1. UNIT OUTCOME PART SLIDE
Higher Maths Polynomials 1
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2. UNIT OUTCOME PART SLIDE NOTE
Higher Maths Polynomials 2
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Introduction to Polynomials
Any expression which still has multiple terms and powers after being simplified is called a Polynomial.
Examples
Polygon means ‘many sides’
2 x x x x + 7 a
9 – 5 a 7 + 3
( 2 x )( 3 x )( x – 8 )
Polynomial means ‘many numbers’
This is a polynomial because it can be multiplied out...

3. 2 x 4 + 7 x 3 + 5x 2 – 4 x + 3 UNIT OUTCOME PART SLIDE NOTE !
Higher Maths Polynomials 3
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Polynomials are normally written in decreasing order of power.
Coefficients and Degree !
Degree
2 x x 3 + 5x 2 – 4 x + 3
Coefficient
Term
Degree of a Polynomial
The ‘number part’ or multiplier in front of each term in the polynomial.
The value of the highest power in the polynomial.
4 x x 6 + 9x 3
3 x x 3 – x 2
is a polynomial of degree 6.
has coefficients 3, 5 and -1

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Roots of Polynomials
The root of a polynomial function is a value of x for which
f (x)
f (x) = 0 .
Example
Find the roots of
g(x) = 3 x 2 – 12
3 x 2 – 12 = 0
3 ( x 2 – 4 ) = 0
3 x 2 – 12 = 0
or...
3 ( x + 2 )( x – 2 ) = 0
3 x 2 = 12
x 2 = 4
x = 0 or
x – 2 = 0
x = ± 2
x = -2
x = 2

5. UNIT OUTCOME PART SLIDE NOTE
Higher Maths Polynomials 5
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Polynomials and Nested Brackets
Polynomials can be rewritten using brackets within brackets. This is known as nested form.
Example
f ( x ) = ax bx cx dx + e
= ( ax bx cx + d ) x + e
= (( ax bx + c ) x + d ) x + e
= (((ax + b ) x + c ) x + d ) x + e
= (((ax + b ) x + c ) x + d ) x + e a
× x
+ b
× x
+ c
× x
+ d
× x
+ e
f (x)

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Evaluating Polynomials Using Nested Form
Nested form can be used as a way of evaluating functions.
Example
Evaluate
g ( x ) = 2 x x 3 – 10 x 2 – 5 x + 7
for
x = 4
= (((2 x + 3 ) x – 10 ) x – 5 ) x + 7
g (4 ) = (((2 × ) × 4 – 10 ) × 4 – 5 ) × =
531 2
× 4
+ 3
× 4 – 1
× 4
– 5
× 4
+ 7
531

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The Loom Diagram x a b c d + + + +
Evaluation of nested polynomials can be shown in a table.
× x
× x
× x
f ( x ) = ax 3 + bx 2 + cx + d
= (( ax + b ) x + c) x + d
f (x) a
× x b
× x
+ c
× x d
(i.e. the answer) + +
Example 2 4 -3 5 -6
h ( x ) = 4 x 3 – 3 x x – 6 8 10 30
Evaluate
h ( x )
for
x = 2 . 4 5 15 24

8. UNIT OUTCOME PART SLIDE NOTE NOTICE ! f ( x ) = 8 x 7 – 6 x 4 + 5
Higher Maths Polynomials 8
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Division and Quotients
quotient
In any division, the part of the answer which has been divided is called the quotient.
remainder
6 r 2 5 3 2
f ( x ) = 8 x 7 – 6 x
Example
4 x 6 – 3 x r 5
Calculate the quotient and remainder for f ( x ) ÷ 2 x.
2 x
8 x 7 – 6 x
cannot be divided by 2 x
NOTICE !
The power of each term in the quotient is one less than the power of the term in the original polynomial.

9. coefficients of quotient
Higher Maths Polynomials 9
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Try evaluating f ( 3)…
NOTE
Investigating Polynomial Division
Example 3 2 -1
-16 7
f ( x) = (2 x2 + 5x – 1)( x – 3) + 4 6 15 -3 2 5 -1 4
= 2 x3 – x2 – 16 x + 7
coefficients of quotient
alternatively we can write
remainder !
NOTICE
f ( x) ÷ ( x – 3)
When dividing f ( x) by ( x – n), evaluating f ( n) in a table gives:
= 2 x x – 1 r 4
• the coefficients of the quotient
quotient
• the remainder
remainder

10. coefficients of quotient
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Synthetic Division
For any polynomial function f ( x) = ax 4 + bx 3 + cx 2 + dx + e ,
f ( x) divided by ( x – n) can be found as follows: n a b c d e + + + + +
× n
× n
× n
× n
coefficients of quotient
This is called Synthetic Division.
remainder

11. UNIT OUTCOME PART SLIDE NOTE ! g( x ) = 3 x 4 – 2 x 2 + x + 4
Higher Maths Polynomials 11
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Examples of Synthetic Division
Missing terms have coefficient zero. !
Example
g( x ) = 3 x 4 – 2 x 2 + x + 4
Find the quotient and remainder for g( x ) ÷ ( x + 2).
Evaluate g ( -2) : -2 3 -2 1 4 -6 12
-20 38
g( x ) ÷ ( x + 2) 3 -6 10
-19 42
= (3 x 3 – 6 x x – 19) with remainder 42
Alternatively, g( x ) = (3 x 3 – 6 x x – 19)( x + 2) + 42

12. UNIT OUTCOME PART SLIDE NOTE
Higher Maths Polynomials 12
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The Factor Theorem
If a polynomial f ( x) can be divided exactly by a factor ( x – h) , then the remainder, given by f ( h), is zero.
Example
Show that ( x – 4) is a factor of f ( x) = 2 x 4 – 9x x 2 – 3 x – 4
Evaluate f ( 4) :
f ( 4) = 0 4 2 -9 5 -3 -4 8 -4 4 4
( x – 4) is a factor of f ( x) 4 -1 1 1
zero remainder
f ( x) = 2 x 4 – 9x x 2 – 3 x – 4
= ( x – 4)(4 x 3 – x 2 + x + 1 ) + 0

13. UNIT OUTCOME PART SLIDE NOTE ! f ( x) = 2 x 3 + 5 x 2 – 28 x – 15
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Factors of -15 :
± 1
NOTE
Factorising with Synthetic Division
± 3
f ( x) = 2 x x 2 – 28 x – 15
Example
Factorise
± 5
± 15
Consider factors of the number term...
Try evaluating f ( 3) :
f ( 3) = 0 3 2 5
-28
-15 6 33 15
( x – 3)
If f ( h) = 0 then ( x – h)
is a factor. ! 2 11 5
is a factor
zero!
f ( x) = 2 x x 2 – 28 x – 15
= ( x – 3 )( 2 x x + 5 )
Evaluate f ( h) by synthetic division for every factor h.
= ( x – 3 )( 2 x + 1 )( x + 5 )

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Finding Unknown Coefficients
Example
( x + 3) is a factor of f ( x) = 2 x 4 + 6x 3 + px x – 15
Find the value of p.
( x + 3) is a factor
Evaluate f (- 3) :
f (- 3) = 0
- 3 p 2 6 4
-15
- 6
- 3p
9p – 12
9p – 27 = 0
9p = 27 p
- 3p + 4
9p – 27 2
p = 3
zero remainder

15. UNIT OUTCOME PART SLIDE NOTE NOTE ! x
Higher Maths Polynomials 15
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Finding Polynomial Functions from Graphs
The equation of a polynomial can be found from its graph by considering the intercepts.
Equation of a Polynomial From a Graph
f (x) = k( x – a )( x – b )( x – c )
f ( x)
with x-intercepts a , b and c d d
NOTE ! d
k can be found by substituting x a b c
( 0 , d )

16. UNIT OUTCOME PART SLIDE NOTE x f (x) = k( x + 2)( x – 1)( x – 5)
Higher Maths Polynomials 16
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Finding Polynomial Functions from Graphs (continued)
Example
f ( x)
Find the function shown in the graph opposite. 30
- 2 x 1 5
f (x) = k( x + 2)( x – 1)( x – 5)
Substitute k back into original function and multiply out...
f (0) = 30
k(0 + 2)(0 – 1)(0 – 5) = 30
f (x) = 3 ( x + 2)( x – 1)( x – 5)
10 k = 30
= 3 x 3 – 12 x 2 – 21x + 30
k = 3

17. UNIT OUTCOME PART SLIDE NOTE x x a a b b f (a) > 0 f (b) < 0
Higher Maths Polynomials 17
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Location of a Root
A root of a polynomial function f ( x) lies between a and b if :
f ( x)
f ( x)
root
root a b
or... x x a b
f (a) > 0
f (b) < 0
f (a) < 0
f (b) > 0
and
and
If the roots are not rational, it is still possible to find an approximate value by using an iterative process similar to trial and error.

18. UNIT OUTCOME PART SLIDE NOTE
Higher Maths Polynomials 18
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Finding Approximate Roots
The approximate root can be calculated by an iterative process:
Example x
f ( x)
root between
f ( x) = x 3 – 4 x 2 – 2 x + 7
Show that f ( x) has a root between 1 and 2. 1 2
- 5 2
1 and 2
1.3
1 and 1.3
f (1) = 2
1.2
0.568
1.2 and 1.3
(above x-axis)
1. 25
0.203
1.25 and 1.3
f (2) = - 5
(below x-axis)
1. 28
1.25 and 1.28
1. 27
0.057
1.27 and 1.28
f ( x) crosses the x-axis between 1 and 2.
1. 275
0.020
1.275 and 1.28
The root is at approximately x = 1.28