HSC Math Ext. 1. - Talent 100 · 2016-09-23 · TALENT 100: HSC SUCCESS. SIMPLIFIE˙˘ Page 2 of 34...

HSC Math Ext. 1.

Polynomials I.

HMX1 Polynomials I © TALENT 100

TALENT 100: HSC SUCCESS. SIMPLIFIED. www.talent-100.com.au

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Definition of a Polynomial

�� = �� + �� + �� + ⋯ + �� + �� + �� = � ��

��

INTRODUCTION TO POLYNOMIALS

DEFINITION AND TERMINOLOGY A polynomial is an algebraic expression of the form:

The above expression is known as the polynomial ��.

• Polynomials are continuous and differentiable at every point.

• The exponents of a polynomial are non-negative integers. Hence, equations containing

5�� or 3�� are not polynomials.

• Coefficients are real numbers (at least for the scope of the extension 1 HSC Course)

Term Symbol Definition Significance

Degree � Highest power of � Determines the general shape

of the curve

Coefficient ��, �� … , �� Constants in front of

powers of �

Leading Coefficient �� Constant in front of highest

power of �

Determines the behavior of the

graph for large values of x.

Constant Term �� Constant term that does

not change even as � varies

Constant term, and is the y

intercept

Zeroes ��, ��, … �� Values of � for which

�� = 0

�-intercepts of the Polynomial.

A polynomial of degree � may

have up to n roots



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TYPES OF POLYNOMIALS Some polynomials have special names:

Degree �� Polynomial �� Special Name

0 �� = �� Constant Polynomial

1 �� = �� + �� Linear Polynomial

2 �� = �� + �� + �� Quadratic Polynomial

3 �� = �"�" + �� + �� + �� Cubic Polynomial

4 �� = �� + �"�" + �� + �� + �� Quartic Polynomial

� �� = �� + �� + ⋯ + �� Monic Polynomial

ROOTS & ZEROES The equation �� = 0 is known as a polynomial equation of degree � where � is the

degree of the polynomial �. The real number $ such that ��$� = 0 is known as a root (or

solution) of the polynomial equation, as well as a zero of the polynomial �. A polynomial

equation may have more than one root (up to � roots, where � is the degree of the

polynomial) or may have none at all (e.g. �� + 1 = 0�

A polynomial has ‘zeroes’ and a polynomial equation has ‘roots’.



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Question 1

Identify which of these following expressions are polynomials. For the polynomials you have

identified, identify the degree, leading coefficient and constant.

a) 2�" + 24� − 15

Since all the powers are non-negative integers, it is a polynomial

It is a polynomial of degree 3, with leading coefficient 2, and constant term −15

b) �& + 3�" − 2��

It is not a polynomial because it contains −2��, which has a negative power.

c) 3�� − 4� + �'(

It is not a polynomial because it contains �'(, which has a fractional power.

d) 2�� + 1�� − 4�� + 3�

2�� + 1�� − 4�� + 3� = �2� + 2�� − � − 12�

= 2�" − 2�� − 24� + 2�� − 2� − 24

= 2�" − 26� − 24

Since all the powers are non-negative integers, it is a polynomial

It is a polynomial of degree 3, with leading coefficient 2 and constant term −24

e) �* + 5+

It is not a polynomial because 5+ is not a non-negative integer power

f) 0

It is a polynomial of degree 0, with leading coefficient 0 and constant term 0



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Question 2

Consider the polynomial �� = �� − 1�� + 1�� + 2�

a) Write down the zeroes of ��

The zeroes are 1, −1 and −2

b) By expanding ��, find the roots of the polynomial equation �" + 2�� − � − 2 = 0

�� = �� − 1�� + 2� = �" − � + 2�� − 2 = �" + 2�� − � − 2

Since the roots of the equation are the zeroes of the polynomial ��, the roots are

� = 1, −1 and −2



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Degree of Product

Given non-zero polynomials �� and ,�� with degree m and n

respectively, then:

deg0�� ∙ ,��2 = 3 + �

OPERATIONS ON POLYNOMIALS Performing operations with polynomials is a relatively straightforward exercise. In order to

perform addition or subtraction, we group like terms and add or subtract coefficients.

Multiplication of two polynomials is performed according to the normal expansion method.

Division however is more complicated and will be looked at in detail in another section.

Degree of Sum and Difference

Given �� and ,�� with degree m and n respectively then:

• If � ≠ 3, then deg0�� ± ,��2 = max �3, ��

• If � = 3, then deg0�� ± ,��2 ≤ �



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Question 3

Given that �� = �� + 2� − 1 and ,�� = �" + �� + 1, state the degree of the

polynomial :�� if

a) :�� = �� + ,��

First, note that the degree of �� is 2 and the degree of ,�� is 3

If :�� = �� + ,��, then the degree of :�� is max�2, 3�. Hence, the degree of

:�� is 3

b) :�� = �� × ,��

If :�� = �� × ,��, then the degree of :�� is 2 + 3. Hence, the degree of :�� is 5

Question 4

Given that �� = 2�� + � and ,�� = � − 1, find

a) �� − ,��

= 2�� + � − �� − 1� = 2�� + � − � + 1 = 2�� + 1

b) �� × ,��

= �2�� + �� − 1� = 2�" − 2�� + �� − � = 2�" − �� − �



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Question 5

Consider the polynomials �� = 2�� + 3� + 1 and ,�� = −2�� + 1

a) Find the degree of �� + ,��

Since the degree of �� and ,�� are equal, the sum has to be evaluated

�� + ,�� = 2�� + 3� + 1 − 2�� + 1 = 2�� − 2�� + 3� + 2 = 3� + 2

�� + ,�� is degree 1

b) State the degree of �� × ,��

The degree will be 2 + 2 = 4

c) Verify the answer to b) via direct expansion of �� × ,��

�� × ,�� = �2�� + 3� + 1��−2�� + 1� = −4�� − 6�" − 2�� + 2�� + 3� + 1 = −4�� − 6�" + 3� + 1

Which is a polynomial of degree 4



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Question 6

Given �� = �" + 3� and ,�� = 3� + 1, find the degree of the following polynomials

a) �� × ,��


b) <��=�


c) ��<,��=�

The degree will be 3 + 1 + 1 = 5

d) <��=� + ��<,��=�

The degree will be max�6,5�. Hence the polynomial is of degree 6



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�

>

�

>

> = �* − 5�" + 4� = �� − 1�� + 1�� − 2�� + 2�

GRAPHING POLYNOMIALS

GENERAL SHAPE OF THE CURVE

POLYNOMIALS OF ODD DEGREE Question 7 (Conceptual)

The graph of two polynomials of odd degree are shown below:

a) What do you notice about the general shapes of the graphs?

They are roughly the same shape and differ by the number of roots.

b) What do you notice about the shape of the graph as � → ±∞ (i.e. the tails of the curve)?

The tails of the curves point in opposite directions

General Shape of a Polynomial

The general shape of a polynomial will be determined by whether the

degree of the polynomial is even or odd.

> = �" − � = �� − 1��

+ 1�



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�

>

�

>

c) By examining the graphs of the cubic below, determine the minimum and maximum

number of roots a cubic can have?

The first graph shows that a cubic can have a maximum of 3 roots, which is consistent

with the fact that cubics are of degree 3

Since the tails point in opposite directions, one of them must touch the �-axis

eventually. So there is a minimum of one root

d) What is the minimum number of roots a polynomial with odd degree can have?

Since the tails of the a polynomial of odd degree go across in opposite diagonals, there

must be at least one real root.



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POLYNOMIALS OF EVEN DEGREE Question 8 (Conceptual)

The graphs of a quadratic and quartic polynomial are shown below:

a) What do you notice about the general shapes of the graphs?

They are roughly the same shape, i.e. their tails go in the same direction. They differ in

the number of roots they have.

b) What do you notice about the shape of the graph as � → ±∞ (i.e. the tails of the curve)?

As � → ±∞, the tails of the curve go in the same direction. In this case, both point

upwards

�

>

�

>

> = �� − 1 > = �� − 5�� + 4



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�

> >

c) By examining the graph below, determine the minimum and maximum number of roots

a quartic can have

The first graph shows that a quartic can have a maximum of 4 roots, which is consistent

with the fact that quartics are of degree 4

Since the tails point in the same direction, it is possible for the tails to both point away

from the �-axis, and there will be no roots (Provided the other parts of the graph do not

intersect)

d) What is the minimum number of roots a polynomial of even degree can have?

Since the tails of polynomials of even degree go in the same direction, the minimum

number of roots is 0



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SUMMARY OF ODD AND EVEN DEGREE FUNCTIONS

Degree Direction of the Tails Maximum Number of Roots

(for Polynomial of degree �)

Minimum Number of

Roots

Odd Opposite Direction � 1

Even Same Direction � 0

BEHAVIOUR FOR LARGE |X|

We have established that �� will affect the general shape of the curve (odd or even �). The

sign of the leading coefficient determines the value as � → ±∞

A� > 0 A� < 0

� → ∞, � even or odd �� → +∞ �� → −∞

� → −∞, � even �� → +∞ �� → −∞

� → −∞, � odd �� → −∞ �� → +∞

Talent Tip: This table does not need to be memorised. The degree of � will determine the

general shape of the curve, and the sign of �� will determine whether or not to reflect that

shape across the �-axis

For large values of |�|, the leading term �� dominates the function



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SHAPE OF SINGLE, DOUBLE AND TRIPLE ROOTS Question 9 (Conceptual)

The graph below shows the curve > = �� + 4�� + 1�� − 3�"

a) What are the zeroes of the polynomial?

The zeroes of �� = �� + 4�� + 1�� − 3�" are −4, −1 and 3

b) Compare the shape of the curve at the zeroes

At � = −4, the graph simply cuts the �-axis. This resembles a straight line

At � = −1, the graph turns, and touches the �-axis. This resembles a quadratic

At � = 3, the graph resembles a cubic

�

>

−4

−1

3



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The general shape of single, double and triple roots are seen below

�

>

Single Root Double Root Triple Root

Behaviour of Multiple Zeroes

• At a single zero, the curve cuts the x-axis, not tangent to it.

• At a zero of even multiplicity, the curve lies tangent to the x-axis

without crossing it

• At a zero of odd multiplicity (≥ 3), the curve lies tangent to the x-axis

and possesses a point of inflection at this root, crossing over the x-axis.



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CURVE SKETCHING METHOD

Talent Tip: Be sure to mark on your graph both the � and > intercepts as well as clearly

indicating the behaviour of the curve at its extremities.

A basic method for sketching polynomial functions:

STEP 1: Determine the general shape of the graph, and its behavior at the

extremities by examining the leading term ��.

• When � is even, the tails of the curve will be on the same side of the

> − ��FG (like a quadratic)

• When � is odd, the tails of the curve will be on different sides (like a

cubic)

• The sign of �� determines the behavior at the extremeties.

Mark the “tails” of the polynomial

STEP 2: Determine the zeroes of the polynomial

STEP 3: Determine the nature of the zeroes, i.e. their multiplicity and hence

their shape

STEP 4: Determine the y-intercept by evaluating ��0�.

STEP 5: Sketch the curve using the above information



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Question 10

Graph the curve > = �� − 4�� − 4�

STEP 1: Determine the general shape of the graph

Expanding, > = �" − 4�� − 4� + 16

The leading coefficient is positive and the degree is odd

Hence, as � → −∞, > → −∞ and as � → +∞, > → +∞.


> = �� − 4�� − 4�

= �� − 2�� + 2�� − 4�

Thus the zeroes are � = −2, 2, 4

STEP 3: Determine the nature of the zeroes

In this case, all the roots are single and thus they just cross the �-axis.



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STEP 4: Determine the y-intercept by evaluating ��0�

��0� = �0 − 4��0 − 4� = −4 × −4 = 16

∴ �0, 16� is the >-intercept.

STEP 5: Sketch the curve



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Question 11

Sketch the graph of > = −5�� − 2�� + 4�


The leading coefficient is negative, and the degree is odd

Hence, as � → −∞, > → ∞ and as � → +∞, > → −∞.


� = 0, 2, −4


In this case, all the roots are single and thus they just cross the �-axis.


��0� = 0

�0, 0� is the >-intercept




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Question 12

Graph the curve > = �"�� − 1�� − 3��


The leading coefficient is positive, and the degree is odd

Hence, as � → −∞, > → −∞ and as � → +∞, > → +∞.


� = 0, 1, 3


There is a triple root at � = 0, so it is cubic in shape.

There is a double root at � = 3, so it is parabolic in shape


��0� = 0

�0, 0� is the >-intercept




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Question 13

Neatly sketch the graphs of the following polynomials, showing any important information

a) > = �� + 4�� − 1�� − 3�

The zeroes are � = −4,1,3. Each of these zeroes is a single zero and so the curve just

cuts the graph at each of these points

The leading coefficient is positive and the degree is odd meaning that when � → −∞,

> → −∞ and when � → +∞, > → +∞.

When � = 0, > = �0 + 4��0 − 1��0 − 3� = 12

The >-intercept is �0, 12�



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b) Graph the curve > = �� − 2� − 15�� + � + 1� (You may assume the graph has a

basic quartic shape)

Factorising, > = �� − 5�� + 3�� + � + 1�

The zeroes are � = −3, 5

Leading coefficient is 1 and polynomial is of even degree, so when � → ∞, > → ∞ and

when � → −∞, > → ∞.

When � = 0, > = �−5��3��1� = −15

HMX1

TALENT 100: HSC SUCCESS. SIMPLIFIED.

POLYNOMIAL

DIVIDING INTEGERS If we are asked to divide a number

I "remainder" r. Alternatively, we can write this as

An example is dividing 3489 by

3489 = 317 · 11 + 2, i.e. the

DIVIDING POLYNOMIALS Likewise if we are asked to divide a polynomial

in the form �(�) = Q(�) ·

remainder polynomial where

,(�) and :(�) are unique for every polynomial division

Polynomials I

HSC SUCCESS. SIMPLIFIED. www.talent

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OLYNOMIAL LONG DIVISION

If we are asked to divide a number R by another number � we can write it as

. Alternatively, we can write this as R = � · I + S where

by 11. Using the normal long division method we deduce that

, i.e. the I = 317 and S = 2

OLYNOMIALS Likewise if we are asked to divide a polynomial �(�) by a polynomial Q(�)

( ) ,(�) + :(�) where ,(�)is the quotient and

remainder polynomial where 0 9 deg :(�) C deg Q(�) − 1. A consequence of this is that

for every polynomial division.

© TALENT 100

www.talent-100.com.au

we can write it as R T U =

where 0 9 S 9 � − 1

normal long division method we deduce that

( ), we can write this

is the quotient and :(�)is the

. A consequence of this is that



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Question 14 (Worked Example)

Use the division algorithm to divide 3�" + 4�� + 8� + 9 by � + 1



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Question 15

Using polynomial long division, divide the following, and express in the form �(�) =

V(�),(�) + :(�)

a) �� + 3�� − � + 6 by � − 3

�W + X�Y − � + Z = (� − X)(�X + X�Y + [Y� + X\) + [[[

Talent Tip: In order to preserve the structure of the long division, it is recommended that

you add a placeholder 0�", such that the columns all contain terms with the same power of

�

Talent Tip: In each line, you are subtracting the multiplied divisor from the dividend.

Remember that when you subtract a negative, it becomes addition



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b) �" + 2�� − 23� − 60 by � + 4

�" + 2�� − 23� − 60 = �� + 4�� − 2� − 15� + 0

c) �� + 2�" − �� + 4� + 10 by �� − � + 2

�� + 2�" − �� + 4� + 10 = �� − � + 2�� + 3�� + �−2� + 10�



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d) 3�" + 3�� + 6� − 1 by 3�� + 1

Talent Tip: If not specified, write the result of the division in the form �� = Q�� ×,�� + :��



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DIVISION THEOREM When �� is divided by Q��, there remains unique polynomials ,�� and :�� such that -

• �� = Q�� × ,�� + :��

• deg :�� < deg Q��, or :�� = 0

REMAINDER THEOREM Question 16 – Proof of Remainder Theorem (Conceptual)

Suppose that the polynomial �� is divided by Q�� = �� − ��

a) Using the division theorem, write an expression for ��

�� = Q�� × ,�� + :�� = �� − ��,�� + :��

b) Determine the degree of the remainder :��, and hence show that �� = �� − ��,�� + S, where S is a constant

Since the divisor is degree 1, the remainder must be degree 0, which is a constant

Hence, �� = �� − ��,�� + S

c) Hence, show that �� = S

When � = �

�� = �� − ��,�� + S = 0 × ,�� + S = S

Remainder Theorem

The remainder S of polynomial �� upon division by a linear divisor

�� − �� is ��.



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Question 17

Let �� = �" − 3�� + 4� + 1

a) Evaluate ��1�

��1� = 1" − 3 × 1� + 4 × 1 + 1 = 1 − 3 + 4 + 1 = 3

b) Hence, find the remainder when �� is divided by �� − 1�

By the remainder theorem, the remainder will be ��1�

Hence, the remainder is 3



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Question 18

Find the remainder when the following polynomials are divided by the given divisor

a) �� = �" − 3�� + 5 is divided by � − 4

By the remainder theorem, the remainder will be ��4�

��4� = 4" − 3 × 4� + 5 = 64 − 48 + 5 = 21

The remainder will be 21

b) �� = �� + �" + 2�� + 3� + 5 is divided by �

� = � − 0, and so by the remainder theorem, the remainder will be ��0�

��0� = 0 + 0 + 0 + 0 + 5 = 5


c) �� = �" + 2�� − 5� − 6 is divided by � + 3

By the remainder theorem, the remainder will be �� + 3�

��−3� = �−3�" + 2 × �−3�� − 5 × �−3� − 6 = −27 + 18 + 15 − 6 = 0




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THE FACTOR THEOREM

Question 19

a) Show that � = 1, ±2 are zeroes of �� = �" − 4� − �� + 4

��1� = 1 − 4 − 1 + 4 = 0

��−2� = −8 + 8 − 4 + 4 = 0

��2� = 8 − 8 − 4 + 4 = 0

Hence � = 1, ±2 are zeroes of ��

b) Hence write �� in factorise form

Since ��1�, ��−2� and ��2� equal 0, then � − 1, � + 2 and � − 2 must be factors of

��

Since the leading coefficient of �� is 1

�� = �� − 1�� − 2�� + 2�

Factor Theorem

�� = 0, then �� − �� is a factor of �� and vice versa



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Question 20

Consider the polynomial �� = 2�" − �� − 5� − 2

a) Show that � + 1 is a factor

If � + 1 is a factor, then ��−1� = 0

��−1� = 2 × �−1�" − �−1�� − 5 × �−1� − 2

= −2 − 1 + 5 − 2

= 0

Hence, � + 1 is a factor

b) Using long division, factorise ��

�� = �� + 1��2�� − 3� − 2� = �� + 1��2� + 1�� − 2�



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Question 21

Given that 3� − 1 is a factor of �� = 3�" − 4�� − 17� + 6, find the other factors of ��

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