Further algebramathsbooks.net/JACPlus Books/11 Advance General/Ch08...Chapter 8 Further algebra 273...

8A Polynomial identities 8B Partial fractions 8C Simultaneous equations8

272

areas oF sTudy

The solution of simultaneous equations arising from the intersection of a line with a parabola, circle or • rectangular hyperbola using algebra

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Further algebra

polynomial identitiesBefore discussing the defi nition of a polynomial identity, it is important to remember some basic defi nitions.

• An algebraic expression is made up of terms. • In the term axn, a is referred to as the coeffi cient of xn. • A constant is a term with no variable beside it.

For example 2x3 + 3 is an algebraic expression made up of two terms. The coeffi cient of x3 is 2. The constant is 3.

A polynomial identity is an identity of the form:

knxn + kn - 1xn - 1 + kn - 2xn - 2 + . . . . . + k1x + k0, n ∈ N

where kn, kn - 1. . . are constants and n is an element of the set of natural numbers N.

The degree of a polynomial is given by the highest value of n. Hence a polynomial of degree 1 is linear, of degree 2 is a quadratic, of degree 3 is a cubic, of degree 4 is a quartic and so on.

Worked example 1

Which of the following are polynomials? Give reasons for your answers.

a x3 + 2x2 + 1 b x + 1x c (2x + 6)5

Think WriTe

a In order for x3 + 2x2 + 1 to be a polynomial, the powers must all be greater than or equal to 0, which they are. The highest power of x is 3.

a x3 + 2x2 + 1 is a polynomial of degree 3 since it has descending powers of x and these powers are all greater than or equal to zero, i.e. n ∈ N.

b In order for x + 1x

to be a polynomial, the

powers must all be greater than or equal to 0, which they are not.

b This is not a polynomial since the second term has a power of -1.

8a

maths Quest 11 advanced General mathematics for the Casio Classpad

273Chapter 8 Further algebra

c In order for (2x + 6)5 to be a polynomial, the powers must all be greater than or equal to 0, which they are. The highest power of x is 5.

c This is a polynomial of degree 5, since when expanded, it has n ∈ N.

Two polynomials are said to be equal if each x-value generates the same y-value. Polynomials are identical if they are of the same degree and corresponding coefficients are equal. Therefore, if:

ax3 + bx2 + cx + d = 2x3 - 4x + 8then a = 2, b = 0, c = -4 and d = 8.

If two polynomials are known to be equal, then the process of equating coefficients can be used to solve problems.

Worked example 2

If 5x3 + 2x2- 7x + 1 = (2 a + b)x3 - ax2 -(b - c) x + 1, then find the values of a, b and c.

Think WriTe

Method 1: Technology-free1 If 5x3 + 2x2- 7x + 1 =

(2a + b)x3 - ax2 - (b - c)x + 1, then the each corresponding term must be equal. Equate the terms.

5x3 = (2a + b)x3

⇒ 5 = 2a + b [1]

2x2 = -ax2

⇒ 2 = -a⇒ -2 = a [2]

-7x = -(b - c)x⇒-7 = -(b - c)⇒7 = b - c [3]

2 Solve these equations using substitution. Substituting a = -2 into equation [1] gives 2(-2) + b = 5b = 9Substituting b = 9 into equation [3] gives9 - c = 7c = 2

3 Write the answer. a = -2, b = 9 and c = 2

Method 2: Technology-enabled

1

2 Write the answer. a = -2, b = 9 and c = 2

On the Main screen, using the soft keyboard, tap:• )• {NEnter the equations as shown.Then press E.

274

Worked example 3

Determine values of a and b if m4 + 4 = (m2 + am + 2)(m2 + bm + 2).

Think WriTe

1 The right-hand side must fi rst be expanded. m4 + 4 = m4 + bm3 + 2m2 + am2 + abm2 + 2am +2m2 + 2bm + 4

= m4 + (b + a)m3 + (4 + ab)m2 + (2a + 2b)m + 42 Equate the coeffi cients.

The coeffi cients of m3, m2, and m are zero.

0m3 = (b + a)m3

⇒ 0 = b + a [1]

0m2 = (4 + ab)m2

⇒ 0 = 4 + ab [2]

0m = (2a +2b)m⇒ 0 = 2a + 2b [3]

3 Solve for a and b. From equation [1],b = -aSubstitute b = -a into equation [2]0 = 4 - a2

a2 = 4a = ±2a = 2 or a = -2andb = -2 b = 2

4 Write the answer. When a = 2, b = -2 and when a = -2, b = 2.

Worked example 4

If x - 4 is a factor of x3 - 6x2 + 2x + 24, fi nd the other factor.

Think WriTe

Method 1: Technology-free

1 Since the expression x3 - 6x2 + 2x + 24 is cubic, the other factor must be a quadratic, hence it is of the form ax2 + bx + c.

x3 - 6x2 + 2x + 24 = (x - 4)(ax2 + bx + c)RHS = ax3 + bx2 + cx - 4ax2 - 4bx - 4c = ax3 + (b - 4a)x2 + (c - 4b)x - 4c

2 Equate the coeffi cients. x3 = ax3

⇒ 1 = a [1]

-6x2 = (b - 4a)x2

⇒ -6 = b - 4a [2]

2x = (c - 4b)x⇒ 2 = c - 4b [3]

3 Solve for a, b and c. Substitute a = 1 into equation [2]-6 = b - 4ab = -2Substitute b = -2 into equation [3]2 = c - 4bc = -6

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Worked example 4



4 Substitute the values for a, b and c into ax2 + bx + c and write the answer.

When a = 1, b = -2 and c = -6 then the quadratic factor of x3 - 6x2 + 2x + 24 is x2 - 2x - 6.


1

2 Write the answer. The quadratic factor of x3 - 6x2 + 2x + 24 is x2 - 2x - 6.

A 1. polynomial identity is an identity of the formknxn + kn - 1 xn - 1 + kn - 2xn - 2 + . . . . . + k1x + k0, n ∈ N

where kn, kn - 1 . . . are constants and n is an element of the set of natural numbers N.The degree of a polynomial is given by the highest value of 2. n.Polynomials are identical if they are of the same degree and corresponding coefficients 3. are equal.If two polynomials are known to be equal, then the process of equating coefficients can 4. be used to solve problems.

rememBer

polynomial identities 1 We 1 For each of the following expressions:

i state whether or not it is a polynomialii if yes to i then give its degree.

a 2x4 + 1 b 23x + x2 + 3 c (3x2 + 2)3 d x x

x

3 2+

2 We 2 Find the values of a, b and c if(2a + b)x3 + (b - c)x2 + (a + 2c)x + 5 = 3x3 - 5x2 + 10x + 5.

3 Find the values of a, b and c ifx3 + 2x2 - 13x + 16 = (x - 2)(ax2 + bx + c) + 6.

4 Find constants a, b given that2x3- 5x2 + 10 = (x - 2)2(2x + a) + bx + c.

5 We 3 Determine the values of a and b ifm4 + 25 = (m2 + am + 5)(m2 + bm + 5).

6 If x2 = a(x + 1)2 + b(x + 1) + c, find the values of a, b and c.

7 If ax3 + bx2 + cx + d = (2x - 1)2 (mx + n), express b in terms of c and d.

exerCise

8a

On the Main screen, tap:• Action• Transformation• factorComplete the entry line as:factor(x3 - 6x2 + 2x + 24)Then press E.

276

8 We 4 If x - 2 is a factor of x3 + 3x2 - 16x + 12, find the other factor.

9 If x + 1 is a factor of x3 - x2 + x + 3, find the other factor.

10 If 2x + 1 is a factor of 2x3 + 7x2 - 7x - 5, find the other factor.

partial fractionsWhen a function is expressed as one polynomial divided by another,

f (x) = g xh x

( )( )

, it is often desirable to express this using partial fractions.

This enables the function to be graphed more easily and also helps with the process of integration (which you will learn about in Mathematical Methods CAS).

proper fractionsIf g(x)and h(x) are both linear functions, then the function can be expressed as a proper fraction in the form:

f (x) = Ab

h x+

( ).

Worked example 5

Express 4 5

3x

x++

-- in the form A

bx

++-- 3

.

Think WriTe

Method 1: Technology-free1 Express the numerator as 4(x - 3) + b; the

value of b must be 17.

4 53

4 3 173

xx

xx

+-

= - +-

( )

2 Write the answer in the form

Ab

x+

- 3.

= +-

417

3x


1


Ab

x+

- 3.

4 53

417

3x

x x+

-= +

-

8B eBookpluseBookplus

Interactivityint-0975

Partial fractions


On the Main screen, tap:• Action• Transformation• propFracComplete the entry line as:

propFrac 4 53

xx

+-

Then press E.


Consider the case where g(x) is a polynomial of degree 1 and h(x) is a polynomial of degree 2.

In this case the function, f xg xh x

( )( )( )

= , is a proper fraction, since the numerator has a smaller

power than the denominator. For every linear factor (ax + b) in the denominator, there will be a partial fraction of the form

f xA

ax b( ) =

+.

For every repeated linear factor of the form (ax + b)2 in the denominator, then the partial

fractions will be of the form f xA

ax bB

ax b( )

( ) ( )=

++

+ 2. On occasions when it is impossible

to express the partial fractions in the form f xA

ax bB

ax b( )

( ) ( )=

++

+ 2, they can be written as

f xA

ax bB

ax bC

ax b( )

( ) ( ) ( )=

++

++

+ 2.

Worked example 6

Express x

x x

++-- --

33 402

in partial fraction form.

Think WriTe


1 Factorise the denominator x2 - 3x - 40.x

x x

xx x

+- -

= +- +

3

3 40

38 52 ( )( )

, x ∈ R\{-5, 8}

2 The denominator has two linear factors so there will be two partial fractions of the

form A

xB

x( ) ( ).

-+

+8 5

xx x

Ax

Bx

+- +

=-

++

38 5 8 5( )( )

3 Express the sum of the two fractions on the right as a single fraction.

xx x

A x B xx x

+- +

= + + -- +

38 5

5 88 5( )( )

( ) ( )( )( )

4 Equate the numerators and simplify. x + 3 = A(x + 5) + B(x - 8)x + 3 = Ax + 5A + Bx - 8Bx + 3 = (A + B)x + 5A - 8B

5 Equate the coefficients to solve for A and B. x = (A + B)x⇒ 1 = A + B⇒ 1 - B = A [1]

3 = 5A - 8B [2]

Substitute equation [1] into equation [2].

⇒ 3 = 5(1 - B) - 8B⇒ 13B = 2

⇒ B = 213

⇒ A = 1113

6 Substitute the values for A and B and write the answer in the form

Ax

Bx( ) ( )-

++8 5

.

x

x x x x+

- -=

-+

+3

3 40

1113 8

213 52 ( ) ( )

,

x ∈R\{-5, 8}

278


1


Ax

Bx( ) ( )

.-

++8 5

x

x x x x+

- -=

-+

+3

3 40

1113 8

213 52 ( ) ( )

,

x ∈R \ {-5, 8}

Worked example 7

Express 2 12 1 2

x

x x

---- ++( )( )

in partial fractions.

Think WriTe


1 The denominator has one linear factor and one repeated linear factor so there will be three partial fractions of the form

Ax

Bx

C

x( ) ( ) ( ).

-+

++

+2 1 1 2

2 1

2 1 2 1 12 2

x

x x

Ax

Bx

C

x

-- +

=-

++

++( )( ) ( ) ( ) ( )

,

x ∈R\{-1, 2}.

2 Express the sum of the three fractions on the right as a single fraction.

2 1

2 1 2

x

x x

-- +

=( )( )

A x B x x C x

x x

( ) ( )( ) ( )

( )( )

+ + - + + -- +

1 2 1 2

2 1

2

2

3 Equate the numerators and simplify. 2x -1 = A(x2 + 2x + 1) + B (x2 - x - 2) + C (x - 2)2x -1 = Ax2 + 2Ax + A + Bx2 - Bx - 2B + Cx - 2C2x -1 = (A + B)x2 + (2A - B + C )x + A - 2B - 2C

4 Equate the coeffi cients to solve for A, B and C.

0x2 = (A + B)x2

⇒ 0 = A + B⇒ A = -B [1]

2x = (2A - B + C)x⇒ 2 = 2A - B + C [2]

-1 = A - 2B - 2C [3]

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Worked example 7


On the Main screen, complete the entry line as:

x

x x

+- -

3

3 402

Highlight the equation and tap:• Interactive• Transformation• expand• Partial Fraction• OK


5 Substitute equation [1] into equations [2] and [3].

-3B + C = 2-3B - 2C = -1

6 Solve these equations simultaneously. 3C = 3⇒ C = 1Hence, -3B + 1 = 2⇒ -3B = 1

⇒ B =-13

⇒ A = 13

7 Substitute the values for A, B and C and write the answer in the form

Ax

Bx

C

x( ) ( ) ( ).

-+

++

+2 1 1 2

2 1

2 1

13 2

13 1

1

12 2

x

x x x x x

-- +

=-

-+

++( )( ) ( ) ( ) ( )

,

x ∈R\{-1, 2}


1


Ax

Bx

C

x( ) ( ) ( )-+

++

+2 1 1 2.

2 1

2 1

13 2

13 1

1

12 2

x

x x x x x

-- +

=-

-+

++( )( ) ( ) ( ) ( )

,

x ∈R\{-1, 2}

Sometimes the denominator may consist of an irreducible quadratic (a quadratic which cannot be factorised using real numbers). These types of functions need to be expressed in partial fractions of the form:

f xA

ax bBx C

cx dx e( ) =

++ +

+ +2.

Worked example 8

Express 5 9 108

2

3

x x

x

++ ++--

in partial fractions.

Think WriTe


1 Factorise the denominator. 5 9 10

8

5 9 10

2 2 4

2

3

2

2

x x

x

x x

x x x

+ +-

= + +- + +( )( )

On the Main screen, complete the entry line as:

2 1

2 1 2

x

x x

-- +( )( )


280

2 The denominator has a linear factor and an irreducible quadratic factor so the partial fractions will be of the form

Ax

Bx C

x x-+ +

+ +2 2 42.

5 9 10

8 2 2 4

2

3 2

x x

x

Ax

Bx C

x x

+ +-

=-

+ ++ +

x ∈R\{2}.

3 Express the sum of the two fractions on the right as a single fraction.

5 9 10

8

2 4 2

2

2

3

2

x x

x

A x x Bx C x

x x

+ +-

=

+ + + + --

( ) ( )( )

( )( 22 2 4+ +x )

4 Equate the numerators and simplify. 5x2 + 9x + 10 = A(x2 + 2x + 4) + (Bx + C) (x - 2) = Ax2 + 2Ax + 4A + Bx2 - 2Bx + Cx - 2C) = (A + B)x2 + (2A - 2B + C )x + 4A -2C

5 Equate the coefficients to solve for A, B and C.

5x2 = (A + B)x⇒ A + B = 5⇒ B = 5 - A [1]

9x = (2A - 2B + C)x⇒ 9 = 2A - 2B + C [2]

10 = 4A - 2C [3]

6 Substitute equation [1] into equation [2] and then subtract equation [3] to solve for A, B and C.

Substituting [1] into [2]2A - 2(5 - A) + C = 9 4A - 10 + C = 9 4A + C = 9Subtracting equation [3] 3C = 9 C = 3 A = 4 B = 1

7 Substitute the values for A, B and C and write the answer in the form

Ax

Bx C

x x-+ +

+ +2 2 42.

5 9 10

8

42

3

2 42

2

3 2

x x

x xx

x xx R

+ +-

=-

+ ++ +

∈, \{ }


1


On the Main screen, complete the entry line as:5 9 10

8

2

3

x x

x

+ +-



2 Write the answer in the formA

xBx C

x x-+ +

+ +2 2 42.

5 9 10

8

42

3

2 42

2

3 2

x x

x xx

x xx R

+ +-

=-

+ ++ +

∈, \{ }

improper fractionsIn the case where g(x) has a higher power than h(x) the function f x

g xh x

( )( )( )

= is an improper

fraction. In this case, division of polynomials needs to be performed fi rst either by long division or synthetic division.

Worked example 9

Express x xx

2 5 21

++ ----

as a partial fraction.

Think WriTe


1 The degree of the denominator is less than the degree of the numerator, so division must be performed fi rst.

x - 1 is the divisor.

2 Divide the numerator by the denominator using long division.

)x x x

x xxx

x- + -

---

+1 5 2

6 26 6

4

62

2

3 Express the answer as partial fractions.x x

xx

xx R

2 5 21

64

11

+ --

= + +-

∈, \{ }


1

2 Write the answer. x xx x

x2 5 2

14

16

+ --

=-

+ + , x ∈R\{1}

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Worked example 9

On the Main screen, complete the entry line as:x x

x

2 5 21

+ --


282

For rational functions of the form f xg xh x

( )( )( )

= :

• If g (x) and h (x) are both linear functions, then the function can be expressed in the

form f x Ab

h x( )

( )= + .

• Where the numerator is a linear function and the denominator is a quadratic which can

be factorised, then the partial fraction will be of the form f xA

ax bB

cx d( ) =

++

+.

• When the denominator has repeated linear factors of the form (ax + b)2 then the partial

fractions will be of the form f xA

ax bB

ax b( )

( ) ( )=

++

+ 2.

On occasions when it is impossible to express the partial fractions in the form

f xA

ax bB

ax b( )

( ) ( )=

++

+ 2, they can be written as

f xA

ax bB

ax bC

ax b( )

( ) ( ) ( )=

++

++

+ 2.

• When the denominator contains an irreducible quadratic then the partial fractions will

be of the form f xA

ax bBx C

cx dx e( ) =

++ +

+ +2.

• In the case where g (x) has a higher power than h (x) the function is an improper fraction so division of polynomials needs to be performed either by long division or synthetic division.

rememBer

partial fractions 1 We5 Express each of the following as the sum of two terms.

a 2 31

xx

-+

b 4 72

xx

+-

c xx+-7

2 1d 3 4

2 2xx

-+

2 We6 Express each of the following as partial fractions.

a x

x x+

+ -6

1 4( )( )b x

x x

-- +

5

5 62c 2 1

8 92

x

x x

-+ -

d 3 2

2 9 72

x

x x

+- +


a x

x

-+

1

2 2( )b x

x x

-- +

4

6 92c

2 14

1 3 2

x

x x

+- +( )( )

d 3 5

2 1 2

x

x x

-- +( )( )


a x x

x x x

2

2

3 18

1 2 5

+ ++ - +( )( )

b x

x x x

2

2

5

3 1

++ +( )

c 2

5 5 22( )( )x x x- + -d x x

x

2

3

5 1

27

+ --


a x xx

2 3 12

+ ++

b x xx

3 2 34

+ --

c 3 2 4 5

6

3 2

2

x x x

x x

+ - ++ +

d xx

3 32 1

+-

exerCise

8B

4

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simultaneous equationsIt is impossible to solve one linear equation with two unknown variables. There must be two equations with the same two unknowns for a solution to be found. Such equations are called simultaneous equations.

There are several different ways to solve simultaneous equations. In this section we consider algebraic solutions of simultaneous equations arising from the intersection of a line with a parabola, circle or rectangular hyperbola.

Worked example 10

Solve simultaneously: y = x and y = x2 + 3x + 1.

Think WriTe


1 Write the equations and label them [1] and [2].

y = x [1]y = x2 + 3x + 1 [2]

2 Substitute equation [1] into equation [2]. Substituting [1] into [2]:x2 + 3x + 1 = x

3 Transpose to make the RHS equal 0 and simplify.

x2 + 3x + 1 - x = 0x2 + 2x + 1 = 0

4 Factorise. (x + 1)2 = 0

5 Solve for x. x + 1 = 0x = -1

6 Substitute -1 instead of x into equation [1]. Substituting -1 into [1]:y = -1

7 Write the answer. Solution set: (-1, -1)


1

2 Write the answer. Solving y = x and y = x2 + 3x +1 for x and y gives x = -1 and y = -1.That is, (-1, -1).

8C


284

Worked example 11

Solve simultaneously: y = x + 1 and x2 + y2 = 4.

Think WriTe



y = x + 1 [1]x2 + y2 = 4 [2]

2 Substitute equation [1] into equation [2]. Substituting [1] into [2]: x2 + (x + 1)2 = 4

3 Expand (x + 1)2, using the perfect square identity and transpose to make the RHS = 0.

x2 + x2 + 2x + 1 - 4 = 0 2x2 + 2x - 3 = 0

4 Solve for x, using the quadratic formula. a = 2, b = 2, c = -3

x =± - × × -

×

= ± +

= ±

= ±

=

-

-

-

-

2 2 4 2 3

2 2

2 4 244

2 284

2 2 74

2 ( )

-- ±1 72

5 Write the two values of x separately. x x1 21 7

21 7

2= + = -- -

,

6 Substitute - +1 7

2 instead of x into

equation [1] and simplify.

y11 7

21

1 72

22

1 7 22

1 72

= + +

= + +

= + +

= +

-

-

-

7 Substitute - -1 7

2 instead of x into

equation [1] and simplify.

y21 7

21

1 72

22

1 7 22

1 72

= - +

= - +

= - +

= -

-

-

-

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Worked example 11



8 Write the answer. (Make sure the values of x and y are matched properly; that is, x1 is placed with y1 and x2 with y2.)

Solution set: - + +

1 72

1 72

, ,

- - -

1 72

1 72

,


1

2 Write the answer. Solving y = x + 1 and x2 + y2 = 4 for x and y gives

x = +- ( )7 12

and y = -- ( )7 12

or

x = -7 12

and y = +7 12

That is,

- -+ -

( ),

( )7 1

2

7 1

2 or

7 1

2

7 1

2

- +

, .

Worked example 12

Solve simultaneously: y = 2x - 1 and yx

==--2

3.

Think WriTe



y = 2x - 1 [1]

yx

=-2

3 [2]

2 Substitute equation [1] into equation [2]. Substituting [1] into [2]:

2 12

3x

x- =

-


286

3 Solve for x:(a) Multiply both sides of the equation by

(x - 3).(b) Expand and make the RHS = 0.(c) Identify the values of a, b and c.

(d) Substitute the values of a, b and c into the quadratic formula and simplify.

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

2x2 - 7x + 1 = 0a = 2, b = -7, c = 1

x =± - - × ×

×

= ± -

= ±

7 7 4 2 1

2 2

7 49 84

7 414

2( )

4 Write the two values of x separately. x x1 27 41

47 41

4= + = -

,

5 Substitute 7 414

+ into [1] and simplify. y1 27 41

41

7 412

22

5 412

= × +

-

= + -

= +

6 Substitute 7 41

4-

into [1] and simplify. y2 27 41

41

7 412

22

5 412

= × -

-

= - -

= -

7 Write the answer (leave it in surd form). Solution set: 7 41

45 41

2+ +

, ,

7 414

5 412

- -

,


1




2 Write the answer. Solving y x yx

= - =-

2 12

3and for x and y gives

x y

x y

= - = - -

= +

- ( ) ( )41 74

41 54

41 74

and

or and == +41 54

That is,- -- -

+ +

( ),

( ),

41 74

41 5

441 7

441 5

4or

.

Simultaneous equations, arising from the intersection of a line with a parabola, circle or a rectangular hyperbola, can be solved using algebra as follows:

Transpose one of the equations (it is better to choose a linear equation) to make either 1. x or y the subject and substitute into the other equation.Simplify the resulting equation (if properly simplifi ed, it will result in a quadratic 2. equation).Solve the quadratic equation to fi nd the value(s) of one variable.3. Substitute the value(s) of the fi rst variable into either of the two equations (preferably 4. into the transposed one) and solve for the second variable.Write the solution set.5.

rememBer

simultaneous equations 1 We 10 Solve each of the following simultaneously.

a y = x, y = x2 + 5x + 4 b y = -x, y = x2 + 3x + 4c y = 2x, y = x2 + 4x + 1 d y = 3x, y = x2 + 8x + 6e y = -2x, y = x2 - 2x - 1 f y = x + 5, y = x2 − 4x + 11

g y + x = 1, y = 3x2 + 2x - 1 h 2x + 3y = 6, yx=

2

2 - 4

i 2x - 4y = 12, y = -2x2 + x + 6 j 2y + 6x = 4, y = 4 - 3x2

2 We 11 Solve each of the following simultaneously.a y = x, y2 + x2 = 1 b y = -x, y2 + x2 = 1c y = 2x, x2 + y2 = 4 d y = 1 - x, 4 = y2 + x2

e x + y = 2, x2 + y2 - 9 = 0 f y - x - 3 = 0, (x + 3)2 + y2 = 16

g y = 1 - x2

, (x - 2)2 + y2 - 1 = 0 h y + 2 = 3x, (x - 1)2 + (y + 3)2 = 4

i 2x + 4y = 4, x2 + (y + 1)2 = 25 j 6x - 3y = 12, (x - 2)2 + (y - 1)2 - 36 = 0

3 We 12 Solve each of the following simultaneously.

a y = 2x, y = 2

1x - b y = x + 1, y = 4

2 - x

c y = 3x - 4, y = 4

1 2- xd y = 1 - 2x, y = 1

2x + -1

exerCise

8C eBookpluseBookplus

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Using substitution to solve simultaneous

equations

288

e 2y - x = 6, y - 3 = 2

1x -f 2x + 4y - 8 = 0, y + 1 = 3

2 - x

g x - 3 = 2y, y = 21x -

h x = 2y - 4, 2

3x -1 - y = 0

i 4x - 3y = 12, 4 - 13 2x -

= y j x + y = 5, 1

4 3- x = y + 2

4 mC Which of the following represent the solution to the pair of simultaneous equations x + y = 6 and y = 3x2 + 12x + 10?

i -( )1

313

6, ii (-1, 7) iii (-1, 1) iv (-4, 10)

A i only B i and ii C ii and iv D ii and iii E i and iv

5 Buttons are to be attached to a shirt as shown on the diagram at right. If we draw a set of axes through the centre of the button, the position of the two holes can be described as the points of intersection of the line y = x with the circle x2 + y2 = 1. The other two holes are positioned at the points of intersection of the line y = -x with the same circle. Find the coordinates of the four holes. Give the answer correct to 2 decimal places.

y = −x

y2 + x2 = 1

y = x

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summary

Polynomials

A polynomial identity is an identity of the form:• knxn + kn - 1xn - 1 + kn - 2xn - 2 + . . . . . + k1x + k0, n ∈ N

where kn, kn - 1. . . are constants and n is contained within the set of natural numbers N.The degree of a polynomial is given by the highest value of • n.Polynomials are identical if they are of the same degree and corresponding coefficients are equal.• If two polynomials are known to be equal, then the process of equating coefficients can be used.•

Partial fractions

For rational functions of the form f xg xh x

( )( )( )

= :

If • g(x) and h(x) are both linear functions, then the function can be expressed in the form f x Ab

h x( )

( )= + .

Where the numerator is a linear function and the denominator is a quadratic which can be factorised, then the •

partial fraction will be of the form f xA

ax bB

cx d( ) =

++

+.

When the denominator has repeated linear factors of the form (• ax + b)2 then the partial fractions will be of

the form f xA

ax bB

ax b( )

( ) ( )=

++

+ 2. On occasions when it is impossible to express the partial fractions in

the form f xA

ax bB

ax b( )

( ) ( )=

++

+ 2, they can be written as f x

Aax b

Bax b

C

ax b( )

( ) ( ) ( )=

++

++

+ 2.

When the denominator contains an irreducible quadratic then the partial fractions will be of the form •

f xA

ax bBx C

cx dx e( )

( )=

++ +

+ +2.

In the case where • g(x) has a higher power than h(x), the function is then an improper fraction so division of polynomials needs to be performed either by long division or synthetic division.

Simultaneous equations

Simultaneous equations, arising from the intersection of a line with a parabola, circle or a rectangular hyperbola, can be solved using algebra as follows:

Transpose one of the equations (it is better to choose a linear equation) to make either • x or y the subject and substitute into the other equation.Simplify the resulting equation (if properly simplified, it will result in a quadratic equation).• Solve the quadratic equation to find the value(s) of one variable.• Substitute the value(s) of the first variable into either of the two equations (preferably into the transposed • one) and solve for the second variable.Write the solution set.•

290

ChapTer revieW

shorT ansWer

1 Determine the values of a and b where x4 + 7x3 +12x2 + x - 1 = (x2 + ax + 1)(x2 + bx - 1).

2 x - 2 is a factor of x3 + x2 - 11x + 10; find the other factor.

3 Express x

x x

-- -

22

8 202 as partial fractions.

4 Express x x x

x

3 2

2

2 1

2

- + --

as partial fractions.

5 Find the coordinates of the points of intersection of

the line y = 5x with the hyperbola yx

=-15

2.

6 Find the coordinates of the points of intersection of the line y = x with the parabola y = 4 - x2.

mulTiple ChoiCe

1 If 2x3 + 7x2 - 16x + 6 = (2x - 1)(ax2 + bx + c), then the values of a, b and c are:A a = 1, b = 4, c = 6 B a = -1, b = -4, c = 6C a = 1, b = 4, c = -6 D a = 1, b = -4, c = 6E a = 1, b = 4, c = -5

2 If 2x2 - 5x - 3 = a(x - b)(x - c) then the values of a, b and c areA a = 2, b = 3, c = 1 B a = 1, b = 3, c = 1

C a = 2, b = 3, c = 12

D a = 2, b = 3, c = -12

E a = 1, b = 3, c = 12

3 If 4 22 4 2 4x

x xA

xB

x+

+ -=

++

-( )( ), then:

A A = 1, B = 3 B A = 3, B = 1C A = 1, B = 4 D A = 4, B = 1E A = 2, B = 3

4 If 5 12

2 4 2 4

2

3 2 2

x x

x x x

Ax

Bx C

x x

+ ++ +

= + ++ +

, then the

values of A, B and C would be:

A A = 1, B = 2, C = 3 B A = 1, B = 3, C = 2C A = 1, B = 2, C = 5 D A = -3, B = 2, C = 5E A = 3, B = 2, C = -5

5 A solution to the pair of simultaneous equations

y = 1.5x and yx

=-

+2320 4

2 71( )

. is:

A 52

32

,

B 5

294

,

C 32

154

,

D 32

94

,

E none of these

6 The equation y = 1 - x4

and (y - 1)2 + (x - 3)2 = 9

are solved simultaneously. When one of the equations is substituted into the other and the resultant equation is transposed to the form ax2 + bx + c = 0, the values of a, b and c are:

A 17

6, -6, 18 B 17

6, -6, 0 C 1

16, -6, 0

D 1

16, -6, 9 E 17

16, -6, 9

exTended response

1 Find the coordinates of the points of intersection of the line y = x with:

a the hyperbola yx

=-

+23 1

6

b the circle (y + 1)2 + x2 - 4 = 0.In each case give the answer correct to 2 decimal places.

2 Consider the design, shown on the diagram at right:If we take the point of intersection of the straight lines to be an origin, the design can be described by the following system of equations:

y = x y = -x

y = 1x

y = - 1

xy = 0x = 0

x2 + y2 = 4

AB

C I

R N

D H

S M

EF

G

P

TKJ

L

Q O



As can be seen from the diagram, there are 20 points of intersection (not counting the centre point).a What is the radius of a circle described by the equation x2 + y2 = 4?b Using the answer to a, state the coordinates of points A, F, K and P.c Find the coordinates of point I by solving an appropriate pair of simultaneous equations algebraically.

Leave the answer in a surd form.d State the coordinates of points C, R and N using symmetry.e Find the coordinates of point H by solving algebraically an appropriate pair of simultaneous equations.f Using the symmetry of the design and your answer to part e, write the coordinates of points D, S and M.

g State the points of intersection of the hyperbola yx

= 1 and the circle.

h Find the coordinates of the points in question g by solving an appropriate pair of equations graphically, using a table of values or one of the iteration methods. Give the answer correct to 2 decimal places.

i State the points of intersection of the hyperbola yx

=-1

and the circle.

j Choose a method and use it to fi nd the coordinates of the points in question i. Give the answer correct to 2 decimal places.

3 A section of a roller coaster track is shown at right. It consists of three parts with the following equations:

AB: h d= +- 316

2 10

BC: h d d= - +316

2 3 16

CDE: h = 0.02d 3 - 1.25d 2 + 25d - 147.56where h is the height of the track above the ground level and d is the horizontal distance from A.a Find the coordinates of point B, by solving a pair of

simultaneous equations algebraically.b The track is closest to the ground when it is 8 metres horizontally from A. What is its height at that

point?c Find the horizontal distance(s) from A, when the car is 6 metres above ground level.d Use a CAS calculator to fi nd the coordinates of point C. e By using a table of values or otherwise, fi nd the

coordinates of point D.f Point E is 30 metres horizontally from A and is the

highest point of this section of the track. Find the maximum height of the track.

g The track runs alongside the amusements pavilion. The roof of the pavilion follows the rule h = 0.4d + 4. As seen from the diagram, the car, while on this section of the track, will be level with the roof four times. Find the height of the car above the ground at each of these four points.

h

dDistance

Hei

ght (

m)

A

B C

D

E

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292

eBookpluseBookplus aCTiviTies

Chapter openerDigital doc

10 Quick Questions: Warm up with ten quick • questions on further algebra. (page 272)

8A Polynomial identitiesTutorial

We4 • int-1063: Watch how to fi nd the quadratic factor of a cubic given the linear factor. (page 274)

8B Partial fractionsTutorials

We7 • int-1064: Watch how to express a linear function divided by a cubic as a partial fraction. (page 278) We9 • int-1065: Watch how to express a cubic divided by a linear function as a partial fraction. (page 281)

Digital doc

WorkSHEET 8.1: Use the bisection and secant • methods as well as the null factor law to solve simple and complex simultaneous equations, and apply learning to worded problems. (page 282)

Interactivity

Partial fractions• int-0975: Consolidate your understanding of how to determine partial fractions. (page 276)

8C Simultaneous equationsTutorial

We11 • int-1066: Watch how to solve simultaneously a linear and an elliptical equation. (page 284)

Digital docs

SkillSHEET 8.1: Practise using substitution to solve • simultaneous equations. (page 287)WorkSHEET 8.2: Practise fi nding solutions to linear • and non-linear simultaneous equations. (page 288)

Chapter reviewDigital doc

Test Yourself: Take the end-of-chapter test to test • your progress. (page 291)

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