11 x1 t15 04 polynomial theorems (2013)
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Polynomial Theorems
Polynomial TheoremsRemainder TheoremIf the polynomial P(x) is divided by (x – a), then the remainder is P(a)
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
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
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
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
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
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
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
e.g. Find the remainder when is divided by (x – 2)
3 25 17 11P x x x x
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
OR 3 19 30P x x x
OR 3 19 30P x x x ( 2) x
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
2x
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
2x
constant constant=constant
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
2x
constant constant=constant
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
2x
constant constant=constant
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
2x
constant constant=constant
15
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?
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
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?
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
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?
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
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
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
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
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
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
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
(ii) Factorise 3 24 16 9 36P x x x x
(ii) Factorise 3 24 16 9 36P x x x x
Constant factors must be a factor of the constant
(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
(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!!!
(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
(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
(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
(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
(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
(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
(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
(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
(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
(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
(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
(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
(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
(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
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?
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?
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?
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)?
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)?
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
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
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
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
2002 Extension 1 HSC Q2c)Suppose where Q(x) is a polynomial.
Find the value of a.
322 23 xQxaxx
2002 Extension 1 HSC Q2c)Suppose where Q(x) is a polynomial.
Find the value of a.
322 23 xQxaxx
32 P
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
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
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
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?
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.
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
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)
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)
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
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)
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)
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
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
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
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
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
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
Exercise 4D; 1bc, 3ac, 4ac, 7acf, 8bdf, 9b, 12ac, 13, 14, 17, 23*
2 11use 2
x
P
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