11X1 T13 04 polynomial theorems
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Transcript of 11X1 T13 04 polynomial theorems
![Page 1: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/1.jpg)
Polynomial Theorems
![Page 2: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/2.jpg)
Polynomial TheoremsRemainder TheoremIf the polynomial P(x) is divided by (x – a), then the remainder is P(a)
![Page 3: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/3.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/4.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/5.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/6.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/7.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/8.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/9.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/10.jpg)
e.g. Find the remainder when is divided by (x – 2)
3 25 17 11P x x x x
![Page 11: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/11.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/12.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/13.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/14.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/15.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/16.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/17.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/18.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/19.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/20.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/21.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/22.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/23.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/24.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/25.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/26.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/27.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/28.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/29.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/30.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/31.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/32.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/33.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/34.jpg)
OR 3 19 30P x x x
![Page 35: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/35.jpg)
OR 3 19 30P x x x ( 2) x
![Page 36: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/36.jpg)
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
![Page 37: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/37.jpg)
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
![Page 38: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/38.jpg)
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
![Page 39: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/39.jpg)
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
2x
![Page 40: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/40.jpg)
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
2x
constant constant=constant
![Page 41: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/41.jpg)
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
2x
constant constant=constant
![Page 42: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/42.jpg)
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
2x
constant constant=constant
![Page 43: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/43.jpg)
OR 3 19 30P x x x ( 2) x
leading term leading term=leading term
2x
constant constant=constant
15
![Page 44: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/44.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/45.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/46.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/47.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/48.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/49.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/50.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/51.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/52.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/53.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/54.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/55.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/56.jpg)
(ii) Factorise 3 24 16 9 36P x x x x
![Page 57: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/57.jpg)
(ii) Factorise 3 24 16 9 36P x x x x
Constant factors must be a factor of the constant
![Page 58: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/58.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/59.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/60.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/61.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/62.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/63.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/64.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/65.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/66.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/67.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/68.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/69.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/70.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/71.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/72.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/73.jpg)
(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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/74.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/75.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/76.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/77.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/78.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/79.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/80.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/81.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/82.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/83.jpg)
2002 Extension 1 HSC Q2c)Suppose where Q(x) is a polynomial.
Find the value of a.
322 23 xQxaxx
![Page 84: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/84.jpg)
2002 Extension 1 HSC Q2c)Suppose where Q(x) is a polynomial.
Find the value of a.
322 23 xQxaxx
32 P
![Page 85: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/85.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/86.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/87.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/88.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/89.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/90.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/91.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/92.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/93.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/94.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/95.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/96.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/97.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/98.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/99.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/100.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/101.jpg)
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: 11X1 T13 04 polynomial theorems](https://reader036.fdocuments.in/reader036/viewer/2022081400/5559c7e3d8b42a93208b4575/html5/thumbnails/102.jpg)
Exercise 4D; 1bc, 3ac, 4ac, 7acf, 8bdf, 9b, 12ac, 13, 14, 17, 23*
2 11use 2
x
P