + 821t3 + 0t2 − 6t

−28t2 + 8−28t2 + 8

0

Restriction: t 6= ±√

147

24 MHR Chapter 2

Section 2.2 Page 52 Question 7Since multiplication is commutative, the divisor and quotient are interchangeable.a) dividend: 255; divisor: 11; quotient: 23; remainder: 2b) dividend: 8y3 + 6y2 − 4y − 5; divisor: 4y + 3; quotient: 2y2 − 1; remainder: −2c) dividend: x2 + x + 3; divisor: x; quotient: x + 1; remainder: 3


Section 2.2 Page 52 Question 9a) i) x2 + x + 1

x − 1)

x3 + 0x2 + 0x − 1x3 − x2

x2 + 0x

x2 − xx − 1x − 1

0

Restriction: x 6= 1

ii) x2 + 2x + 4

x − 2)

x3 + 0x2 + 0x − 8x3 − 2x2

2x2 + 0x

2x2 − 4x4x − 84x − 8

0

Restriction: x 6= 2

iii) x2 + 3x + 9

x − 3)

x3 + 0x2 + 0x − 27x3 − 3x2

3x2 + 0x

3x2 − 9x9x − 279x − 27

0

Restriction: x 6= 3x2 + x + 1 = 10(x2) + 11(x) + 12(1)b)

x2 + 2x + 4 = 20(x2) + 21(x) + 22(1); The coefficients are a geometric sequence.

x2 + 3x + 9 = 30(x2) + 31(x) + 32(1)

c)x2 + 4x + 16

x − 4)

x3 + 0x2 + 0x − 64x3 − 4x2

4x2 + 0x

4x2 − 16x16x − 6416x − 64

0Restriction: x 6= 4

x2 + 5x + 25

x − 5)

x3 + 0x2 + 0x − 125x3 − 5x2

5x2 + 0x

5x2 − 25x25x − 12525x − 125

0Restriction: x 6= 5


x4 + x3 + x2 + x + 1

x − 1)

x5 + 0x4 + 0x3 + 0x2 + 0x − 1

x5 − x4

x4 + 0x3

x4 − x3

x3 + 0x2

x3 − x2

x2 + 0x

x2 − xx − 1x − 1

0

Thus,

(x5 − 1) ÷ (x − 1) = x4 + x3 + x2 + x + 1

for x 6= 1.

b)x4 + 2x3 + 4x2 + 8x + 16

x − 2)

x5 + 0x4 + 0x3 + 0x2 + 0x − 32

x5 − 2x4

2x4 + 0x3

2x4 − 4x3

4x3 + 0x2

4x3 − 8x2

8x2 + 0x

8x2 − 16x16x − 3216x − 32

0

Thus,

(x5−32)÷ (x−2) = x4+2x3+4x2+8x+16

for x 6= 2.



a) 3x + 2

2x2 + 0x + 1)

6x3 + 4x2 + 4x + 3

6x3 + 0x2 + 3x

4x2 + x + 3

4x2 + 0x + 2

x + 1

Thus, (6x3 + 4x2 + 4x + 3) ÷ (2x2 + 1) = 3x + 2 +x + 1

2x2 + 1for x ∈


a)x − 4

x + 3)

x2 − x − kx2 + 3x

−4x − k−4x − 12−k + 12

Set the remainder equal to 0 and solve for k.

−k + 12 = 0−k = −12k = 12

x + 3 divides x2 − x − k evenly if k = 12.

b) 3y + 2

2y − 1)

6y2 + y − k6y2 − 3y

4y − k4y − 2−k + 2


−k + 2 = 0−k = −2k = 2

2y − 1 is a factor of 6y2 + y − k if k = 2.

c) 4x + 13

x − 1)

4x2 + 9x + k

4x2 − 4x13x + k

13x − 13k + 13


k + 13 = 3

k = −10

x − 1 divides 4x2 + 9x + k with a remainder of 3 ifk = −10.

d)x2 + 3x + 1

2x + 1)

2x3 + 7x2 + 5x − k2x3 + x2

6x2 + 5x

6x2 + 3x

2x − k2x + 1

−k − 1

Set the remainder equal to −5 and solve for k.

−k − 1 = −5−k = −4k = 4

2x+ 1 divides 2x3 + 7x2 + 5x− k with a remainder of−5 if k = 4.


The area of the triangle is given by A =12bh, where h is the height and b is the base. Thus, b =

2Ah

.

A = 6x2 − 5x − 4 andh = 3x − 4

Thus,

b =2(6x2 − 5x − 4)

3x − 4

=2(3x − 4)(2x + 1)

3x − 4= 2(2x + 1)

= 4x + 2

The base of the triangle is represented by 4x + 2.



The area of the trapezoid is given by A =12h(a + b), where h is the height and a and b are the bases. Thus, h =

2Aa + b

.

A = 12y2 − 11y + 2a = 5y − 3b = 3y + 1

Thus,

h =2(12y2 − 11y + 2)(5y − 3) + (3y + 1)

=2(3y − 2)(4y − 1)

8y − 2

=2(3y − 2)(4y − 1)

2(4y − 1)= 3y − 2

The height of the trapezoid is represented by 3y − 2.


The expression5x2 + 14x − 3

x + 3is defined for all real numbers except x = −3. The expression 5x − 1 is defined for all

real numbers.


The amount, A, of the investment, P , after n months can be expressed as A = P (1+ i)n, where i is the interest rate perperiod, expressed as a decimal.

a) Determine the amount for P = 500, i = 0.01, and n = 10.

A(10) = 500(1 + 0.01)10.= 552.31

After 10 months, the investment is worth approximately $552.21.b) Determine the amount for P = 500, i = 0.01, and n = 25.

A(10) = 500(1 + 0.01)25.= 641.22

After 25 months, the investment is worth approximately $641.22.c) Determine the amount for P = 500, i = 0.01, and n = 12(10) or 120.

A(120) = 500(1 + 0.01)120.= 1650.19

After 10 years, the investment is worth approximately $1650.19.

28 MHR Chapter 2


a) At the end of the 22nd month, Selena withdraws $211.37 and the account is exhausted.

Month Principal Interest Savings Withdrawal Balance1 $10,000.00 $100.00 $10,100.00 $500.00 $9,600.002 $9,600.00 $96.00 $9,696.00 $500.00 $9,196.003 $9,196.00 $91.96 $9,287.96 $500.00 $8,787.964 $8,787.96 $87.88 $8,875.84 $500.00 $8,375.845 $8,375.84 $83.76 $8,459.60 $500.00 $7,959.606 $7,959.60 $79.60 $8,039.19 $500.00 $7,539.197 $7,539.19 $75.39 $7,614.59 $500.00 $7,114.598 $7,114.59 $71.15 $7,185.73 $500.00 $6,685.739 $6,685.73 $66.86 $6,752.59 $500.00 $6,252.59

10 $6,252.59 $62.53 $6,315.11 $500.00 $5,815.1111 $5,815.11 $58.15 $5,873.27 $500.00 $5,373.2712 $5,373.27 $53.73 $5,427.00 $500.00 $4,927.0013 $4,927.00 $49.27 $4,976.27 $500.00 $4,476.2714 $4,476.27 $44.76 $4,521.03 $500.00 $4,021.0315 $4,021.03 $40.21 $4,061.24 $500.00 $3,561.2416 $3,561.24 $35.61 $3,596.85 $500.00 $3,096.8517 $3,096.85 $30.97 $3,127.82 $500.00 $2,627.8218 $2,627.82 $26.28 $2,654.10 $500.00 $2,154.1019 $2,154.10 $21.54 $2,175.64 $500.00 $1,675.6420 $1,675.64 $16.76 $1,692.40 $500.00 $1,192.4021 $1,192.40 $11.92 $1,204.32 $500.00 $704.3222 $704.32 $7.04 $711.37 $500.00 $211.37

b)

Selena's GIC

$0.00

$2,000.00

$4,000.00

$6,000.00

$8,000.00

$10,000.00

$12,000.00

0 5 10 15 20 25

Time (months)

Bal

ance

c) The x-intercept (Time in months) lies somewhere between 22.4 and 22.5. This defines the time when the balancein the account is $0.



x2 − 1x − 1

=(x − 1)(x + 1)

x − 1a) i)

= x + 1, x 6= 1

x3 − 1x − 1

=(x − 1)(x2 + x + 1)

x − 1ii)

= x2 + x + 1, x 6= 1

x4 − 1x − 1

=(x − 1)(x3 + x2 + x + 1)

x − 1iii)

= x3 + x2 + x + 1, x 6= 1

x5 − 1x − 1

=(x − 1)(x4 + x3 + x2 + x + 1)

x − 1iv)

= x4 + x3 + x2 + x + 1, x 6= 1

xn − 1x − 1

=(x − 1)(xn−1 + xn−2 + . . . + x2 + x + 1)

x − 1b)

= xn−1 + xn−2 + . . . + x2 + x + 1, x 6= 1

c) Answers may vary.


a) This is a geometric series with a common ratio of x.

b) Since the series 1+ x+ x2 + x3 + . . .+ xk results when simplifying the quotientxk+1 − 1x − 1

, the formula for the sum

of the geometric series with 1 as the first term and x as the common ratio isxk+1 − 1x − 1

, x 6= 1.c) i) At the end of 1 month, the total value of the investment is S1 = 1.

ii) At the end of 2 months, the total value of the investment is S2 = 1 + (1 + r).

iii) At the end of 3 months, the total value of the investment is S3 = 1 + (1 + r) + (1 + r)2 or(1 + r)3 − 1(1 + r) − 1

.

iv) At the end of k months, the total value is Sk = 1 + (1 + r) + (1 + r)2 + . . . + (1 + r)k−1 or(1 + r)k − 1

r.

d) Use k = 120 and r = 0.01.

Sk =(1 + r)k − 1

r

S120 =(1 + 0.01)120 − 1

0.01

=(1.01)120 − 1

0.01.= 230.038 69

The total value of the investment is approximately $230.04.e) Each of the 1000 dollars accumulates in the same manner as the $1 in part d). The total value of this investment is

$1000 × 230.038 69 or $230 038.69.


a) No. When dividing a polynomial by a linear polynomial, the remainder will be a constant (lesser degree).b) No. Yes. When dividing a polynomial by a quadratic polynomial, the remainder can be either a linear polynomial

or a constant (lesser degree).c) The division of polynomials is considered complete when the degree of the remainder is less than the degree of the

divisor.


a) Every three months, Mario transfers $150 ×(1.002)3 − 1

0.002or $450.90.

b) At the end of seven years, the total value of Mario’s GIC is $450.90 ×(1.013)28 − 1

0.013or $15 112.18.

30 MHR Chapter 2

2.3 The Remainder Theorem

Practise


P (x) = 2x2 − 3x − 2a)P (1) = 2(1)2 − 3(1) − 2

= 2 − 3 − 2= −3

P (x) = 2x2 − 3x − 2b)P (0) = 2(0)2 − 3(0) − 2

= 0 − 0 − 2= −2

P (x) = 2x2 − 3x − 2c)P (2) = 2(2)2 − 3(2) − 2

= 8 − 6 − 2= 0

P (x) = 2x2 − 3x − 2d)P (−2) = 2(−2)2 − 3(−2) − 2

= 8 + 6 − 2= 12


P (x) = x2 − 5x − 3a)P (2) = 22 − 5(2) − 3

= 4 − 10 − 3= −9

The remainder is −9.

P (x) = 2x2 + x − 10b)P (2) = 2(2)2 + 2 − 10

= 8 + 2 − 10= 0

The remainder is 0.

P (x) = x3 + 2x2 − 8x + 1c)P (2) = 23 + 2(2)2 − 8(2) + 1

= 8 + 8 − 16 + 1= 1

The remainder is 1.

P (x) = x3 − 3x2 + 5x − 2d)P (2) = 23 − 3(2)2 + 5(2) − 2

= 8 − 12 + 10 − 2= 4

The remainder is 4.

P (x) = 2x3 + 3x2 − 9x − 10e)P (2) = 2(2)3 + 3(2)2 − 9(2) − 10

= 16 + 12 − 18 − 10= 0

The remainder is 0.

P (x) = 3x3 − 12x + 2f)P (2) = 3(2)3 − 12(2) + 2

= 24 − 24 + 2= 2

The remainder is 2.


P (x) = x2 + 2x + 4a)

P (2) = (2)2 + 2(2) + 4

= 4 + 4 + 4

= 12

The remainder is 12.

P (n) = 4n2 + 7n − 5b)P (−3) = 4(−3)2 + 7(−3) − 5

= 36 − 21 − 5= 10


P (x) = x3 + 2x2 − 3x + 1c)P (1) = 13 + 2(1)2 − 3(1) + 1

= 1 + 2 − 3 + 1= 1

The remainder is 1.

P (x) = x3 + 6x2 − 3x2 − x + 8d)P (−1) = (−1)3 + 6(−1)2 − 3(−1)2 − (−1) + 8

= −1 + 6 − 3 + 1 + 8= 11


2.3 The Remainder Theorem MHR 31

P (w) = 2w3 + 3w2 − 5w + 2e)P (−3) = 2(−3)3 + 3(−3)2 − 5(−3) + 2

= −54 + 27 + 15 + 2= −10


P (y) = y3 − 8f)P (−2) = (−2)3 − 8

= −8 − 8= −16


P (x) = x3 + x2 + 3g)

P (−4) = (−4)3 + (−4)2 + 3= −64 + 16 + 3= −45


P (x) = 1 − x3h)P (1) = 1 − 13

= 1 − 1= 0

The remainder is 0.

P (m) = m4 − 2m3 + m2 + 12m − 6i)P (2) = 24 − 2(2)3 + 22 + 12(2) − 6

= 16 − 16 + 4 + 24 − 6= 22


P (t) = 2 − 3t + t2 + t3j)P (4) = 2 − 3(4) + 42 + 43

= 2 − 12 + 16 + 64= 70


P (y) = 2y4 − 3y2 + 1k)P (3) = 2(3)4 − 3(3)2 + 1

= 162 − 27 + 1= 136


P (x) = 2 − x + x2 − x3 − x4l)P (−2) = 2 − (−2) + (−2)2 − (−2)3 − (−2)4

= 2 + 2 + 4 + 8 − 16= 0

The remainder is 0.

P (x) = 3x2 −√

2x + 3m)

P (−√

2) = 3(−√

2)2 −√

2(−√

2) + 3

= 6 + 2 + 3

= 11


P (r) = 2r4 − 4r2 − 9n)

P (√

3) = 2(√

3)4 − 4(√

3)2 − 9= 18 − 12 − 9= −3



P (x) = kx2 + 3x + 1a)

P (−2) = 3k(−2)2 + 3(−2) + 1 = 3

4k − 5 = 34k = 8

k = 2

P (x) = x3 + x2 + kx − 17b)P (2) = 3

23 + 22 + k(2) − 17 = 32k − 5 = 3

2k = 8

k = 4

P (x) = x3 + 4x2 − x + kc)P (1) = 3

13 + 4(1)2 − 1 + k = 3k = −1

P (x) = x3 + kx2 + x + 2d)

P (−1) = 3(−1)3 + k(−1)2 + (−1) + 2 = 3

k = 3

32 MHR Chapter 2


P (x) = px3 − x2 + qx − 2P (1) = 0

p(1)3 − 12 + q(1) − 2 = 0p − 1 + q − 2 = 0

p + q = 3 (1)

P (−2) = −18p(−2)3 − (−2)2 + q(−2) − 2 = −18

−8p − 4 − 2q − 2 = −188p + 2q = 12

4p + q = 6 (2)

Subtract (1) from (2).

3p = 3

p = 1 (3)

Substitute (3) into (1).

1 + q = 3

q = 2

A check by substitution confirms p = 1 and q = 2.



Let P (x) = 2x3 + 4x2 − kx + 5.

P (−3) = 2(−3)3 + 4(−3)2 − k(−3) + 5= −54 + 36 + 3k + 5= 3k − 13 (1)

Let Q(y) = 6y3 − 3y2 + 2y + 7.

Q

(

12

)

= 6(

12

)3

− 3(

12

)2

+ 2(

12

)

+ 7

=34−

34+ 1 + 7

= 8 (2)

Equate (1) and (2).

3k − 13 = 83k = 21

k = 7


P (x) = kx3 − 3x2 + 5x − 8 (1)P (2) = 22

k(2)3 − 3(2)2 + 5(2) − 8 = 228k − 12 + 10 − 8 = 22

8k = 32

k = 4 (2)


P (x) = 4x3 − 3x2 + 5x − 8Determine the remainder of P (x) when divided by x + 1.

P (−1) = 4(−1)3 − 3(−1)2 + 5(−1) − 8= −4 − 3 − 5 − 8= −20




A(h) = h2 + 0.5ha)

A

(

72

)

=(

72

)2

+12

(

72

)

=494

+74

=564

= 14


b) When the height is72

, the area is 14.


P (n) = 6n2 − 5n + 8a)

P

(

−12

)

= 6(

−12

)2

− 5(

−12

)

+ 8

=64+

52+

162

=32+

52+

162

= 12



h(d) = 0.0003d2 + 2a)

h(500) = 0.0003(500)2 + 2

= 77


h(d) = 0.0003d2 + 2b)

h(−500) = 0.0003(−500)2 + 2= 77


- 400 - 200 200 400d

20

40

60

80

100h

c) The results are equal.d) At 500 m either side of the centre point, the cable is 77 m above the roadway.


a) −0.017d + 0.45d − 50

)

−0.017d2 + 1.3d + 2.5−0.017d2 + 0.85d

0.45d + 2.5

0.45d − 22.525

b) When the horizontal distance is 50 m, the height is 25 m.

c) −0.017d − 0.06d − 80

)

−0.017d2 + 1.3d + 2.5−0.017d2 + 1.36d

−0.06d + 2.5−0.06d − 4.8

−2.3

d) Since the hammer cannot have a negative height,it must have landed closer than 80 m from thethrower.


0.5t + 3

t − 2)

0.5t2 + 2t + 0

0.5t2 − t3t + 0

3t − 66

Q(t) = 0.5t + 3, R = 6. Q(t) is the average number of ticketsissued between 2 h and time t. R is the number of ticketsissued in the first 2 h.

b) Answers will vary.c) The formula is impossible. It would require

issuing a negative number of tickets in eachof the last 3 h.

34 MHR Chapter 2


a) Answers will vary.b)

0.1t + 10.5

t − 5)

0.1t2 + 10t + 25

0.1t2 − 0.5t10.5t + 25

10.5t − 52.577.5

Q(t) = 0.1t + 10.5, R = 77.5. Q(t) represents theaverage number of trees sold from December 5 today t. R is the number of trees sold by December 5.


f (1) = f (−2)13 + 6(1)2 + k(1) − 4 = (−2)3 + 6(−2)2 + k(−2) − 4

1 + 6 + k = −8 + 24 − 2k3k = 9

k = 3

c) No. The demand for trees falls sharply after December 24.


P (x) = x2 + 5x + 7

P (−k) = 3(−k)2 + 5(−k) + 7 = 3

k2 − 5k + 7 = 3k2 − 5k + 4 = 0

(k − 1)(k − 4) = 0k = 1 or 4


a) x−a is a factor of bx2+cx+d and ba2+ca+d = 0.

P (x) = bx2 + cx + db)

P (a) = 0

b(a)2 + c(a) + d = 0

ba2 + ca + d = 0

a =−c ±

√

c2 − 4bd2b

Section 2.3 Page 60 Question 22a) Quotient: −Q(x); Remainder: R.b) Answers will vary.

x2 + 2x + 3x − 1

= x + 3 +6

x − 1x2 + 2x + 3−x + 1

= −x − 3 +6

−x + 1

P (x) = (x − b)Q(x) + Rc)P (x)x − b

= Q(x) +R

x − b

−P (x)x − b

= −(

Q(x) +R

x − b

)

P (x)b − x

= −Q(x) +R

b − x

Section 2.3 Page 61 Question 23Answers will vary.

P (x) = (x + 1)(x − 3) − 4a)= x2 − 2x − 7

Check.

P (3) = (3 + 1)(3 − 3) − 4= −4

Q(x) = (x2 + x + 1)(x + 2) + 3b)

= x3 + 2x2 + x2 + 2x + x + 2 + 3

= x3 + 3x2 + 3x + 5

Check.

Q(−2) = (−2)3 + 3(−2)2 + 3(−2) + 5= −8 + 12 − 6 + 5= 3

R(x) = x3(2x − 1) + 1c)= 2x4 − x3 + 1

Check.

R

(

12

)

= 2(

12

)4

−(

12

)3

+ 1

=18−

18+ 1

= 1



x2 − 1x2 + 0x + 1

)

x4 + 0x3 + 0x2 + 0x − 1x4 + 0x3 + x2

−x2 + 0x − 1−x2 + 0x − 1

0

Quotient: x2 − 1; Remainder: 0

b)x + 2

x2 + 0x + 1)

x3 + 2x2 + 5x + 1

x3 + 0x2 + x

2x2 + 4x + 1

2x2 + 0x + 2

4x − 1

Quotient: x + 2; Remainder: 4x − 1

c)x6 + x4 + x2 + 1

x2 + 0x − 1)

x8 + 0x7 + 0x6 + 0x5 + 0x4 + 0x3 + 0x2 + 0x − 1

x8 + 0x7 − x6

x6 + 0x5 + 0x4

x6 + 0x5 − x4

x4 + 0x3 + 0x2

x4 + 0x2 − x2

x2 + 0x − 1x2 + 0x − 1

0

Quotient: x6 + x4 + x2 + 1; Remainder: 0; Restriction: x 6= ±1

d)2x − 3

x2 − x + 0)

2x3 − 5x2 + 6x − 72x3 − 2x2 + 0x

−3x2 + 6x − 7−3x2 + 3x + 0

3x − 7

Quotient: 2x − 3; Remainder: 3x − 7; Restriction: x 6= 0, 1


The remainder upon division of P (x) by x2 + a can be found by replacing x2 with −a wherever it occurs in P (x). Theremainder upon division of P (x) by x2 + bx can be found by replacing x2 with −bx wherever it occurs in P (x). Theconjectures can be justified using the division statements P (x) = Q(x)(x2+a)+R(x) and P (x) = Q(x)(x2+bx)+R(x).

36 MHR Chapter 2

2.4 The Factor Theorem

Practise


P (x) = x3 − 3x2 + 4x − 2a)P (1) = 13 − 3(1)2 + 4(1) − 2

= 1 − 3 + 4 − 2= 0

Since P (1) = 0, x − 1 is a factor of P (x).

P (x) = 2x3 − x2 − 3x − 2b)P (1) = 2(1)3 − 12 − 3(1) − 2

= 2 − 1 − 3 − 2= −4

Since P (1) 6= 0, x − 1 is not a factor of P (x).

P (x) = 3x3 − x − 3c)P (1) = 3(1)3 − (1) − 3

= 3 − 1 − 3= −1

Since P (1) 6= 0, x − 1 is not a factor of P (x).

P (x) = 2x3 + 4x2 − 5x − 1d)P (1) = 2(1)3 + 4(1)2 − 5(1) − 1

= 2 + 4 − 5 − 1= 0



P (x) = x3 + 2x2 + 2x + 1a)

P (−1) = (−1)3 + 2(−1)2 + 2(−1) + 1= −1 + 2 − 2 + 1= 0

Since P (−1) = 0, x + 1 is a factor of P (x).

P (x) = x3 − 3x2 + 4x − 4b)P (2) = 23 − 3(2)2 + 4(2) − 4

= 8 − 12 + 8 − 4= 0


P (m) = m3 − 3m2 + m − 3c)P (3) = 33 − 3(3)2 + 3 − 3

= 27 − 27 + 3 − 3= 0

Since P (3) = 0, m − 3 is a factor of P (m).

P (x) = x3 + 7x2 + 17x + 15d)

P (−3) = (−3)3 + 7(−3)2 + 17(−3) + 15= −27 + 63 − 51 + 15= 0


P (x) = 4x2 − 9x − 9e)P (3) = 4(3)2 − 9(3) − 9

= 36 − 27 − 9= 0


P (x) = 6x2 − 11x − 17f)P (−1) = 6(−1)2 − 11(−1) − 17

= 6 + 11 − 17= 0


P (y) = 2y3 − 5y2 + 2y + 1g)P (1) = 2(1)3 − 5(1)2 + 2(1) + 1

= 2 − 5 + 2 + 1= 0

Since P (1) = 0, y − 1 is a factor of P (y).

P (x) = x3 − 6x − 4h)P (−2) = (−2)3 − 6(−2) − 4

= −8 + 12 − 4= 0


2.4 The Factor Theorem MHR 37


P (x) = 2x3 + x2 + 2x + 1a)

P

(

−12

)

= 2(

−12

)3

+(

−12

)2

+ 2(

−12

)

+ 1

= −14+

14− 1 + 1

= 0

Since P

(

−12

)

= 0, 2x + 1 is a factor of P (x).

P (x) = 2x3 − 3x2 − 2x + 3b)

P

(

32

)

= 2(

32

)3

− 3(

32

)2

− 2(

32

)

+ 3

=274

−274

− 3 + 3

= 0

Since P

(

32

)

= 0, 2x − 3 is a factor of P (x).

P (y) = 3y3 + 8y2 + 3y − 2c)

P

(

13

)

= 3(

13

)3

+ 8(

13

)2

+ 3(

13

)

− 2

=19+

89+ 1 − 2

= 0

Since P

(

13

)

= 0, 3y − 1 is a factor of P (y).

P (n) = 6n3 − 7n2 + 1d)

P

(

−13

)

= 6(

−13

)3

− 7(

−13

)2

+ 1

= −29−

79+ 1

= 0

Since P

(

−13

)

= 0, 3n + 1 is a factor of P (n).

P (x) = 8x2 − 2x − 1e)

P

(

12

)

= 8(

12

)2

− 2(

12

)

− 1

= 2 − 1 − 1= 0

Since P

(

12

)


P (x) = 3x3 − x2 − 3x + 1f)

P

(

13

)

= 3(

13

)3

−(

13

)2

− 3(

13

)

+ 1

=19−

19− 1 + 1

= 0

Since P

(

13

)



a) The possible roots are ±1,±12,±3,±

32

.

P (x) = 2x3 − 9x2 + 10x − 3P (1) = 2(1)3 − 9(1)2 + 10(1) − 3

= 2 − 9 + 10 − 3= 0 0

1 2 -9 10 -3

2

2

-7

-7

3

3

Since P (1) = 0, x − 1 is one factor of P (x). Synthetic division yields another factor of 2x2 − 7x + 3.

P (x) = (x − 1)(2x2 − 7x + 3)= (x − 1)(2x − 1)(x − 3)

So, 2x3 − 9x2 + 10x − 3 = (x − 1)(2x − 1)(x − 3).b) The possible roots are ±1,±

12,±

14,±3,±

32,±

34

.

P (y) = 4y3 − 7y − 3P (−1) = 4(−1)3 − 7(−1) − 3

= −4 + 7 − 3= 0 0

-1 4 0 -7 -3

4

-4

-4

4

-3

3

38 MHR Chapter 2

Since P (−1) = 0, y + 1 is one factor of P (y). Synthetic division yields another factor of 4y2 − 4y − 3.

P (y) = (y + 1)(4y2 − 4y − 3)= (y + 1)(2y + 1)(2y − 3)

So, 4y3 − 7y − 3 = (y + 1)(2y + 1)(2y − 3).c) The possible roots are ±1,±

13,±2,±

23,±3,±6.

P (x) = 3x3 − 4x2 − 17x + 6P (−2) = 3(−2)3 − 4(−2)2 − 17(−2) + 6

= −24 − 16 + 34 + 6= 0

0

-2 3 -4 -17 6

3

-6

-10

20

3

-6

Since P (−2) = 0, x + 2 is one factor of P (x). Synthetic division yields another factor of 3x2 − 10x + 3.

P (x) = (x + 2)(3x2 − 10x + 3)= (x + 2)(3x − 1)(x − 3)

So, 3x3 − 4x2 − 17x + 6 = (x + 2)(3x − 1)(x − 3).d) The possible roots are ±1,±

13,±2,±

23,±4,±

43,±8,±

83

.

P (x) = 3x3 − 2x2 − 12x + 8P (2) = 3(2)3 − 2(2)2 − 12(2) + 8

= 24 − 8 − 24 + 8= 0

0

2 3 -2 -12 8

3

6

4

8

-4

-8

Since P (2) = 0, x − 2 is one factor of P (x). Synthetic division yields another factor of 3x2 + 4x − 4.

P (x) = (x − 2)(3x2 + 4x − 4)= (x − 2)(3x − 2)(x + 2)

So, 3x3 − 2x2 − 12x + 8 = (x − 2)(3x − 2)(x + 2).e) The possible roots are ±1,±

12,±2,±3,±

32,±4,±6,±12.

P (x) = 2x3 + 13x2 + 23x + 12

P (−1) = 2(−1)3 + 13(−1)2 + 23(−1) + 12= −2 + 13 − 23 + 12= 0

0

-1 2 13 23 12

2

-2

11

-11

12

-12

Since P (−1) = 0, x + 1 is one factor of P (x). Synthetic division yields another factor of 2x2 + 11x + 12.

P (x) = (x + 1)(2x2 + 11x + 12)

= (x + 1)(2x + 3)(x + 4)

So, 2x3 + 13x2 + 23x + 12 = (x + 1)(2x + 3)(x + 4).


f) The possible roots are ±1,±12,±2,±5,±

52,±10.

P (x) = 2x3 − 3x2 + 3x − 10P (2) = 2(2)3 − 3(2)2 + 3(2) − 10

= 16 − 12 + 6 − 10= 0 0

2 2 -3 3 -10

2

4

1

2

5

10

Since P (2) = 0, x − 2 is one factor of P (x). Synthetic division yields another factor of 2x2 + x + 5.

P (x) = (x − 2)(2x2 + x + 5)

2x2 + x + 5 cannot be factored. So, 2x3 + 13x2 + 23x + 12 = (x − 2)(2x2 + x + 5).g) The possible roots are ±1,±

12,±

13,±

16,±3,±

32,±5,±

52,±

53,±

56,±15,±

152

.

P (x) = 6x3 − 11x2 − 26x + 15P (3) = 6(3)3 − 11(3)2 − 26(3) + 15

= 162 − 99 − 78 + 15= 0 0

3 6 -11 -26 15

6

18

7

21

-5

-15

Since P (3) = 0, x − 3 is one factor of P (x). Synthetic division yields another factor of 6x2 + 7x − 5.

P (x) = (x − 3)(6x2 + 7x − 5)= (x − 3)(2x − 1)(3x + 5)

So, 6x3 − 11x2 − 26x + 15 = (x − 3)(2x − 1)(3x + 5).h) The possible roots are ±1,±

12,±

14,±2.

P (y) = 4y3 + 8y2 − y − 2P (−2) = 4(−2)3 + 8(−2)2 − (−2) − 2

= −32 + +32 + 2 − 2= 0 0

-2 4 8 -1 -2

4

-8

0

0

-1

2

Since P (−2) = 0, y + 2 is one factor of P (y). Synthetic division yields another factor of 4y2 − 1.

P (y) = (y + 2)(4y2 − 1)= (y + 2)(2y − 1)(2y + 1)

So, 4y3 + 8y2 − y − 2 = (y + 2)(2y − 1)(2y + 1).i) The coefficients suggest a factor by grouping strategy.

P (x) = 4x3 + 3x2 − 4x − 3= x2(4x + 3) − 1(4x + 3)= (4x + 3)(x2 − 1)= (4x + 3)(x − 1)(x + 1)

j) The possible roots are ±1,±12,±

13,±

16,±5,±

52,±

53,±

56

.

P (w) = 6w3 + 16w2 − 21w + 5

P

(

13

)

= 6(

13

)3

+ 16(

13

)2

− 21(

13

)

+ 5

=29+

169

− 7 + 5

= 0

40 MHR Chapter 2

Since P

(

13

)

= 0, 3w − 1 is one factor of P (w). Long division yields another factor of 2w2 + 6w − 5. Since

b2 − 4ac = 76, which is not a perfect square, 2w2 + 6w − 5 cannot be factored further.

So, 6w3 + 16w2 − 21w + 5 = (3w − 1)(2w2 + 6w − 5).



For this solution, the model A(t) = 3t2 + 2t defines the total amount of forest cut down after t years.a) A(2) = 3(2)2 + 2(2) or 16. A(2) is the number of millions of hectares of forest cut down in 2002.

A(5) − A(2) = 3(5)2 + 2(5) −[

3(2)2 + 2(2]

b)

= 85 − 16= 69

A(5) − A(2) is the number of millions of hectares of forest cut down between 2002 and 2005.

A(5) − A(2)5 − 2

=3(5)2 + 2(5) −

[

3(2)2 + 2(2]

3c)

=85 − 16

3

=693

= 23

A(5) − A(2)5 − 2

is the average area of forest cut down between 2002 and 2005.

d)A(t) − A(2)

t − 2is the average area of forest cut down between 2002 and 2000 + t.

e) Answers may vary.


f (x) − f (b) = f (x) − f (3)a) i)= x2 − 32

= x2 − 9= (x − 3)(x + 3)

f (x) − f (b)x − b

=f (x) − f (3)

x − 3ii)

=(x − 3)(x + 3)

x − 3= x + 3, x 6= 3

f (x) − f (b) = f (x) − f (2)b) i)= x2 + 2x − 1 − (22 + 2(2) − 1)= x2 + 2x − 1 − 7= x2 + 2x − 8= (x + 4)(x − 2)

f (x) − f (b)x − b

=f (x) − f (3)

x − 2ii)

=(x + 4)(x − 2)

x − 2= x + 4, x 6= 2

f (x) − f (b) = f (x) − f (−1)c) i)= 2x3 + 4x2 − 5x − 7 − (2(−1)3 + 4(−1)2 − 5(−1) − 7)= 2x3 + 4x2 − 5x − 7= (x + 1)(2x2 + 2x − 7)

f (x) − f (b)x − b

=f (x) − f (−1)

x − (−1)ii)

=(x + 1)(2x2 + 2x − 7)

x + 1= 2x2 + 2x − 7, x 6= 1



a)f (x)x − b

= Q(x) +R(x)x − b

c) Q(x) =f (x) − f (b)

x − b

f (x)x − b

= Q(x) +R(x)x − b

b)

f (x) = Q(x)(x − b) + R(x)R(x) = f (x) −Q(x)(x − b)R(b) = f (b) −Q(b)(b − b)

= f (b)

The division statement can be rewritten asf (x)x − b

= Q(x) +f (b)x − b

.

d) Yes; reasons will vary. The numerator is always divisible by the denominator, x − b.

e)f (x) − f (b)

x − bdefines the slope of the line from the point (b, f (b)) to the point (x, f (x)).


For this solution, the model M (t) = 5t3 + 2t2 + 10t defines the total emissions after t years.

M (t) = 5t3 + 2t2 + 10ta)

M (4) = 5(4)3 + 2(4)2 + 10(4)

= 320 + 32 + 40

= 392

The mass of emissions in the year 2004 is 392 million tonnes.

M (7) −M (4) = 5(7)3 + 2(7)2 + 10(7) −[

5(4)3 + 2(4)2 + 10(4)]

b)

= 1715 + 98 + 70 − 392= 1491

The total mass of emissions from 2004 to 2007 is 1491 million tonnes.

M (7) −M (4)7 − 4

=1491

3c)

= 497

The average mass of emissions from 2004 to 2007 is 497 million tonnes per year.

d)M (t) −M (1)

t − 1represents the average mass emitted over the period from 2001 to 2000 + t, where t > 1.


P (x) = x4 + 4x3 − 7x2 − 34x − 24a)P (−1) = (−1)4 + 4(−1)3 − 7(−1)2 − 34(−1) − 24

= 1 − 4 − 7 + 34 − 24= 0

x + 1 is one factor of P (x). Synthetic division yields another factor of x3 + 3x2 − 10x − 24.

Q(x) = x3 + 3x2 − 10x − 24Q(−2) = (−2)3 + 3(−2)2 − 10(−2) − 24

= −8 + 12 + 20 − 24= 0

x + 2 is also a factor of P (x). Synthetic division yields another factor of x2 + x − 12. Summarizing,

P (x) = (x + 1)Q(x)

= (x + 1)(x + 2)(x2 + x − 12)= (x + 1)(x + 2)(x + 4)(x − 3)

So, x4 + 4x3 − 7x2 − 34x − 24 = (x + 1)(x + 2)(x + 4)(x − 3).

42 MHR Chapter 2

b) The expression can be factored using a strategy similar to a). Factoring by grouping can also be employed.

P (x) = x5 + 3x4 − 5x3 − 15x2 + 4x + 12= x4(x + 3) − 5x2(x + 3) + 4(x + 3)= (x + 3)(x4 − 5x2 + 4)= (x + 3)(x2 − 1)(x2 − 4)= (x + 3)(x + 1)(x − 1)(x + 2)(x − 2)

So, x5 + 3x4 − 5x3 − 15x2 + 4x + 12 = (x + 3)(x + 1)(x − 1)(x + 2)(x − 2).

c) Factor by grouping.

P (x) = 8x3 + 4x2 − 2x − 1= 4x2(2x + 1) − 1(2x + 1)= (2x + 1)(4x2 − 1)= (2x + 1)(2x + 1)(2x − 1)= (2x + 1)2(2x − 1)

So, 8x3 + 4x2 − 2x − 1 = (2x + 1)2(2x − 1).

d) Factor by grouping.

P (x) = 8x3 − 12x2 − 2x + 3= 4x2(2x − 3) − 1(2x − 3)= (2x − 3)(4x2 − 1)= (2x − 3)(2x + 1)(2x − 1)

So, 8x3−12x2−2x+3 = (2x−3)(2x+1)(2x−1).


P (x) = x4 + 6x3 + 11x2 + 6x

= x(x3 + 6x2 + 11x + 6)

= xQ(x)

Q(−1) = (−1)3 + 6(−1)2 + 11(−1) + 6= −1 + 6 − 11 + 6= 0

To this point, x and x + 1 are factors of P (x). Synthetic division of Q(x) by x + 1 reveals

P (x) = xQ(x)

= x(x + 1)(x2 + 5x + 6)

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

So, x4 + 6x3 + 11x2 + 6x = x(x + 1)(x + 2)(x + 3). The expression is the product of four consecutive integers.


P (x) = 4x4 − 3x3 − 2x2 + kx − 9a)P (3) = 0

4(3)4 − 3(3)3 − 2(3)2 + k(3) − 9 = 0324 − 81 − 18 + 3k − 9 = 0

3k = −216k = −72

P (x) = kx3 − 10x2 + 2x + 3b)P (3) = 0

k(3)3 − 10(3)2 + 2(3) + 3 = 027k − 90 + 9 = 0

27k = 81

k = 3


P (2) = 0 ⇐⇒ x − 2 is a factor.a)P (−4) = 0 ⇐⇒ x + 4 is a factor.

P

(

−13

)

= 0 ⇐⇒ 3x + 1 is a factor.

b) No. A third-degree polynomial can have at most three factors.



P (x) = 6x3 + mx2 + nx − 5P (−1) = 0

6(−1)3 + m(−1)2 + n(−1) − 5 = 0−6 + m − n − 5 = 0

m − n = 11 (1)P (1) = −4

6(1)3 + m(1)2 + n(1) − 5 = −46 + m + n − 5 = −4

m + n = −5 (2)Add (1) and (2).

2m = 6

m = 3 (3)


3 + n = −5n = −8

A check by substitution confirms m = 3 and n = −8.


Factor by grouping.P (x) = x3 − ax2 + bx2 − abx + cx − ac

= x2(x − a) + bx(x − a) + c(x − a)= (x − a)(x2 + bx + c)

Thus, x − a is a factor of x3 − ax2 + bx2 − abx + cx − ac.


P (x) = x3 − 1a) i)P (1) = 13 − 1

= 0

x − 1 is a factor of x3 − 1. Synthetic division revealsanother factor of x2 + x + 1.

x3 − 1 = (x − 1)(x2 + x + 1)

P (x) = x3 + 1ii)

P (−1) = (−1)3 + 1= 0

x + 1 is a factor of x3 + 1. Synthetic division revealsanother factor of x2 − x + 1.

x3 + 1 = (x + 1)(x2 − x + 1)

P (x) = x3 − 27iii)P (3) = 33 − 27

= 0

x − 3 is a factor of x3 − 27. Synthetic division revealsanother factor of x2 + 3x + 9.

x3 − 27 = (x − 3)(x2 + 3x + 9)

P (x) = x3 + 64iv)

P (−4) = (−4)3 + 64= 0

x + 4 is a factor of x3 + 64. Synthetic division revealsanother factor of x2 − 4x + 16.

x3 + 64 = (x + 4)(x2 − 4x + 16)

44 MHR Chapter 2

P (x) = 8x3 − 1v)

P

(

12

)

= 8(

12

)3

− 1

= 1 − 1= 0

2x− 1 is a factor of 8x3 − 1. Synthetic division revealsanother factor of 4x2 + 2x + 1.

8x3 − 1 = (2x − 1)(4x2 + 2x + 1)

P (x) = 64x3 + 1vi)

P

(

−14

)

= 64(

−14

)3

+ 1

= −1 + 1= 0

4x+1 is a factor of 64x3+1. Synthetic division revealsanother factor of 16x2 − 4x + 1.

64x3 + 1 = (4x + 1)(16x2 − 4x + 1)

b) The expressions have the same signs. x + y is a factor of x3 + y3. The other factor is x2 − xy + y2.c) The expressions have the same signs. x − y is a factor of x3 − y3. The other factor is x2 + xy + y2.

P (x) = 8x3 + 125d)

= (2x)3 + (5)3

= (2x + 5)((2x)2 − (2x)(5) + (5)2)= (2x + 5)(4x2 − 10x + 25)

Q(x) = 27x3 − 64= (3x)3 − (4)3

= (3x − 4)((3x)2 + (3x)(4) + (4)2)= (3x − 4)(9x2 + 12x + 16)

P (x) = x6 + y6e)

= (x2)3 + (y2)3

= (x2 + y2)(x4 − x2y2 + y4)

f) Answers may vary. x − y is not a factor of x3n + y3n; x + y is a factor of x3n + y3n if and only if n is odd.g) Answers may vary. x − y is a factor of x3n − y3n; x + y is a factor of x3n − y3n if and only if n is even.


a) Both x + 1 and x − 1 are factors of x100 − 1.

P (x) = x100 − 1P (±1) = (±1)100 − 1

= 1 − 1= 0

c) Let P (x) = xn − 1. Since P (1) = 0, x − 1 is afactor of every polynomial of the form xn − 1.

d) Let P (x) = xn + 1. Since P (−1) = 0 for all oddn, x+ 1 is a factor of every polynomial of the formxn + 1, where n is odd.

P (x) = x99 + 1b)

P (−1) = (−1)99 + 1= −1 + 1= 0

x + 1 is a factor of x99 + 1.

P (1) = (1)99 + 1

= 1 + 1

= 2

Since P (1) 6= 0, x − 1 is not a factor of x99 + 1.



V (r) =43π(r3 + 9r2 + 27r + 27)

V (−3) =43π((−3)3 + 9(−3)2 + 27(−3) + 27)

=43π(−27 + 81 − 81 + 27)

= 0

Since V (−3) = 0, r + 3 is a factor of V (r). Synthetic division of V (r) by r + 3 reveals another factor of r2 + 6r + 9.

V (r) =43π(r + 3)(r2 + 6r + 9)

=43π(r + 3)(r + 3)(r + 3)

=43π(r + 3)3

The radius of the larger sphere is r + 3.


P (x) = x4 + y4a) i)

P (−y) = (−y)4 + y4

= y4 + y4

= 2y4

x + y is not a factor of x4 + y4.

P (y) = (y)4 + y4

= 2y4

x − y is not a factor of x4 + y4.

P (x) = x4 − y4ii)P (−y) = (−y)4 − y4

= y4 − y4

= 0

x + y is a factor of x4 − y4.P (y) = (y)4 − y4

= 0

x − y is a factor of x4 − y4.

P (x) = x5 + y5iii)

P (−y) = (−y)5 + y5

= −y5 + y5

= 0

x + y is a factor of x5 + y5.

P (y) = y5 + y5

= 2y5

x − y is a not factor of x5 + y5.

P (x) = x5 − y5iv)P (−y) = (−y)5 − y5

= −y5 − y5

= −2y5

x + y is not a factor of x5 − y5.P (y) = y5 − y5

= 0


P (x) = x6 + y6v)

P (−y) = (−y)6 + y6

= y6 + y6

= 2y6

x + y is not a factor of x6 + y6.

P (y) = y6 + y6

= 2y6

x − y is not a factor of x6 + y6.

P (x) = x6 − y6vi)P (−y) = (−y)6 − y6

= y6 − y6

= 0

x + y is a factor of x6 − y6.P (y) = y6 − y6

= 0


46 MHR Chapter 2

P (x) = x7 + y7vii)

P (−y) = (−y)7 + y7

= −y7 + y7

= 0

x + y is a factor of x7 + y7.

P (y) = y7 + y7

= 2y7

x − y is a not factor of x7 + y7.

P (x) = x7 − y7viii)P (−y) = (−y)7 − y7

= −y7 − y7

= −2y7

x + y is not a factor of x7 − y7.P (y) = y7 − y7

= 0


b) x + y is a factor of xn + yn if n is odd.

x − y is a factor of xn − yn if n is odd.

x + y and x − y are factors of xn − yn if n is even.

Neither x + y nor x − y is a factor of xn + yn if n is even.

c) Since x8 − y8 is of the form xn − yn, where n is even, both x + y and x − y are factors.d) Since x11 + y11 is of the form xn + yn, where n is odd, only x + y is a factor.


If there is more than one zero, there must be three zeros. Since they are all integers, the factors must be of the formx − m or m − x, where m is an integer. When these factors are multiplied, a must be ±1.


P (x) = ax3 + bx2 + cx + da)

P (1) = 0

a(1)3 + b(1)2 + c(1) + d = 0

a + b + c + d = 0

If x − 1 is a factor of P (x), a + b + c + d = 0.b) i) Since 3 + 5 − 6 − 2 = 0, x − 1 is a factor of 3x3 + 5x2 − 6x − 2.

ii) Since 2 − 9 − 1 − 8 6= 0, x − 1 is not a factor of 2x3 − 9x2 − x − 8.iii) Since −5 + 4 + 1 = 0, x − 1 is a factor of −5x3 + 4x + 1.


x + y is a factor of P (x) if and only if P (−y) = 0.

P (x) = x2(y2 − 1) − y2(1 + x2) + x2 + y2

P (−y) = (−y)2(y2 − 1) − y2(1 + (−y)2) + (−y)2 + y2

= y2(y2 − 1) − y2(1 + y2) + y2 + y2

= y4 − y2 − y2 − y4 + y2 + y2

= 0


3x2 − 1 −2x2

=1

x2(3x4 − x2 − 2)a)

=1

x2(3x2 + 2)(x2 − 1)

=1

x2(3x2 + 2)(x − 1)(x + 1)

2x2 − xy − 3y2 = (2x − 3y)(x + y)b)


P (x) = x3 − 6x +11x

−6

x3c)

=1

x3(x6 − 6x4 + 11x2 − 6)

=1

x3((x2)3 − 6(x2)2 + 11(x2) − 6)

Since P (1) = 0, x2 − 1 is a factor of P (x). Use synthetic division.

P (x) =1

x3(x2 − 1)((x2)2 − 5(x2) + 6)

=1

x3(x − 1)(x + 1)(x2 − 3)(x2 − 2) 0

1 1 -6 11 -6

1

1

-5

-5

6

6

P (y) = 15 − 23x

y+ 9

x2

y2−

x3

y3d)

=1

y3(15y3 − 23xy2 + 9x2y − x3)

Since P (x) = 0, y − x is a factor of P (y). Use synthetic division.

P (y) =1

y3(y − x)(15y2 − 8xy + x2)

=1

y3(y − x)(3y − x)(5y − x) 0

15 -23 9 -

15

15

-8

-8

x x x x² ³

x ²

x ³x ²x

x

48 MHR Chapter 2

2.5 Roots of Polynomial Equations

Practise


(x + 1)(x − 4)(x + 5) = 0a)x = −1, 4, or −5

(x − 2)(x − 7)(x + 6) = 0b)x = 2, 7, or −6

x(x + 3)(x − 8) = 0c)x = 0, −3, or 8

(x + 6)(x − 3)2 = 0d)x = −6 or 3 (double)


P (x) = 0a)

x3 + 3x2 − x − 3 = 0x2(x + 3) − 1(x + 3) = 0

(x + 3)(x2 − 1) = 0x = −3, −1, or 1

Check.

P (−3) = (−3)3 + 3(−3)2 − (−3) − 3= −27 + 27 + 3 − 3= 0

P (−1) = (−1)3 + 3(−1)2 − (−1) − 3= −1 + 3 + 1 − 3= 0

P (1) = 13 + 3(1)2 − 1 − 3= 1 + 3 − 1 − 3= 0

The solutions are −3, −1, and 1.

P (x) = 0b)

x3 − 3x2 − 4x + 12 = 0x2(x − 3) − 4(x − 3) = 0

(x − 3)(x2 − 4) = 0x = 3,±2

Check.

P (3) = 33 − 3(3)2 − 4(3) + 12= 27 − 27 − 12 + 12= 0

P (−2) = (−2)3 − 3(−2)2 − 4(−2) + 12= −8 − 12 + 8 + 12= 0

P (2) = 23 − 3(2)2 − 4(2) + 12= 8 − 12 − 8 + 12= 0

The solutions are 3 and ±2.

P (t) = t3 + 2t2 − 7t + 4c)P (1) = 13 + 2(1)2 − 7(1) + 4

= 1 + 2 − 7 + 4= 0

t − 1 is a factor of P (t). Synthetic division reveals thefactor t2 + 3t − 4.

P (t) = 0

(t − 1)(t2 + 3t − 4) = 0(t − 1)(t + 4)(t − 1) = 0

t = 1 or −4Check.

P (−4) = (−4)3 + 2(−4)2 − 7(−4) + 4= −64 + 32 + 28 + 4= 0

The solutions are −4 and 1.

P (y) = 0d)

y3 − 3y2 − 16y + 48 = 0y2(y − 3) − 16(y − 3) = 0

(y − 3)(y2 − 16) = 0y = 3,±4

Check.

P (3) = 33 − 3(3)2 − 16(3) + 48= 27 − 27 − 48 + 48= 0

P (−4) = (−4)3 − 3(−4)2 − 16(−4) + 48= −64 − 48 + 64 + 48= 0

P (4) = 43 − 3(4)2 − 16(4) + 48= 64 − 48 − 64 + 48= 0

The solutions are 3 and ±4.

2.5 Roots of Polynomial Equations MHR 49

P (a) = a3 − 4a2 + a + 6e)P (2) = 23 − 4(2)2 + 2 + 6

= 8 − 16 + 8= 0

a − 2 is a factor of P (a). Synthetic division reveals thefactor a2 − 2a − 3.

P (a) = 0

(a − 2)(a2 − 2a − 3) = 0(a − 2)(a − 3)(a + 1) = 0

a = 2, 3, or − 1Check.

P (3) = 33 − 4(3)2 + 3 + 6= 27 − 36 + 9= 0

P (−1) = (−1)3 − 4(−1)2 − 1 + 6= −1 − 4 − 1 + 6= 0

The solutions are 2, 3, and −1.

P (x) = x3 − 4x2 − 3x + 18f)P (−2) = (−2)3 − 4(−2)2 − 3(−2) + 18

= −8 − 16 + 6 + 18= 0

x + 2 is a factor of P (x). Synthetic division reveals thefactor x2 − 6x + 9.

P (x) = 0

(x + 2)(x2 − 6x + 9) = 0(x + 2)(x − 3)3 = 0

x = −2, 3 (double)Check.

P (3) = 33 − 4(3)2 − 3(3) + 18= 27 − 36 − 9 + 18= 0

The solutions are −2 and 3 (double).


P (x) = 0a)

2x3 − 3x2 − 2x = 0x(2x2 − 3x − 2) = 0x(2x + 1)(x − 2) = 0

x = 0, −12, or 2

Check.

P (0) = 2(0)3 − 3(0)2 − 2(0)= 0

P

(

−12

)

= 2(

−12

)3

− 3(

−12

)2

− 2(

−12

)

= −14−

34+ 1

= 0

P (2) = 2(2)3 − 3(2)2 − 2(2)= 16 − 12 − 4= 0

The solutions are 0, −12, and 2.

P (x) = 0b)

3x3 − 10x2 + 3x = 0x(3x2 − 10x + 3) = 0x(3x − 1)(x − 3) = 0

x = 0,13, or 3

Check.

P (0) = 3(0)3 − 10(0)2 + 3(0)= 0

P

(

13

)

= 3(

13

)3

− 10(

13

)2

+ 3(

13

)

=19−

109

+ 1

= 0

P (3) = 3(3)3 − 10(3)2 + 3(3)= 81 − 90 + 9= 0

The solutions are 0,13, and 3.

50 MHR Chapter 2

P (z) = 0c)

9z3 − 4z = 0z(9z2 − 4) = 0

z(3z + 2)(3z − 2) = 0

z = 0, −23, or

23

P (0) = 9(0)3 − 4(0)= 0

Check.

P

(

−23

)

= 9(

−23

)3

− 4(

−23

)

= −7227

+83

= −83+

83

= 0

P

(

23

)

= 9(

23

)3

− 4(

23

)

=7227

−83

= 0

The solutions are 0, −23, and

23

.

P (x) = 0d)

16x3 + 8x2 + x = 0

x(16x2 + 8x + 1) = 0

x(4x + 1)2 = 0

x = 0, −14

(double)

Check.

P (0) = 16(0)3 + 8(0)2 + (0)

= 0

P

(

−14

)

= 16(

−14

)3

+ 8(

−14

)2

+(

−14

)

= −1664

+816

−14

= 0

The solutions are 0 and −14

(double).


P (x) = 0a) i)

x3 − 8x = 0x(x2 − 8) = 0

x = 0

x2 − 8 = 0x2 = 8

x = ±2√

2

The roots are 0, ±2√

2.ii) The roots are 0 and ±2

√2.= ±2.83.

P (x) = x3 − 10x + 3b) i)P (3) = 33 − 10(3) + 3

= 27 − 30 + 3= 0

x − 3 is a factor of P (x). Synthetic division reveals an-other factor of x2 + 3x − 1. Use the quadratic formula todetermine the remaining roots.

x =−3 ±

√

32 − 4(1)(−1)2(1)

=−3 ±

√13

2

The roots are 3,−3 ±

√13

2.

ii) The roots are 3 and approximately 0.30 and −3.30.


P (x) = x3 − 6x2 + 6x + 8c) i)P (4) = 43 − 6(4)2 + 6(4) + 8

= 64 − 96 + 24 + 8= 0

x − 4 is a factor of P (x). Synthetic division revealsanother factor of x2−2x−2. Use the quadratic formulato determine the remaining roots.

x =−(−2) ±

√

(−2)2 − 4(1)(−2)2(1)

=2 ±

√12

2

=2 ± 2

√3

2

= 1 ±√

3

The roots are 4, 1 ±√

3.ii) The roots are 4 and approximately 2.73 and −0.73.

P (x) = x3 + 3x2 − 15x − 25d) i)P (−5) = (−5)3 + 3(−5)2 − 15(−5) − 25

= −125 + 75 + 75 − 25= 0

x + 5 is a factor of P (x). Synthetic division reveals an-other factor of x2 − 2x − 5. Use the quadratic formula todetermine the remaining roots.

x =−(−2) ±

√

(−2)2 − 4(1)(−5)2(1)

=2 ±

√24

2

= 1 ±√

6

The roots are −5, 1 ±√

6.ii) The roots are −5 and approximately 3.45 and −1.45.

P (v) = v3 + 5v2 − 18e) i)P (−3) = (−3)3 + 5(−3)2 − 18

= −27 + 45 − 18= 0

v + 3 is a factor P (v). Synthetic division reveals anotherfactor of v2 + 2v − 6. Use the quadratic formula to de-termine the remaining roots.

v =−2 ±

√

22 − 4(1)(−6)2(1)

=−2 ±

√28

2

=−2 ± 2

√7

2

= −1 ±√

7

The roots are −3, −1 ±√

7.ii) The roots are −3 and approximately 1.65 and −3.65.

P (x) = x(x + 4)(x + 1) − 4f) i)= x3 + 5x2 + 4x − 4

P (−2) = (−2)3 + 5(−2)2 + 4(−2) − 4= −8 + 20 − 8 − 4= 0

x + 2 is a factor of P (x). Synthetic division reveals an-other factor of x2 + 3x− 2. Use the quadratic formula todetermine the remaining roots.

x =−3 ±

√

32 − 4(1)(−2)2(1)

=−3 ±

√17

2

The roots are −2,−3 ±

√17

2.

ii) The roots are −2 and approximately 0.56 and −3.56.


a) The roots are −4.31, −1, and 0.81.

52 MHR Chapter 2

b) The roots are −4, −1.69, and 1.19.

c) The root is −1.

d) The roots are −1.09, 1.57, and 3.51.

e) The roots are ±1.32.

f) The roots are −0.75, 1.16, and 4.59.


g) The roots are −2.65, −1, 0.27, and 1.38.

h) The roots are −3.58, −0.79, and 0.88.



From the diagram, the dimensions of the box are determined to be x by 60 − 2x by60 − 2x

2cm. Let V (x) be the

volume of the box, in cubic centimetres, in terms of the height, x, x ∈ (0, 30). Determine the volume of the box.

V (x) = x(60 − 2x)(

60 − 2x2

)

= x(30 − x)(60 − 2x)= 2x2 − 120x2 + 1800x (1)

Determine x for a volume of 8000 cm3.

V (x) = 8000

2x2 − 120x2 + 1800x = 8000x3 − 60x2 + 900x − 4000 = 0 (2)

GRAPHING CALCULATOR

Polynomials (1) and (2) appear in the adjacent graphic. The Zero operation of the graphing calculator reveals that 10is the only root of x3 − 60x2 + 900x − 4000 in the interval (0, 30). The value of x is 10 cm.

54 MHR Chapter 2


a) Let the point P be the general point on the function f (x) =√

4 − x2. T

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