Post on 27-Mar-2015
1Higher Maths 2 1 1 Polynomials
Any expression which still has multiple terms and powers after being simplified is called a Polynomial.
Introduction to Polynomials
2
Polynomial
means ‘many
numbers’
2 x
4 + 6 x
3 + 5 x
2 + 4 x + 7
Examples
9 – 5 a
7 +a3
( 2 x + 3 )( 3 x + 1 )( x – 8 )
Polygon means
‘many sides’
This is a polynomial
because it can be
multiplied out...
Higher Maths 2 1 1 Polynomials
2 x
4 + 7 x
3 + 5x
2 – 4 x + 3
3
Coefficients and Degree
The value of the highest
power in the polynomial.
4 x
5 + 2 x
6 + 9x
3 is a polynomial of
degree 6.
Coefficien
t
3 x
4 + 5 x
3 – x
2
has coefficients 3, 5
and -1
Degree
Term
The ‘number part’ or
multiplier in front of
each term in the
polynomial.
Degree of a Polynomial
Polynomials are normally
written in decreasing order
of power.
Higher Maths 2 1 1 Polynomials
Roots of Polynomials
4
The root of a polynomial function is a
value of xfor which
f (x)f (x) = 0 .
Find the roots of
g(x) = 3 x
2 – 12
3 x
2 – 12 = 0
3 x
2 = 12
x
2 = 4
x = ± 2
3 ( x
2 – 4 ) = 0
3 ( x + 2 )( x – 2 ) = 0
x + 2 = 0 x – 2 = 0
x = 2 x = -2
3 x
2 – 12 = 0
or...
or
Example
Higher Maths 2 1 1 Polynomials
Polynomials and Nested BracketsPolynomials can be rewritten using brackets within brackets. This is known as nested form.Example
5
f ( x ) = ax 4 + bx
3 + cx 2 + dx + e
= ( ax 3 + bx
2 + cx + d ) x + e
= (( ax 2 + bx + c ) x + d ) x + e
= (((ax + b ) x + c ) x + d ) x + e
× xa + b + c + e× x × x+ d f (x)
= (((ax + b ) x + c ) x + d ) x + e
× x
Higher Maths 2 1 1 Polynomials
Evaluating Polynomials Using Nested Form
Example
6
g ( x ) = 2 x 4 + 3 x
3 – 10 x 2 – 5 x + 7
= (((2 x + 3 ) x – 10 ) x – 5 ) x +
7
2
Evaluate
for
x = 4
g (4 ) = (((2 × 4 + 3 ) × 4 – 10 ) × 4 – 5 ) × 4 + 7
=
× 4
531
+ 3 1– 0× 4 – 5× 4 × 4 + 7 531
Nested form can be used as a way of evaluating functions.
Higher Maths 2 1 1 Polynomials
The Loom Diagram
7
Evaluation of nested polynomials can be shown in a table.f ( x ) = ax
3 + bx 2 + cx + d
= (( ax + b ) x + c) x + d
× x b × x + c × x+ d+a
a b c dx
× x
+
× x
+
× x
+ +
Exampleh ( x ) = 4 x
3 – 3 x 2 + 5 x –
6Evaluate
h ( x ) for
x = 2 .
2 4 -3 5 -6
4 5 15 24
8 10 30
f (x)(i.e. the
answer)
Higher Maths 2 1 1 Polynomials
Division and Quotients
8
Example
f ( x ) = 8 x 7 – 6 x
4 + 5
Calculate the quotient
and remainder for f ( x ) ÷ 2 x.
4 x 6 – 3 x
3 r 52 x 8 x
7 – 6 x 4 + 5
5 326 r 2
quotient remaind
erIn any division, the part of the answer which has been divided is called the quotient.
cannot be divided by 2
xThe power of each term in the quotient is one less
than the power of the term in the original
polynomial.
Higher Maths 2 1 1 Polynomials
Investigating Polynomial Division
9
Example
f ( x) = (2 x2 + 5x – 1)( x – 3) + 4
= 2 x3 – x2 – 16 x + 7
f ( x) ÷ ( x – 3)
= 2 x2 + 5x – 1 r 4
alternatively we can write
quotient
Try evaluating f ( 3)… 3 2 -1 -16 7
2 5 -1 4
6 15 -3
When dividing f ( x) by ( x –
n),evaluating f ( n) in a table gives:• the coefficients of the quotient • the remainder
remainder
coefficients of quotient
remainder
Higher Maths 2 1 1 Polynomials
Synthetic Division
10
a b c dn
× n
+ + + +e+
× n × n × n
coefficients of quotient
remainder
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:
This is called Synthetic Division.
Higher Maths 2 1 1 Polynomials
= (3 x 3 – 6 x
2 + 10x – 19) with remainder 42
Examples of Synthetic Division
11
Exampleg( x ) = 3 x
4 – 2 x 2 + x + 4
Find the quotient and remainder for g( x ) ÷
( x + 2).-2 3 0 -2 1
3 -6 10 -19
-6 12 -20
4
42
38
Evaluate g ( -2) :
Missing terms have coefficient zero.
g( x ) ÷ ( x + 2)
Alternatively, g( x ) = (3 x 3 – 6 x
2 + 10x – 19)( x + 2) + 42
Higher Maths 2 1 1 Polynomials
The Factor Theorem
12
If a polynomial f ( x) can be divided exactly by a factor (
x – h) , then the remainder, given by f ( h), is zero.
ExampleShow that ( x – 4) is a factor of f ( x) = 2 x
4 – 9x 3 + 5 x
2 – 3
x – 44 2 -9 5 -3
4 -1 1 1
8 -4 4-4
0
4Evaluate f ( 4) :
( x – 4) is a
factor of f
( x)
zero remainde
r
f ( 4) = 0
f ( x) = 2 x 4 – 9x
3 + 5 x 2 – 3 x – 4
= ( x – 4)(4 x 3 – x
2 + x + 1 ) + 0
Higher Maths 2 1 1 Polynomials
Factorising with Synthetic Division
13
Factorise
Try evaluating f
( 3) :
± 1± 3± 5
±
15
Example
Evaluate f ( h) by syntheticdivision for every
factor h.
3 2 5 -28 -15
2 11 5 0
6 33 15
f ( x) = 2 x 3 + 5 x
2 – 28 x – 15
( x – 3)
f ( 3) = 0
is a factorIf f ( h) = 0
then ( x –
h)is a factor.
= ( x – 3 )( 2 x 2 + 11 x + 5 )
f ( x) = 2 x 3 + 5 x
2 – 28 x – 15
= ( x – 3 )( 2 x + 1 )( x + 5 )
Consider factors of the number term...
Factors of -15 :
zero!
Higher Maths 2 1 1 Polynomials
9p – 27
Finding Unknown Coefficients
14
( x + 3) is a factor of f ( x) = 2 x 4 + 6x
3 + px 2 + 4 x –
15
Example
Find the value of
p.Evaluate f (- 3) :- 3 2 6 p 4
2 0 p - 3p + 4
- 6 0 - 3p
-15
9p – 12
( x + 3) is a factor
f (- 3) = 0
9p – 27 = 0
9p = 27
p = 3 zero remainde
r
Higher Maths 2 1 1 Polynomials
dd
d
Finding Polynomial Functions from Graphs
15
The equation of a polynomial can be found from
its graph by considering the intercepts.
ba c
f (x) = k( x – a )( x – b )( x – c )Equation of a
Polynomial From a Graph
k can be found
by substituting( 0 , d )
with x-intercepts a , b and cf ( x)
Higher Maths 2 1 1 Polynomials
Finding Polynomial Functions from Graphs (continued)
16
Example
f ( x)
- 2
1 5
30Find the function
shown in the graph
opposite.
f (x) = k( x + 2)( x – 1)( x – 5)
f (0) = 30
k(0 + 2)(0 – 1)(0 – 5) = 30
10 k = 30
k = 3
f (x) = 3 ( x + 2)( x – 1)( x – 5)
= 3 x 3 – 12 x
2 – 21x + 30
Substitute k back into
original function and multiply out...
Higher Maths 2 1 1 Polynomials
Location of a Root
17
f ( x)
ba
f (a) > 0 f (b) < 0
A root of a polynomial function f ( x) lies between a
and b if :
and
or...
f ( x)
ba
f (a) < 0 f (b) > 0and
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.
root root
Higher Maths 2 1 1 Polynomials
Finding Approximate Roots
18
Example
Show that f ( x) has a
root between 1 and 2.
f ( x) = x
3 – 4 x
2 – 2 x + 7
f (1) = 2
f (2) = - 5
(above x-axis)
(below x-axis)
f ( x) crosses the x-
axis between 1 and
2.
f ( x)x root
between1 21 and 22 - 5
1 and 1.31.3 - 0.163
1.25 and 1.31. 25 0.203
1.25 and 1.281. 28 - 0.016
1.27 and 1.281. 27 0.0571.275 and 1.281. 275 0.020
The approximate root can be calculated by an iterative process:
1.2 and 1.31.2 0.568
The root is at approximately x
= 1.28
Higher Maths 2 1 1 Polynomials