Higher Maths 2.1.1 - Polynomials

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.


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


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


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.


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)


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.


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


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.


= (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


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


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!


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


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)


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...


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


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


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Higher Maths 2.1.1 - Polynomials