Engineering Mathematics 4 Episode 2 docx

5.3 Brackets and factorisation When two or more terms in an algebraic expression contain a common factor, then this factor can be shown outside of a bracket.. a C bis now a common factor

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Use the table to determine:

(a) the number of millimetres in 9.5 inches,

(b) a speed of 50 miles per hour in

kilometres per hour,

(c) the number of miles in 300 km,

(d) the number of kilograms in 30 pounds

weight,

(e) the number of pounds and ounces in

42 kilograms (correct to the nearest

ounce),

(f) the number of litres in 15 gallons, and

(g) the number of gallons in 40 litres

2.2 kg D 13.64 kg

(e) 42 kg D 42 ð 2.2 lb D 92.4 lb

0.4 lb D 0.4 ð 16 oz D 6.4 oz D 6 oz, correct

to the nearest ounce

Thus 42 kg D 92 lb 6 oz, correct to the

near-est ounce

(f) 15 gallons D 15 ð 8 pints D 120 pints

120 pints D 120

1.76 litres D 68.18 litres(g) 40 litres D 40 ð 1.76 pints D 70.4 pints

70.4 pints D 70.4

8 gallons D 8.8 gallons

Now try the following exercise

Exercise 15 Further problems conversion

tables and charts

1 Currency exchange rates listed in a

news-paper included the following:

Calculate (a) how many Italian euros

£32.50 will buy, (b) the number ofCanadian dollars that can be purchasedfor £74.80, (c) the pounds sterling whichcan be exchanged for 14 040 yen, (d) thepounds sterling which can be exchangedfor 1754.30 Swedish kronor, and (e) theAustralian dollars which can be boughtfor £55

[(a) 48.10 euros (b) $179.52(c) £75.89 (d) £132.40(e) 148.50 dollars]

2 Below is a list of some metric to imperialconversions

Length 2.54 cm D 1 inch

1.61 km D 1 mileWeight

Capacity 1 litre D 1.76 pints

Use the list to determine (a) the number

of millimetres in 15 inches, (b) a speed of

35 mph in km/h, (c) the number of metres in 235 miles, (d) the number ofpounds and ounces in 24 kg (correct tothe nearest ounce), (e) the number of kilo-grams in 15 lb, (f) the number of litres in

kilo-12 gallons and (g) the number of gallons

in 25 litres



(a) 381 mm(c) 378.35 km (d) 52 lb 13 oz(b) 56.35 km/h(e) 6.82 kg (f) 54.55 l(g) 5.5 gallons





3 Deduce the following information fromthe BR train timetable shown in Table 4.3:(a) At what time should a man catch atrain at Mossley Hill to enable him

to be in Manchester Piccadilly by8.15 a.m.?

(b) A girl leaves Hunts Cross at8.17 a.m and travels to ManchesterOxford Road How long does thejourney take What is the averagespeed of the journey?

(c) A man living at Edge Hill has to be

at work at Trafford Park by 8.45 a.m

It takes him 10 minutes to walk to

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Table 4.3 Liverpool, Hunt’s Cross and Warrington ! Manchester

Reproduced with permission of British Rail

his work from Trafford Park

sta-tion What time train should he catch

from Edge Hill?

The statementv D u C at is said to be a formula

forvin terms of u, a and t

v, u, a and t are called symbols.

The single term on the left-hand side of theequation,v, is called the subject of the formulae.

Provided values are given for all the symbols in

a formula except one, the remaining symbol can

be made the subject of the formula and may beevaluated by using a calculator

Problem 16 In an electrical circuit thevoltage V is given by Ohm’s law, i.e

V D IR Find, correct to 4 significant figures,the voltage when I D 5.36 A and

R D14.76

V D IR D

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Hence, voltage V = 79.11 V, correct to 4

signifi-cant figures.

Problem 17 The surface area A of a hollow

cone is given by A D rl Determine, correct

to 1 decimal place, the surface area when

Problem 19 The power, P watts, dissipated

in an electrical circuit may be expressed by

Hence power, P = 8.46 W, correct to 3

signifi-cant figures.

Problem 20 The volume V cm3 of a right

circular cone is given by V D 1

3r

2h Giventhat r D 4.321 cm and h D 18.35 cm, find

the volume, correct to 4 significant figures

are masses, d their distance apart and G is a

constant Find the value of the force giventhat G D 6.67 ð 1011, m1D7.36, m2 D15.5and d D 22.6 Express the answer in standardform, correct to 3 significant figures

Problem 22 The time of swing t seconds,

of a simple pendulum is given by

˛ D0.0043 and t D 80

R D R0

DD

Hence, resistance, R= 19.6 Z , correct to 3 nificant figures.

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sig-Now try the following exercise

Exercise 16 Further problems on

evalua-tion of formulae

1 A formula used in connection with gases

P D1500, V D 5 and T D 200

[R D 37.5]

2 The velocity of a body is given by

vDu C at The initial velocity u is

mea-sured when time t is 15 seconds and

found to be 12 m/s If the

accelera-tion a is 9.81 m/s2 calculate the final

3 Find the distance s, given that s D 12gt2,

time t D 0.032 seconds and acceleration

due to gravity g D 9.81 m/s2

[0.00502 m or 5.02 mm]

4 The energy stored in a capacitor is given

by E D 12CV2 joules Determine the

energy when capacitance C D 5 ð

106 farads and voltage V D 240V

7 Velocity D frequency ð wavelength

Find the velocity when the frequency is

1825 Hz and the wavelength is 0.154 m

1R2 C

1R3 when R1D5.5 ,

R2 D7.42 and R3D12.6

[2.526 ]

9 Power D force ð distance

time Find thepower when a force of 3760 N raises an

14 Energy, E joules, is given by the formula

E D 12LI2 Evaluate the energy when

L D5.5 and I D 1.2 [E D 3.96 J]

15 The current I amperes in an a.c circuit

is given by I D p V

R2CX2 Evaluate thecurrent when V D 250, R D 11.0 and

ab

c

de[vD7.327]

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Assignment 1

This assignment covers the material

con-tained in Chapters 1 to 4 The marks for

each question are shown in brackets at

the end of each question.

1 Simplify (a) 22

3ł3

13(b) 1

4

7ð2

14 ł

1

3C

15

C2 724(9)

2 A piece of steel, 1.69 m long, is cut

into three pieces in the ratio 2 to 5 to

6 Determine, in centimetres, the lengths

of the three pieces (4)

3 Evaluate 576.29

19.3

(a) correct to 4 significant figures

(b) correct to 1 decimal place (2)

4 Determine, correct to 1 decimal places,

11 Convert the following decimal numbers

to binary, via octal:

(a) 479 (b) 185.2890625 (6)

12 Convert (a) 5F16 into its decimal alent (b) 13210 into its hexadecimalequivalent (c) 1101010112 into its hex-

14 Evaluate the following, each correct to 2decimal places:

(a)

36.22ð0.56127.8 ð 12.83

3

2p

15 If 1.6 km D 1 mile, determine the speed

of 45 miles/hour in kilometres per hour

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Algebra

5.1 Basic operations

Algebra is that part of mathematics in which the

relations and properties of numbers are investigated

by means of general symbols For example, the area

of a rectangle is found by multiplying the length

by the breadth; this is expressed algebraically as

A D l ð b, where A represents the area, l the length

and b the breadth

The basic laws introduced in arithmetic are

Problem 1 Evaluate: 3ab 2bc C abc

Replacing p, q and r with their numerical valuesgives:

4p2qr3D422

12

32

of the positive terms gives:

5x 8x D− 3x

Alternatively3x C 2x C x C 7x D 3x C 2x x 7x

D− 3x

Problem 4 Find the sum of 4a, 3b, c, 2a,

5b and 6cEach symbol must be dealt with individually.For the ‘a’ terms: C4a 2a D 2a

For the ‘b’ terms: C3b 5b D 2bFor the ‘c’ terms: Cc C6c D 7cThus

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The algebraic expressions may be tabulated as

shown below, forming columns for the a’s, b’s, c’s

Problem 6 Subtract 2x C 3y 4z from

x 2y C 5z

x 2y C 5z2x C 3y 4zSubtracting gives: − x − 5y Y9z

(Note that C5z 4z D C5z C 4z D 9z)

An alternative method of subtracting algebraic

expressions is to ‘change the signs of the bottom

line and add’ Hence:

x 2y C 5z

2x 3y C 4zAdding gives: − x − 5y Y9z

Problem 7 Multiply 2a C 3b by a C b

Each term in the first expression is multiplied by a,

then each term in the first expression is multiplied

by b, and the two results are added The usual layout

Adding gives: 2a2 Y5ab Y3b2

Problem 8 Multiply 3x 2y2C4xy by

2x 5y

3x 2y 2 C 4xy 2x 5y Multiplying

by 2x ! 6x 2

4xy 2 C 8x 2 y Multiplying

by 5y ! 20xy2 15xy C 10y3Adding gives: 6x2− 24xy2

Exercise 17 Further problems on basic

[6a213ab C 3ac 5b2Cbc]

10 Simplify (i) 3a ł 9ab (ii) 4a2b ł2a

i 13b ii 2ab

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am(v) an D 1

8 and c D 2Using the second law of indices,

Using the second law of indices gives:

p4

D2332 D 108

Problem 14 Simplify: x

2y3Cxy2xy

Algebraic expressions of the forma C b

c can be splitinto a

x2y

xy2xy D

x2yxy

xy2

xy

xyxy

y −1

Problem 16 Simplify: p31/2q24Using the third law of indices gives:

p3ð1/2q2ð4Dp 3=2/ q8

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Problem 17 Simplify: mn

23

m1/2n1/44

The brackets indicate that each letter in the bracket

must be raised to the power outside Using the third

law of indices gives:

mn23

m1/2n1/44 D m1ð3n2ð3

m1/2ð4n1/4ð4 D m3n6

m2n1Using the second law of indices gives:

It is usual to express the answer in the same form

as the question Hence

d23e22f1/25Dd1e0f9/2

Dd1f9/2 since e0D1from the sixth law of indices

x2y1/2px 3

y2

x5y31/2 D x2y1/2x1/2y2/3

x5/2y3/2Using the first and second laws of indices gives:

from the fifth and sixth laws of indices

Exercise 18 Further problems on laws of

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5.3 Brackets and factorisation

When two or more terms in an algebraic expression

contain a common factor, then this factor can be

shown outside of a bracket For example

ab C ac D ab C c

which is simply the reverse of law (v) of algebra on

page 34, and

6px C 2py 4pz D 2p3x C y 2z

This process is called factorisation.

Problem 21 Remove the brackets and

simplify the expression:

3aY3b − 2c − 4d

Problem 22 Simplify:

a22a ab a3b C a

When the brackets are removed, both 2a and ab

in the first bracket must be multiplied by 1 andboth 3b and a in the second bracket by a Thus

a22a ab a3b C a

Da22a C ab 3ab a2Collecting similar terms together gives: 2a 2abSince 2a is a common factor, the answer can beexpressed as: − 2a 1 Yb /

Problem 23 Simplify: a C ba b

Each term in the second bracket has to be multiplied

by each term in the first bracket Thus:

a C ba b D aa b C ba b

Da2ab C ab b2

Da2− b2

a bMultiplying by a ! a2 CabMultiplying by b ! ab b2

Problem 24 Simplify: 3x 3y2

2x 3y2 D2x 3y2x 3y

D2x2x 3y 3y2x 3y

D4x26xy 6xy C 9y2

D4x2− 12xyY9y2

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Alternatively, 2x 3y

2x 3yMultiplying by 2x ! 4x2 6xy

Multiplying by 3y ! 6xy C 9y2

Adding gives: 4x2 12xy C 9y2

Problem 25 Remove the brackets from the

expression: 2[p23q C r C q2]

In this problem there are two brackets and the ‘inner’

one is removed first

Hence, 2[p23q C r C q2]

D2[p23q 3r C q2]

D2p2− 6q − 6rY2q2

Problem 26 Remove the brackets and

simplify the expression:

2a [3f24a b 5a C 2bg C 4a]

Removing the innermost brackets gives:

i.e − 11aY36b or 36b − 11a

(see law (iii), page 34)Problem 27 Simplify:

since 6x is common to both terms

Problem 28 Factorise: (a) xy 3xz(b) 4a2C16ab3 (c) 3a2b 6ab2C15ab

For each part of this problem, the HCF of the termswill become one of the factors Thus:

of x y Thus:

ax y C bx y D x − y / aYb /

Problem 30 Factorise:

2ax 3ay C 2bx 3by

ais a common factor of the first two terms and b acommon factor of the last two terms Thus:2ax 3ay C 2bx 3by

Da2x 3y C b2x 3y

2x 3y is now a common factor thus:

a2x 3y C b2x 3y

D 2x − 3y / aYb /

Alternatively, 2x is a common factor of the originalfirst and third terms and 3y is a common factor ofthe second and fourth terms Thus:

2ax 3ay C 2bx 3by

D2xa C b 3ya C b

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a C bis now a common factor thus:

brac-kets and factorisation

In Problems 1 to 9, remove the brackets and

simplify where possible:

(ii) 2xyy C 3x C 4x2

12 (i) ay C by C a C b (ii) px C qx C py C qy(iii) 2ax C 3ay 4bx 6by

Problem 33 Simplify: a C 5a ð 2a 3a

The order of precedence is brackets, multiplication,then subtraction Hence

a C5a ð 2a 3a D 6a ð 2a 3a

D12a2− 3a

or 3a 4a −1/

Problem 34 Simplify: a C 5a ð 2a 3a

The order of precedence is brackets, multiplication,then subtraction Hence

a C5a ð 2a 3a D a C 5a ð a

Da C 5a2

Da − 5a2 or a 1− 5a /

Problem 35 Simplify: a ł 5a C 2a 3a

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The order of precedence is division, then addition

and subtraction Hence

Problem 36 Simplify: a ł 5a C 2a 3a

The order of precedence is brackets, division and

subtraction Hence

a ł 5a C 2a 3a D a ł 7a 3a

D a7a 3a D

1

7− 3a

Problem 37 Simplify:

3c C 2c ð 4c C c ł 5c 8c

The order of precedence is division, multiplication,

addition and subtraction Hence:

The order of precedence is brackets, division,

mul-tiplication and addition Hence,

3c C 2c ð 4c C c ł 5c 8c

D3c C 2c ð 4c C c ł 3c

D3c C 2c ð 4c C c

3cNow c

D 2a 34a C5 ð 6 3a

D 2a 34a C30 3a

D 2a4a

34aC30 3a

D 1

2

34aC30 3a

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and changing ‘of’ to ‘ð’ gives:

1

3ð3p C 4p ð 2p

i.e pY8p2 or p 1 Y8p /

funda-mental laws and precedence

Simplify the following:

1 2x ł 4x C 6x

1

2C6x

2 2x ł 4x C 6x

15

6 2y C 4 ł 6y C 34 5y

23y C12 13y

7 3 ł y C 2 ł y C 1

5

1

2C3y

5.5 Direct and inverse proportionality

An expression such as y D 3x contains two

vari-ables For every value of x there is a corresponding

value of y The variable x is called the independent

variable and y is called the dependent variable.

When an increase or decrease in an independent

variable leads to an increase or decrease of the same

proportion in the dependent variable this is termed

direct proportion If y D 3x then y is directly

proportional to x, which may be written as y ˛ x

or y D kx, where k is called the coefficient of

proportionality (in this case, k being equal to 3).

When an increase in an independent variableleads to a decrease of the same proportion in thedependent variable (or vice versa) this is termed

inverse proportion If y is inversely proportional

to x then y ˛ 1

x or y D k/x Alternatively, k D xy,that is, for inverse proportionality the product of thevariables is constant

Examples of laws involving direct and inverseproportional in science include:

(i) Hooke’s law, which states that within the

elastic limit of a material, the strain ε duced is directly proportional to the stress, ,producing it, i.e ε ˛ or ε D k

pro-(ii) Charles’s law, which states that for a given

mass of gas at constant pressure the volume V

is directly proportional to its thermodynamictemperature T, i.e V ˛ T or V D kT.(iii) Ohm’s law, which states that the current I

flowing through a fixed resistor is directlyproportional to the applied voltage V, i.e

I ˛ V or I D kV

(iv) Boyle’s law, which states that for a gas

at constant temperature, the volume V of afixed mass of gas is inversely proportional

to its absolute pressure p, i.e p ˛ 1/V or

p D k/V, i.e pV D k

Problem 42 If y is directly proportional to

xand y D 2.48 when x D 0.4, determine(a) the coefficient of proportionality and(b) the value of y when x D 0.65

(a) y ˛ x, i.e y D kx If y D 2.48 when x D 0.4,2.48 D k0.4

Hence the coefficient of proportionality,

k D 2.480.4 D6.2(b) y D kx, hence, when x D 0.65,

y D 6.20.65 D 4.03

Problem 43 Hooke’s law states that stress

is directly proportional to strain ε withinthe elastic limit of a material When, for mildsteel, the stress is 25 ð 106 Pascals, the strain

is 0.000125 Determine (a) the coefficient of

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proportionality and (b) the value of strain

when the stress is 18 ð 106 Pascals

(a) ˛ ε, i.e D kε, from which k D /ε Hence

the coefficient of proportionality,

k D 25 ð 10

60.000125 D200×10 9 pascals

(The coefficient of proportionality k in this case

is called Young’s Modulus of Elasticity)

Problem 44 The electrical resistance R of a

piece of wire is inversely proportional to the

cross-sectional area A When A D 5 mm2,

R D7.02 ohms Determine (a) the coefficient

of proportionality and (b) the cross-sectional

area when the resistance is 4 ohms

(a) R ˛ 1

A, i.e R D k/A or k D RA Hence,

when R D 7.2 and A D 5, the coefficient of

Problem 45 Boyle’s law states that at

constant temperature, the volume V of a

fixed mass of gas is inversely proportional to

its absolute pressure p If a gas occupies a

volume of 0.08 m3 at a pressure of

1.5 ð 106 Pascals determine (a) the

coefficient of proportionality and (b) the

volume if the pressure is changed to

Exercise 21 Further problems on direct

and inverse proportionality

1 If p is directly proportional to q and

p D37.5 when q D 2.5, determine (a) theconstant of proportionality and (b) thevalue of p when q is 5.2

[(a) 15 (b) 78]

2 Charles’s law states that for a given mass

of gas at constant pressure the volume

is directly proportional to its namic temperature A gas occupies a vol-ume of 2.25 litres at 300 K Determine(a) the constant of proportionality, (b) thevolume at 420 K, and (c) the temperaturewhen the volume is 2.625 litres

thermody-[(a) 0.0075 (b) 3.15 litres (c) 350 K]

3 Ohm’s law states that the current flowing

in a fixed resistor is directly proportional

to the applied voltage When 30 volts

is applied across a resistor the currentflowing through the resistor is 2.4 ð 103amperes Determine (a) the constant ofproportionality, (b) the current when thevoltage is 52 volts and (c) the voltagewhen the current is 3.6 ð 103 amperes

(a) 0.00008 (b) 4.16 ð 103 A(c) 45 V

4 If y is inversely proportional to x and

y D15.3 when x D 0.6, determine (a) thecoefficient of proportionality, (b) thevalue of y when x is 1.5, and (c) the value

of x when y is 27.2

[(a) 9.18 (b) 6.12 (c) 0.3375]

5 Boyle’s law states that for a gas at stant temperature, the volume of a fixedmass of gas is inversely proportional toits absolute pressure If a gas occupies avolume of 1.5 m3 at a pressure of 200 ð

con-103 Pascals, determine (a) the constant ofproportionality, (b) the volume when thepressure is 800 ð 103 Pascals and (c) thepressure when the volume is 1.25 m3

(a) 300 ð 103 (b) 0.375 m2(c) 240 ð 103 Pa

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Further algebra

6.1 Polynomial division

Before looking at long division in algebra let us

revise long division with numbers (we may have

forgotten, since calculators do the job for us!)

(1) 16 divided into 2 won’t go

(2) 16 divided into 20 goes 1

(3) Put 1 above the zero

(4) Multiply 16 by 1 giving 16

(5) Subtract 16 from 20 giving 4

(6) Bring down the 8

(7) 16 divided into 48 goes 3 times

(8) Put the 3 above the 8

Below are some examples of division in algebra,which in some respects, is similar to long divisionwith numbers

(Note that a polynomial is an expression of the form

fx D a C bx C cx2Cdx3C Ð Ð Ð

and polynomial division is sometimes required

when resolving into partial fractions — seeChapter 7)

Problem 1 Divide 2x2Cx 3 by x 1

2x2Cx 3 is called the dividend and x 1 the

divisor The usual layout is shown below with the

dividend and divisor both arranged in descendingpowers of the symbols

2x C 3

x 1

2x2 C x 32x2 2x3x 33x 3 .Dividing the first term of the dividend by the firstterm of the divisor, i.e 2x

2

x gives 2x, which is putabove the first term of the dividend as shown Thedivisor is then multiplied by 2x, i.e 2xx 1 D2x2 2x, which is placed under the dividend asshown Subtracting gives 3x 3 The process isthen repeated, i.e the first term of the divisor,

x, is divided into 3x, giving C3, which is placedabove the dividend as shown Then 3x 1 D3x 3 which is placed under the 3x 3 Theremainder, on subtraction, is zero, which completesthe process

Thus 2x2 Yx −3/ ÷ x −1/= 2xY 3/

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[A check can be made on this answer by multiplying

(1) xinto 3x3 goes 3x2 Put 3x2 above 3x3

(6) Subtract(7) xinto xy2 goes y2 Put y2 above dividend(8) y2x C y D xy2Cy3

(9) SubtractThus x

2a2 2ab b22a b

4a3 6a2b C5b34a3 2a2b

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x 1

7 Divide 3x3C2x25x C 4 by x C 2

3x24x C 3 2

6.2 The factor theorem

There is a simple relationship between the factors of

a quadratic expression and the roots of the equation

obtained by equating the expression to zero

For example, consider the quadratic equation

x2C2x 8 D 0

To solve this we may factorise the quadratic

expression x2C2x 8 giving x 2x C 4

Hence x 2x C 4 D 0

Then, if the product of two numbers is zero, one

or both of those numbers must equal zero Therefore,either x 2 D 0, from which, x D 2

x2C2x 8 D 0 we could deduce at once that (x 2)

is a factor of the expression x2C2x8 We wouldn’tnormally solve quadratic equations this way — butsuppose we have to factorise a cubic expression (i.e.one in which the highest power of the variable is3) A cubic equation might have three simple linearfactors and the difficulty of discovering all thesefactors by trial and error would be considerable It is

to deal with this kind of case that we use the factor

theorem This is just a generalised version of what

we established above for the quadratic expression.The factor theorem provides a method of factorisingany polynomial, fx, which has simple factors

A statement of the factor theorem says:

‘if x =a is a root of the equation f x / =0, then x − a / is a factor of f x /’

The following worked problems show the use of thefactor theorem

Problem 6 Factorise x37x 6 and use it

to solve the cubic equation: x37x 6 D 0

We have a choice now We can divide x37x 6

by x 3 or we could continue our ‘trial and error’

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by substituting further values for x in the given

expression — and hope to arrive at fx D 0

Let us do both ways Firstly, dividing out gives:

x2C3x C 2 factorises ‘on sight’ as x C 1x C 2

Therefore

x3− 7x −6= x −3/ xY 1/ xY 2/

A second method is to continue to substitute values

of x into fx

Our expression for f3 was 3373 6 We

can see that if we continue with positive values of

xthe first term will predominate such that fx will

x C2 is a factor (also as shown above)

To solve x3 7x 6 D 0, we substitute the

factors, i.e

x 3x C 1x C 2 D 0

from which, x =3, x =− 1 and x =−2

Note that the values of x, i.e 3, 1 and 2, are

all factors of the constant term, i.e the 6 This can

give us a clue as to what values of x we should

consider

Problem 7 Solve the cubic equation

x32x25x C 6 D 0 by using the factor

theorem

Let fx D x32x25x C 6 and let us substitute

simple values of x like 1, 2, 3, 1, 2, and so on

f1 D 1321251 C 6 D 0,

hence x 1 is a factorf2 D 2322252 C 6 6D 0

f3 D 3323253 C 6 D 0,

hence x 3 is a factorf1 D 1321251 C 6 6D 0f2 D 2322252 C 6 D 0,

hence x C 2 is a factorHence, x32x25x C 6 D x 1x 3x C 2Therefore if x32x25x C 6 D 0

then x 1x 3x C 2 D 0

from which, x =1, x =3 and x =−2

Alternatively, having obtained one factor, i.e

x 1 we could divide this into x32x25x C 6

Summarising, the factor theorem provides us with

a method of factorising simple expressions, and analternative, in certain circumstances, to polynomialdivision

Exercise 23 Further problems on the

fac-tor theorem

Use the factor theorem to factorise the sions given in problems 1 to 4

expres-1 x2C2x 3 [x 1x C 3]

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5 Use the factor theorem to factorise

x3C4x2Cx 6 and hence solve the cubic

6.3 The remainder theorem

Dividing a general quadratic expression

(ax2Cbx C c) by (x p), where p is any whole

number, by long division (see Section 6.1) gives:

The remainder, c C b C app D c C bp C ap2 or

ap2Cbp C c This is, in fact, what the remainder

theorem states, i.e.

‘if ax2 YbxYc /is divided by x − p /,

the remainder will be ap2 YbpYc’

If, in the dividend (ax2Cbx C c), we substitute p

for x we get the remainder ap2Cbp C c

For example, when 3x24x C 5 is divided by

x 2 the remainder is ap2Cbp C c, (where a D 3,

Similarly, when 4x27x C 9 is divided by x C 3,the remainder is ap2 Cbp C c, (where a D 4,

b D 7, c D 9 and p D 3) i.e the remainderis: 432C73 C 9 D 36 C 21 C 9 D 66Also, when x2C3x 2 is divided by x 1,the remainder is 112C31 2 D 2

It is not particularly useful, on its own, to knowthe remainder of an algebraic division However,

if the remainder should be zero then (x p) is afactor This is very useful therefore when factorisingexpressions

For example, when 2x2Cx 3 is divided by

x 1, the remainder is 212C11 3 D 0, whichmeans that x 1 is a factor of 2x2Cx 3

In this case the other factor is 2x C 3, i.e

2x2Cx 3 D x 12x 3

The remainder theorem may also be stated for a

cubic equation as:

sub-For example, when 3x3C2x2x C4 is divided

by x 1, the remainder is: ap3Cbp2 Ccp C d(where a D 3, b D 2, c D 1, d D 4 and p D 1),i.e the remainder is: 313C212C11 C 4 D

3 C 2 1 C 4 D 8.

Similarly, when x37x6 is divided by x3,the remainder is: 133C032736 D 0, whichmeans that x 3 is a factor of x37x 6.Here are some more examples on the remaindertheorem