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INTRODUCTION TO POLYNOMIALS
DEFINITION AND TERMINOLOGY
A polynomial is an algebraic expression of the form:
Definition of a Polynomial
= + + + ⋯ + + + = 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
Degree
Definition
Significance
Highest power of Determines the general shape
of the curve
Coefficient
Leading Coefficient
Constant Term
Zeroes
, … , , , … Constants in front of
powers of Constant in front of highest
Determines the behavior of the
power of graph for large values of x.
Constant term that does
Constant term, and is the y
not change even as varies
intercept
Values of for which
-intercepts of the Polynomial.
= 0
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
= 1
= + 2
= + + 3
= " " + + + 4
= + " " + + + Quartic Polynomial
= + + ⋯ + Monic Polynomial
Constant Polynomial
Linear Polynomial
Quadratic Polynomial
Cubic 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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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.
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 ≤ Division however is more complicated and will be looked at in detail in another section.
Degree of Product
Given non-zero polynomials and , with degree m and n
respectively, then:
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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 max2, 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
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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) × ,
The degree will be 3 + 1 = 4
b) <=
The degree will be 3 + 3 = 6
c) <,=
The degree will be 3 + 1 + 1 = 5
d) <= + <,=
The degree will be max6,5. Hence the polynomial is of degree 6
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GRAPHING POLYNOMIALS
GENERAL SHAPE OF THE CURVE
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.
POLYNOMIALS OF ODD DEGREE
Question 7 (Conceptual)
The graph of two polynomials of odd degree are shown below:
>
>
> = " − > = * − 5 " + 4
= − 1
= − 1 + 1 − 2 + 2
+ 1
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
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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:
>
>
> = − 1
> = − 5 + 4
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
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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 Minimum Number of
(for Polynomial of degree ) Roots
Odd
Opposite Direction
1
Even
Same Direction
0
BEHAVIOUR FOR LARGE |X|
For large values of ||, the leading term dominates the function
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
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SHAPE OF SINGLE, DOUBLE AND TRIPLE ROOTS
Question 9 (Conceptual)
The graph below shows the curve > = + 4 + 1 − 3"
>
−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
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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
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
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.
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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 → +∞, > → +∞.
STEP 2: Determine the zeroes of the polynomial
> = − 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 − 40 − 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
STEP 1: Determine the general shape of the graph
The leading coefficient is negative, and the degree is odd
Hence, as → −∞, > → ∞ and as → +∞, > → −∞.
STEP 2: Determine the zeroes of the polynomial
= 0, 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.
STEP 4: Determine the y-intercept by evaluating 0.
0 = 0
0, 0 is the >-intercept
STEP 5: Sketch the curve
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Question 12
Graph the curve > = " − 1 − 3
STEP 1: Determine the general shape of the graph
The leading coefficient is positive, and the degree is odd
Hence, as → −∞, > → −∞ and as → +∞, > → +∞.
STEP 2: Determine the zeroes of the polynomial
= 0, 1, 3
STEP 3: Determine the nature of the zeroes
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
STEP 4: Determine the y-intercept by evaluating 0.
0 = 0
0, 0 is the >-intercept
STEP 5: Sketch the curve
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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 + 40 − 10 − 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, > = −531 = −15
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POLYNOMIAL LONG DIVISION
DIVIDING INTEGERS
If we are asked to divide a number R by another number we can write it as R T U =
I "remainder" r.. Alternatively, we can write this as R = · I + S where 0 9 S 9 − 1
An example is dividing 3489 by 11. Using the normal long division method we deduce that
3489 = 317 · 11 + 2,, i.e. the I = 317 and S = 2
DIVIDING POLYNOMIALS
( ), we can write this
Likewise if we are asked to divide a polynomial () by a polynomial Q()
( ) · ,() + :() where ,()is
in the form () = Q()
is the quotient and :()is the
remainder polynomial where 0 9 deg :() C deg Q() − 1.. A consequence of this is that
,() and :() are unique for every polynomial division.
division
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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 + XY − + Z = ( − X)(X + XY + [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
= − , + S, where S is a constant
:,
and
hence
show
that
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
The remainder will be 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
The remainder will be 0
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THE FACTOR THEOREM
Factor Theorem
= 0, then − is a factor of and vice versa
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
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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 = + 12 − 3 − 2
= + 12 + 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 TALENT 100: HSC SUCCESS. SIMPLIFIED.
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