11 x1 t15 04 polynomial theorems (2013)

102
Polynomial Theorems

Upload: nigel-simmons

Post on 15-Jun-2015

553 views

Category:

Education


2 download

TRANSCRIPT

Page 1: 11 x1 t15 04 polynomial theorems (2013)

Polynomial Theorems

Page 2: 11 x1 t15 04 polynomial theorems (2013)

Polynomial TheoremsRemainder TheoremIf the polynomial P(x) is divided by (x – a), then the remainder is P(a)

Page 3: 11 x1 t15 04 polynomial theorems (2013)

Polynomial TheoremsRemainder TheoremIf the polynomial P(x) is divided by (x – a), then the remainder is P(a)

Proof: ( ) ( ) ( )P x A x Q x R x

Page 4: 11 x1 t15 04 polynomial theorems (2013)

Polynomial TheoremsRemainder TheoremIf the polynomial P(x) is divided by (x – a), then the remainder is P(a)

Proof: ( ) ( ) ( )P x A x Q x R x

let ( ) ( )A x x a

Page 5: 11 x1 t15 04 polynomial theorems (2013)

Polynomial TheoremsRemainder TheoremIf the polynomial P(x) is divided by (x – a), then the remainder is P(a)

Proof: ( ) ( ) ( )P x A x Q x R x

let ( ) ( )A x x a

( ) ( ) ( )P x x a Q x R x

Page 6: 11 x1 t15 04 polynomial theorems (2013)

Polynomial TheoremsRemainder TheoremIf the polynomial P(x) is divided by (x – a), then the remainder is P(a)

Proof: ( ) ( ) ( )P x A x Q x R x

let ( ) ( )A x x a

( ) ( ) ( )P x x a Q x R x

( ) ( ) ( )P a a a Q a R a

Page 7: 11 x1 t15 04 polynomial theorems (2013)

Polynomial TheoremsRemainder TheoremIf the polynomial P(x) is divided by (x – a), then the remainder is P(a)

Proof: ( ) ( ) ( )P x A x Q x R x

let ( ) ( )A x x a

( ) ( ) ( )P x x a Q x R x

( ) ( ) ( )P a a a Q a R a

( )R a

Page 8: 11 x1 t15 04 polynomial theorems (2013)

Polynomial TheoremsRemainder TheoremIf the polynomial P(x) is divided by (x – a), then the remainder is P(a)

Proof: ( ) ( ) ( )P x A x Q x R x

let ( ) ( )A x x a

( ) ( ) ( )P x x a Q x R x

( ) ( ) ( )P a a a Q a R a

( )R anow degree ( ) 1

( ) is a constantR x

R x

Page 9: 11 x1 t15 04 polynomial theorems (2013)

Polynomial TheoremsRemainder TheoremIf the polynomial P(x) is divided by (x – a), then the remainder is P(a)

Proof: ( ) ( ) ( )P x A x Q x R x

let ( ) ( )A x x a

( ) ( ) ( )P x x a Q x R x

( ) ( ) ( )P a a a Q a R a

( )R anow degree ( ) 1

( ) is a constantR x

R x

( ) ( )

( )R x R a

P a

Page 10: 11 x1 t15 04 polynomial theorems (2013)

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

Page 11: 11 x1 t15 04 polynomial theorems (2013)

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

Page 12: 11 x1 t15 04 polynomial theorems (2013)

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P

Page 13: 11 x1 t15 04 polynomial theorems (2013)

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

Page 14: 11 x1 t15 04 polynomial theorems (2013)

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

Page 15: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

Page 16: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

Page 17: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P

Page 18: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

Page 19: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

Page 20: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

Page 21: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

Page 22: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x

Page 23: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x22x

Page 24: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x22x 19 30x

Page 25: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x

2x

22x 19 30x

Page 26: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x

2x

22x 19 30x 22 4x x

Page 27: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x

2x

22x

15x

19 30x 22 4x x

Page 28: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x

2x

22x

15x

19 30x 22 4x x

30

Page 29: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x

2x

22x

15x

19 30x 22 4x x

30

15

Page 30: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x

2x

22x

15x

19 30x 22 4x x

30

15

15 30x

Page 31: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x

2x

22x

15x

19 30x 22 4x x

30

15

15 30x 0

Page 32: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x

2x

22x

15x

19 30x 22 4x x

30

15

15 30x 0

2( ) ( 2) 2 15P x x x x

Page 33: 11 x1 t15 04 polynomial theorems (2013)

Factor Theorem

If (x – a) is a factor of P(x) then P(a) = 0

e.g. Find the remainder when is divided by (x – 2)

3 25 17 11P x x x x

3 25 17 11P x x x x

3 22 5 2 17 2 2 11P 19

remainder when ( ) is divided by ( 2) is 19P x x

e.g. (i) Show that (x – 2) is a factor of and hence factorise P(x).

3 19 30P x x x

32 2 19 2 30P 0

( 2) is a factorx

3 22 0 19 30x x x x

2x

3 22x x

2x

22x

15x

19 30x 22 4x x

30

15

15 30x 0

2( ) ( 2) 2 15P x x x x ( 2)( 5)( 3)x x x

Page 34: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x

Page 35: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

Page 36: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

Page 37: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

Page 38: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

Page 39: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

Page 40: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

Page 41: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

Page 42: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

Page 43: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

Page 44: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have?

Page 45: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have? 15x

Page 46: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have? 15x

How many x do you need?

Page 47: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have? 15x

How many x do you need? 19x

Page 48: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have? 15x

How many x do you need? 19x

How do you get from what you have to what you need?

Page 49: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have? 15x

How many x do you need? 19x

How do you get from what you have to what you need? 4x

Page 50: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have? 15x

How many x do you need? 19x

How do you get from what you have to what you need? 4x

4 2 ?x

Page 51: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have? 15x

How many x do you need? 19x

How do you get from what you have to what you need? 4x

4 2 ?x

Page 52: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have? 15x

How many x do you need? 19x

How do you get from what you have to what you need? 4x

4 2 ?x

2x

2( ) ( 2) 2 15P x x x x ( 2)( 5)( 3)x x x

Page 53: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have? 15x

How many x do you need? 19x

How do you get from what you have to what you need? 4x

4 2 ?x

2x

Page 54: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have? 15x

How many x do you need? 19x

How do you get from what you have to what you need? 4x

4 2 ?x

2x

2( ) ( 2) 2 15P x x x x

Page 55: 11 x1 t15 04 polynomial theorems (2013)

OR 3 19 30P x x x ( 2) x

leading term leading term=leading term

2x

constant constant=constant

15

If you where to expand out now, how many x would you have? 15x

How many x do you need? 19x

How do you get from what you have to what you need? 4x

4 2 ?x

2x

2( ) ( 2) 2 15P x x x x ( 2)( 5)( 3)x x x

Page 56: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Page 57: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Page 58: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

Page 59: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Page 60: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the form

Page 61: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

Page 62: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 4

Page 63: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 4

Page 64: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 41 2 3 4 6 9 12 18 36= , , , , , , , ,2 2 2 2 2 2 2 2 2

Page 65: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 41 2 3 4 6 9 12 18 36= , , , , , , , ,2 2 2 2 2 2 2 2 2

Page 66: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 41 2 3 4 6 9 12 18 36= , , , , , , , ,2 2 2 2 2 2 2 2 2they could be negative too

Page 67: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 41 2 3 4 6 9 12 18 36= , , , , , , , ,2 2 2 2 2 2 2 2 2they could be negative too

3 24 4 4 16 4 9 4 36P

Page 68: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 41 2 3 4 6 9 12 18 36= , , , , , , , ,2 2 2 2 2 2 2 2 2they could be negative too

3 24 4 4 16 4 9 4 36P 0

Page 69: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 41 2 3 4 6 9 12 18 36= , , , , , , , ,2 2 2 2 2 2 2 2 2they could be negative too

3 24 4 4 16 4 9 4 36P 0

( 4) is a factorx

Page 70: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 41 2 3 4 6 9 12 18 36= , , , , , , , ,2 2 2 2 2 2 2 2 2they could be negative too

3 24 4 4 16 4 9 4 36P 0

( 4) is a factorx

3 2( ) 4 16 9 364

P x x x xx

Page 71: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 41 2 3 4 6 9 12 18 36= , , , , , , , ,2 2 2 2 2 2 2 2 2they could be negative too

3 24 4 4 16 4 9 4 36P 0

( 4) is a factorx

3 2( ) 4 16 9 364

P x x x xx

24x

Page 72: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 41 2 3 4 6 9 12 18 36= , , , , , , , ,2 2 2 2 2 2 2 2 2they could be negative too

3 24 4 4 16 4 9 4 36P 0

( 4) is a factorx

3 2( ) 4 16 9 364

P x x x xx

24x 9

Page 73: 11 x1 t15 04 polynomial theorems (2013)

(ii) Factorise 3 24 16 9 36P x x x x

Constant factors must be a factor of the constant

Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

of course they could be negative!!!

Fractional factors must be of the formfactors of the constant

factors of the leading coefficient

1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , ,4 4 4 4 4 4 4 4 41 2 3 4 6 9 12 18 36= , , , , , , , ,2 2 2 2 2 2 2 2 2they could be negative too

3 24 4 4 16 4 9 4 36P 0

( 4) is a factorx

3 2( ) 4 16 9 364

P x x x xx

24x 9 4 2 3 2 3x x x

Page 74: 11 x1 t15 04 polynomial theorems (2013)

2004 Extension 1 HSC Q3b)Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers.

When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1. (i) What is the value of b?

Page 75: 11 x1 t15 04 polynomial theorems (2013)

2004 Extension 1 HSC Q3b)Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers.

When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1.

111 P

(i) What is the value of b?

Page 76: 11 x1 t15 04 polynomial theorems (2013)

2004 Extension 1 HSC Q3b)Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers.

When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1.

111 P

11b

(i) What is the value of b?

Page 77: 11 x1 t15 04 polynomial theorems (2013)

2004 Extension 1 HSC Q3b)Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers.

When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1.

111 P

11b

(i) What is the value of b?

(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?

Page 78: 11 x1 t15 04 polynomial theorems (2013)

2004 Extension 1 HSC Q3b)Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers.

When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1.

111 P

11b

13 P

(i) What is the value of b?

(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?

Page 79: 11 x1 t15 04 polynomial theorems (2013)

2004 Extension 1 HSC Q3b)Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers.

When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1.

111 P

11b

13 P

(i) What is the value of b?

(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?

14 ba

Page 80: 11 x1 t15 04 polynomial theorems (2013)

2004 Extension 1 HSC Q3b)Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers.

When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1.

111 P

11b

13 P

(i) What is the value of b?

(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?

14 ba

3124

aa

Page 81: 11 x1 t15 04 polynomial theorems (2013)

2004 Extension 1 HSC Q3b)Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers.

When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1.

111 P

11b

13 P

(i) What is the value of b?

(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?

14 ba

3124

aa

8331 xxQxxxP

Page 82: 11 x1 t15 04 polynomial theorems (2013)

2004 Extension 1 HSC Q3b)Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers.

When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1.

111 P

11b

13 P

(i) What is the value of b?

(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?

14 ba

3124

aa

8331 xxQxxxP

83 xxR

Page 83: 11 x1 t15 04 polynomial theorems (2013)

2002 Extension 1 HSC Q2c)Suppose where Q(x) is a polynomial.

Find the value of a.

322 23 xQxaxx

Page 84: 11 x1 t15 04 polynomial theorems (2013)

2002 Extension 1 HSC Q2c)Suppose where Q(x) is a polynomial.

Find the value of a.

322 23 xQxaxx

32 P

Page 85: 11 x1 t15 04 polynomial theorems (2013)

2002 Extension 1 HSC Q2c)Suppose where Q(x) is a polynomial.

Find the value of a.

322 23 xQxaxx

32 P

3222 23 a

Page 86: 11 x1 t15 04 polynomial theorems (2013)

2002 Extension 1 HSC Q2c)Suppose where Q(x) is a polynomial.

Find the value of a.

322 23 xQxaxx

32 P

3222 23 a

316 a

Page 87: 11 x1 t15 04 polynomial theorems (2013)

2002 Extension 1 HSC Q2c)Suppose where Q(x) is a polynomial.

Find the value of a.

322 23 xQxaxx

32 P

3222 23 a

316 a19a

Page 88: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

(i) Why is the most general form of R(x) given by R(x) = ax + b?

Page 89: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

Page 90: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

Page 91: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

P(x) = (x + 1)(x – 4)Q(x) + R(x)

Page 92: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4)

Page 93: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5

Page 94: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5

(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)

Page 95: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

54 R

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5

(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)

Page 96: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

54 R

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5

(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)

54 ba

Page 97: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

54 R

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5

(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)

54 ba 51 P

Page 98: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

54 R

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5

(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)

54 ba 51 P

5 ba

Page 99: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

54 R

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5

(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)

54 ba 51 P

5 ba

2 105

aa

Page 100: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

54 R

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5

(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)

54 ba 51 P

5 ba

2 105

aa

3b

Page 101: 11 x1 t15 04 polynomial theorems (2013)

1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x).

54 R

(i) Why is the most general form of R(x) given by R(x) = ax + b?

The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.

(ii) Given that P(4) = – 5 , show that R(4) = – 5

P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5

(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)

54 ba 51 P

5 ba

2 105

aa

3b 32 xxR

Page 102: 11 x1 t15 04 polynomial theorems (2013)

Exercise 4D; 1bc, 3ac, 4ac, 7acf, 8bdf, 9b, 12ac, 13, 14, 17, 23*

2 11use 2

x

P