Chapter 6 Polynomial PYQ

CHAPTER 6 : POLYNOMIALS

LIM HWEE CHENG
KOLEJ MATRIKULASI JOHOR Jabatan Matematik

Chapter 6
Polynomials

Q1 : 2003 / 2004

Express + + as a sum of partial fraction.
( + )( + )

5 2+17 +17 = + +
( +2)( +1)2 +1 ( +1)2
+2

= ( +1)2+ ( +2)( +1)+ ( +2)
( +2)( +1)2

5 2 + 17 + 17 = ( + 1)2 + ( + 2)( + 1) + ( + 2)

When = −1

5 (−1)2 + 17(−1) + 17 = (0) + (0) + (−1 + 2)

= 5

When = −2
5 (−2)2 + 17(−2) + 17 = (−2 + 1)2 + (0) + (0)
20 − 34 + 17 =
= 3

When = 0
17 = + 2 + 2
17 = 3 + 2 + 2(5)
2 = 4
= 2

Lim Hwee Cheng pg. 1

Chapter 6
Polynomials

Therefore, 5 2+17 +17 = 3+ 2 + 5
( +2)( +1)2 +1 ( +1)2
+2

Q2 : 2003 / 2004

If ( − ) and ( + ) are factors of the expression − + + − ,
determine a and b. Hence, factorise the expression completely.

Let ( ) = − + + −

Given ( − ) is a factor of P(x), ∴ (1) = 0
(1) − (1) + (1) + (1) − = 0

+ = 14 − − − − − − − (1)

Given ( + 2) is another factor of P(x), ∴ (−2) = 0
(−2) − (−2) + (−2) + (−2) − = 0

4 − 2 = −100
2 − = −50 − − − − − − − (2)

Eq(1) + Eq(2)

3 = −36
= −12

From Eq(1) : = 14 − = 14 − (−12) = 26
Therefore ( ) = − − 12 2 + 26 −

Lim Hwee Cheng pg. 2

Chapter 6
Polynomials

Let the Divisor = ( − 1)( + 2) = 2 + − 2

2 + − 2 4 2 − 10 + 6

̅4̅ ̅ ̅2̅̅−̅̅̅6̅̅ ̅ ̅3̅̅−̅̅̅1̅2̅̅ ̅ ̅2̅̅+̅̅̅2̅6̅̅ ̅ ̅−̅̅̅1̅̅2̅
4 4 + 4 3 − 8 2

̅̅−̅̅1̅̅0̅ ̅ ̅3̅̅−̅̅̅4̅̅ ̅ ̅2̅̅+̅̅̅2̅6̅̅ ̅ ̅
−10 3 − 10 2 + 20

6̅̅ ̅ ̅2̅̅+̅̅̅6̅̅ ̅ ̅−̅̅̅1̅̅2̅
6 2 + 6 − 12
̅̅̅̅̅̅̅̅̅̅̅0̅̅̅̅̅̅̅̅̅̅̅̅

Therefore, ( ) = ( ) ( ) + ( ) = ( − 1)( + 2)(4 2 − 10 + 6)
= ( − 1)( + 2). 2 (2 2 − 5 + 3)
= 2( − 1)( + 2)(2 − 3)( − 1)
= 2(2 − 3)( + 2)( − 1)2

Q3: 2004 /2005

Given ( + ) is one factor of ( ) = − − − . Factorise completely ( ),

and express + as a sum of partial fraction
( )

Lim Hwee Cheng pg. 3

Chapter 6
Polynomials

Use the long division

−2 2 − 5 + 3

x+3 −2 3 − 11 2 − 12 + 9
−2 3 − 6 2
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅−̅̅5̅̅ ̅ ̅2̅̅−̅̅̅1̅2̅̅ ̅ ̅̅̅̅̅̅̅̅̅̅̅̅̅

−5 2 − 15
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅3̅̅ ̅ ̅+̅̅̅9̅̅̅

3 + 9
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅0̅

( ) = (−2 2 − 5 + 3)( + 3)
= (1 − 2 )( + 3)( + 3)
= (1 − 2 )( + 3)2

+ = + = + +
( − )( + ) ( + )
( ) − +

+ = ( + ) + ( − )( + )+ ( − )
( − )( + ) ( − )( + )

13 + 18 = ( + ) + ( − )( + ) + ( − )

When = 1 : 13 (1) + 18 = (1 + 2 + (0) + (0)
22
2 3)

49 = 49

24

= 2

When = −3 : 13(−3) + 18 = (0) + (0) + (1 − 2(−3))
−21 = 7
= −3

Lim Hwee Cheng pg. 4

Chapter 6
Polynomials

When = 0 : 13(0) + 18 = 9 + 3 +
18 = 9(2) + 3 + (−3)
3 = 3
B =1

Thus, + = + −
( − )( + ) ( + )
− +

Q4 : 2004 / 2005

A polynomial has the form ( ) = − − + , with x real and p, q constants.
When ( ) is divided by ( − ) the remainder is ( − ). Find the values of p and q, and
factorise ( ) completely if 2 is one of the roots.

Given ( ) = 2 3 − 3 2 − +
( ) = − ( )

2(1)3 − 3(1)2 − (1) + = 2 − 4
− + = −1 − − − − − −(1)

Given 2 is one of the roots. ∴ ( ) =
2(2)3 − 3(2)2 − (2) + = 0
−2 + = −4 − − − − − − − −(2)

Solve for Eq(1) - Eq (2) :
= 3

∴ = −1 + = −1 + 3 = 2

Therefore, ( ) = 2 3 − 3 2 − 3 + 2

Lim Hwee Cheng pg. 5

Chapter 6
Polynomials

2 2 + − 1
2 3 − 3 2 − 3 + 2
x - 2 2 3 − 4 2

2 − 3
2 − 2

− + 2
− + 2
̿̿̿̿̿̿̿…̿̿̿̿̿̿̿

Therefore ( ) = (2 2 + − 1)( − 2)
= (2 − 1)( + 1)( − 2)

Q5 : 2005 / 2006
Polynomial ( ) = + − + has ( + ) as a factor and leaves a remainder 12
when divided by ( − ). Determine the values of a and b .

Given ( ) = 2 3 + 2 − +

Since ( + 1) is a factor of ( ) , ∴ (−1) = 0
2(−1)3 + (−1)2 − (−1) + = 0
+ = 1 − − − − − − − (1)

Given (3) = 12
2(3)3 + (3)2 − (3) + = 12
9 + = −39 − − − − − − − −(2)

Eq(2) – Eq(1) : 8 = −40
= −5

From Eq(1) : = 1 − = 1 − (−5) = 6

Lim Hwee Cheng pg. 6

Chapter 6
Polynomials

Q6 : 2005 / 2006
Two factor of the polynomial ( ) = + + − are ( + ) and ( − ) .

Determine the values of a and b and find the third factor of the polynomial. Hence express

− − as a sum of partial fraction
( )

Given ( ) = 3 + 2 + − 6
Given ( + 1) is a factor of P(x) ∴ (−1) = 0

(−1) = −1 + − − 6 = 0
− = 7--------(1)

4 + 2 = −2 (2) = 8 + 4 + 2 − 6 = 0
2 + = −1--------(2)
when = 2
Eq (2) + Eq (1) 3 = 6 = −5
= 2

( ) = 3 + 2 2 − 5 − 6

x+3
x2 − x − 2 x3 + 2x2 − 5x − 6

x3 − x2 − 2x
3x2 − 3x − 6
3x3 − 3x − 6
0

The third factor is + 3

Lim Hwee Cheng pg. 7

Chapter 6
Polynomials

Q7 : 2006 / 2007

Find the values of A, B, C and D for the expression − + − in the form of the partial
+

fractions + + ++ where A, B, C and D are constants.


Given 4 3−3 2+6 −27 = + + +
4+9 2 2 2+9

4 3 − 3 2 + 6 − 27 = ( )( 2 + 9) + ( 2 + 9) + ( + ) 2

when = 0

−27 = 9

= −3

Compare the coefficient:

3: 4 = +

2: −3 = + ⇒ = −3 − = −3 − (−3) = 0

: 6 = 9 ⇒ = 2
3

= 4 − = 4 − 2 = 10
3 3

4 3−3 2+6 −27 = 2 − 3 + 10
4+9 2 3 2 3( 2+9)

Q8: 2006/2007
(a) Show that ( − ) is a factor of the polynomial ( ) = − − + . Hence,

factorize ( ) completely.
(b) If ( ) = + + leaves remainder 1, 25 and 1 on division by ( − ), ( + )

and ( − ) respectively, find the values of a, b and c. Hence, show that ( ) has two
equal real roots

(a) (3) = (3)3 − 2(3)2 − 5(3) + 6 = 27 − 18 − 15 + 6 = 0
( − 3) is a factor of ( )

Lim Hwee Cheng pg. 8

Chapter 6
Polynomials

2 + − 2
− 3 ̅ ̅3̅̅−̅̅̅̅2̅ ̅ ̅2̅̅−̅̅̅5̅̅ ̅ ̅̅+̅̅̅6̅

3 − 3 2

2 − 5
2 − 3

−2 + 6
−2 + 6

Thus,

( ) = ( − 3)( 2 + − 2)
= ( − 3)( − 1)( + 2)

(b) Given ( ) = 2 + +

(1) = 1
(1)2 + (1) + = 1
+ + = 1 − − − − − − − (1)

(−1) = 25
(−1)2 + (−1) + = 25
− + = 25 − − − − − − − (2)

(2) = 1
(2)2 + (2) + = 1

4 + 2 + = 1 − − − − − − − (3)

Lim Hwee Cheng pg. 9

Chapter 6
Polynomials

Eq(2) – Eq(1)
−2 = 24
= −12

Eq(3) – Eq(1)
3 + = 0
3 − 12 = 0
3 = 12
= 4

From Eq(1) :
4 − 12 + = 1
= 9

Therefore, ( ) = 4 2 − 12 + 9

Discriminant : 2 − 4 = (−12)2 − 4(4)(9)
= 144 − 144
=0

Therefore, ( ) has two equal roots.

Lim Hwee Cheng pg. 10

Chapter 6
Polynomials

Q9 : 2007 /2009

Express + in partial fractions
( + )( − + )

2 + 1 +
( + 2)( 2 − 2 + 4) ≡ + 2 + 2 − 2 + 4

= ( 2−2 +4)+ ( + )( +2)
( +2)( 2−2 +4)

2 + 1 = ( 2 − 2 + 4) + ( + )( + 2)

When = −2 Comparing the coefficient 2
−3 = 12 + = 0
= 1
= − 1
4
4

Comparing the coefficient 0
4 + 2 = 1
2 = 2
= 1

2 +1 ≡ −14 + 14 +1
( +2)( 2−2 +4) +2 2−2 +4

2 +1 ≡ − 1 + +4
( +2)( 2−2 +4) 4( +2) 4( 2−2 +4)

Lim Hwee Cheng pg. 11

Chapter 6
Polynomials

Q10 : 2007/2008
(a) Find a cubic polynomial ( ) = ( + )( + )( + ) satisfying the following

conditions: the coefficient of is 1, Q(-1) = 0 , Q(2) = 0 and Q(3) = −8
(b) A polynomial ( ) = − + + has a factor ( + ) and a remainder

( + ) when divided by ( + ). Find the value of a and b. Hence, factorise ( )
completely

(a) ( ) = ( + )( + )( + )
(−1) = 0 ⇒ + 1 is a factor
(2) = 0 ⇒ − 2 is a factor
( ) = ( + )( − )( + )

When (3) = −8
−8 = (3 + 1)(3 − 2)(3 + )
−8 = (4)(1)(3 + )
−2 = (3 + )
= −5

∴ ( ) = ( + 1)( − 2)( − 5)
∴ ( ) = ( 2 − − 2)( − 5)
∴ ( ) = 3 − 6 2 + 3 + 10

(b) ( ) = 3 − 4 2 + + 18

(−2) = 0 (−1) = 2(−1) + 18 = 16

−8 − 16 − 2 + 18 = 0 − − 4 − + 18 = 16

8 + 2 = 2 + = −2 -------- (2)

4 + = 1--------(1)

Lim Hwee Cheng pg. 12

Chapter 6
Polynomials

Eq(1) – Eq(2) :
3 = 3
= 1 , = −3

∴ ( ) = 3 − 4 2 − 3 + 18

x2 − 6x + 9
x + 2 x3 − 4x2 − 3x + 18

x3 + 2x2
− 6x2 − 3x
− 6x2 −12x
9x + 18
9x + 18

( ) = ( + 2)( 2 − 6 + 9) ⇒ ( ) = ( + 2)( − 3)2

Q11: 2008 / 2009

Express + + in partial fractions
( − )( + )

5 2 + 3 + 8 5 2 + 3 + 8
(1 − 2)(1 + ) = (1 − )(1 + )2 = (1 − ) + (1 + ) + (1 + )2

(1 + )2 + (1 − )(1 + ) + (2 − )
= (1 − )(1 + )2

5 2 + 3 + 8 = (1 + )2 + (1 − )(1 + ) + (1 −

Lim Hwee Cheng pg. 13

Chapter 6
Polynomials

When = 1

5(1)2 + 3(1) + 8 = (1 + 1) + (0) + (0)

16 = (4)
= 4

when = −1
5(−1)2 + 3(−1) + 8 = (0) + (0) + (1 − (−1))

10 = 2
= 5

when = 0
5(0)2 + 3(0) + 8 = + + C

8 = + +

8 = 4 + + 5
= −1

Therefore, 5 2+3 +8 = 4 − 1 + 5
(1− 2)(1+ ) (1− ) (1+ ) (1+ )2

Lim Hwee Cheng pg. 14

Chapter 6
Polynomials

Q12 : 2008 /2009

Polynomial ( ) = − + + can be divided exactly by ( − − ).
Find the values of m and n. Using these values of m and n, factorise the polynomial
completely. Hence, solve the equation − + + − = using the
polynomial ( )

Given ( ) = 3 − 8 2 + + 6
Since P(x) is divided exactly by D(x) = 2 − 2 − 3 = ( + 1)( − 3) , thus ( + 1) and ( − 3)
are the factors of P(x).

Since ( + 1) is one of the factor of P(x), ∴ (−1) = 0
(−1) = (−1)3 − 8(−1)2 + (−1) + 6 = 0

− − 8 − + 6 = 0

= − − 2 ------------------------- (1)

Since ( − 3) is one of the factor of P(x), ∴ (3) = 0
(3) = (3)3 − 8(3)2 + (3) + 6 = 0

27 − 72 + 3 + 6 = 0
27 + 3 = 66 − − − − − − − (2)

Substitute Eq(1) into Eq(2)

27(− − 2) + 3 = 66

−27 − 54 + 3 = 66

−24 = 120

= 120 = −5

−24

From Eq(1) : = −(−5) − 2 = 3

Therefore, ( ) = 3 3 − 8 2 − 5 + 6

Lim Hwee Cheng pg. 15

Chapter 6
Polynomials

3x − 2
x2 − 2x − 3 3x3 − 8x2 − 5x + 6

3x3 − 6x2 − 9x

−2 2 + 4 + 6

−2 2 + 4 + 6

0

( ) = 3 3 − 8 2 − 5 + 6
= ( + 1)( − 3)(3 − 2)

x−2
3x3 − 8x2 − 5x + 6 3x4 −14x3 +11x2 +16x −12

3x4 −8x3 + 5x2 + 6x
- 6x3+16x2 +10x −12
- 6x3+16x2 +10x −12
0

( ) = 3 4 − 14 3 + 11 2 + 16 − 12
= ( + 1)( − 3)(3 − 2)( − 2)

To solve P( x ) = 0

( + 1)( − 3)(3 − 2)( − 2) = 0
2

= −1 , = 3 , = 3 , = 2

Lim Hwee Cheng pg. 16

Chapter 6
Polynomials

Q13: 2009 / 2010

Express − in partial fractions.
( − )( + + )

4 − 3 +
( − 2)( 2 + 2 + 2) = ( − 2) + ( 2 + 2 + 2)

4 − 3 ( 2 + 2 + 2) + ( + )( − 2)
( − 2)( 2 + 2 + 2) =
( − 2)( 2 + 2 + 2)

4 − 3 = ( 2 + 2 + 2) + ( + )( − 2)

When = 2 :

4(2) − 3 = (10) + (0)

5 = 10
= 1

2

When = 0 :

4(0) − 3 = 2 − 2
1

−3 = 2 (2) − 2
= 2

When = 1

4(1) − 3 = (5) + ( + )(−1)
1

1 = 5 (2) − − 2
1

= − 2

Therefore, 4 −3 = 1 + −21 +2 = 1 + − +4
( −2)( 2+2 +2) 2 2+2 +2 2( −2) 2( 2+2 +2)

( −2)

Lim Hwee Cheng pg. 17

Chapter 6
Polynomials

Q14 : 2009 /2010

Given a polynomial ( ) = + + − has factors ( + ) and ( − ).
(a) Find the value of the constants and .
(b) Factorize ( ) completely.
(c) Obtain the solution set for ( ) <

(a) P(x) = 2x3 + ax2 + bx − 30
P(−2) = 2(−8) + a(4) + b(−2) − 30 = 0
−16 + 4a − 2b − 30 = 0
4a − 2b = 46
2a − b = 23........(1)

P(5) = 2(125) + 25a + 5b − 30 = 0
250 + 25a + 5b − 30 = 0
25a + 5b = −220
5a + b = −44.......... .(2)

Eq (1)+ Eq (2)

7a = −21
a = −3

2(−3) − b = 23
b = −29

(b) pg. 18
2x + 3

x2 − 3x −10 2x3 − 3x2 − 29x − 30

(-) 2x3 − 6x2 − 20x

Lim Hwee Cheng

Chapter 6
Polynomials

3x2 − 9x − 30
3x2 − 9x − 30

P(x) = (x + 2)(x − 5)(2x + 3)

(c) P(x) = (x + 2)(x − 5)(2x + 3)  0 choose negative region

x+ 3 −2 −3/2 5
x−5 −+ + +
2x + 3 −− − +
−− + +
−+ − +

Solution set : { : < −2 − 3 < < 5 }

2

Lim Hwee Cheng pg. 19

Chapter 6
Polynomials

Q15: 2010/2011

A polynomial ( ) = + ( + ) + ( + ) + has a factor ( + ).
(a) Express q in terms of p.
(b) Write ( ) in terms of p and x. Determine the quotient when ( ) is divided by
( + ).
(c) Hence, find the value of p if = is one of the roots for ( ) = . Using the
value of p, factorize ( ) completely.

f (x) = px3 + (p + q)x2 + (p + 2q)x +1

a) f (−1) = 0

(−1)3 p + (p + q)(−1)2 + (p + 2q)(−1)+1 = 0

− p + p + q − p − 2q +1 = 0
q =1− p

b) f (x) = px3 + (p + (1− p))x2 + (p + 2(1− p))x +1
= px3 + x2 + (2 − p)x +1
px2 + (1− p)x +1
x +1 px3 + x2 + (2 − p)x +1

(-) px3 + px2

(1− p)x2 + (2 − p)x
(1− p)x2 + (1− p)x

x +1
x +1

Quotient Q(x) = px2 + (1− p)x +1

Lim Hwee Cheng pg. 20

Chapter 6
Polynomials

c) f (3) = 0
p(3)3 + (3)2 + (2 − p)(3)+1 = 0

27 p + 9 + 6 − 3 p +1 = 0
24 p = −16
p=−2
3

To factorize completely,

f (x) = (x +1) − 2 x2 + 1 + 2 x + 1
3  3 

= (x +1)(x − 3) − 2 x − 1 

 3 3

= − 1 (x +1)(x − 3)(2x +1)

3

Q16: 2010 /2011

Dividing ( ) = + + by ( + ) and ( − ) give a remainder of −12 and −16.
Determine the values of and .

Given (−1) = −12
(−1)2 + (−1) + = −12
− + = −13 − − − − − − − (1)

Given (1) = −16
(1)2 + (1) + = −16
+ = −17 − − − − − − − −(2)

Solve Eq(1) + Eq(2) : 2 = −30

= −15

From Eq(2) : = −17 − (−15) = −2

Lim Hwee Cheng pg. 21

Chapter 6
Polynomials

Q17: 2011 / 2012

Express − in the form of partial fractions.
( − )

6 − 13
(3 − 4)2 = (3 − 4) + (3 − 4)2

= (3 −4)+
(3 −4)2

6 − 13 = (3 − 4) +

When x = 4
3

6 (4) − 13 = (0) +

3

= −5

When x = 0

6(0)−13 = −4 + (−5)
= 2

Therefore, 6x −13 = 2 − 5
(3x − 4)2 (3x − 4) (3x − 4)2

Lim Hwee Cheng pg. 22

Chapter 6
Polynomials

Q18: 2011/2012
The polynomial ( ) = − + + , where and are constants, has a factor
( − ) and leaves a remainder of when it is divided by ( − ).

(a) Find the values of and .
(b) Factorise ( ) completely by using the values of and obtained from part (a).

Hence, find the real roots of ( ), where and are not equal to zero.

(a)
Given ( ) = 3 − 2 2 + +

Given P(x) has a factor, ∴ p(2) = 0

8 − 8 + 2a + b = 0
2a + b = 0................(1)

Given ( ) = 3
3 − 2 2 + 2 + = 3
2 = . . . . . . . . . . . . . . . . . . . . . . . (2)

Substitute Eq(2) into Eq(1)

2 + 2 = 0
( + 2) = 0
= 0, = −2
= 0, = 4

When a =0 , b = 0, ∴ ( ) = 3 − 2 2
When = −2 , = 4 , ∴ ( ) = 3 − 2 2 − 2 + 4

(b) For a =0 , b = 0 ( ) = 3 − 2
= 2( − 2)

= ( )( )( − 2)

Lim Hwee Cheng pg. 23

For = −2 , = 4 ( ) = 3 − 2 2 − 2 + 4 Chapter 6
= ( − 2)( 2 − 2) Polynomials
= ( − 2)( + √2)( − √2)
LONG DIVISION

To find the roots, let
p(x) = 0
(x − 2)(x + 2)(x − 2) = 0
x = 2, x = − 2, x = 2

Q19: 2012 / 2013

Express − + − in the form of partial fractions.
− +

Since 2 3−7 2+17 −19 is improper fraction, use long division to transform fraction in standard
2 2−7 +6

form.

Therefore,

2x3 − 7x2 +17x −19 = x + 11x −19 6
2x2 − 7x +16 2x2 − 7x +

= x + 11x −19 2)

(2x − 3)(x −

Lim Hwee Cheng pg. 24

Chapter 6
Polynomials

11 −9 = +
2 2−7 +6 2 −3 −2

= ( −2)+ (2 −3)

(2 −3)( −2)

11 − 9 = ( − 2) + (2 − 3)

Solve for constant A & B :

= 2, = 3

= 0, −19 = −2 − 3
−19 = −2 − 9
−10 = −2
= 5

Therefore, x + 11x − 9 = x + 5 + 3
2x2 − 7x + 6 2x −3 x − 2

Lim Hwee Cheng pg. 25

Chapter 6
Polynomials

Q20: 2012 /2013
The polynomial ( ) = + + − has a factor ( − ) and a remainder 15
when divided by ( + ).

(a) Find the values of and
(b) Factorise ( ) completely and find all zeroes of ( )

(a) P(x) = 2x3 + ax2 + bx − 24

P(2) = 2(2)3 + 4a + 2b − 24 = 0
4a + 2b = 8
2a + b = 4........(1)
P(−3) = 2(−3)3 + a(−3)2 + b(−3) − 24 = 15
− 54 + 9a − 3b − 24 −15 = 0
9a − 3b = 93
3a − b = 31........(2)
(1) + (2)
a = 7,b = −10

b)
2x2 +11x +12

x − 2 2x3 + 7x2 −10x − 24

( )− 2x3 − 4x2

11x2 −10x

− (11x2 − 22x)

12x − 24

− (12x − 24)

0

To find all the zeros of P(x) , let P(x) = 0

Lim Hwee Cheng pg. 26

Chapter 6
Polynomials

(x − 2)(2x2 +11x +12)

(x − 2)(x + 4)(2x + 3)

x = 2,−4,− 3
4

The zeors : −4, 2 , −3
4

Q21: 2013 / 2014

Express in partial fractions form.
+ +

Since is improper fraction, use long division to transform fraction in standard form.
+ +

− −
+ + = + ( + )( + )

− −
( + )( + ) = + + +

( + ) + ( + )
= ( + )( + )

− − = ( + ) + ( + )

When = −2 ; − (− ) − = ( ) + (− + )
− = −
= −4

Lim Hwee Cheng pg. 27

Chapter 6
Polynomials

When = −1 ; −3(−1) − 2 = (−1 + 2) + (0)
3 − 2 =
= 1

Therefore, 2 = 1 + −
2+3 +2 + +

Q22 : 2013 / 2014

A cubic polynomial ( ) has a remainder 3 and 1 when divided by ( − ) and ( − )
respectively.

(a) Let ( ) be a linear factor such that ( ) = ( − )( − ) ( ) + +
, where and are constants. Find the remainder when ( ) is divided by
( − )( − ).

(b) Use the values of and from part (a) to determine ( ) if the coefficient of
for ( ) is 1 and ( ) = . Hence, solve for if ( ) = − .

(a)
Given (1) = 3 & (2) = 1 , ( ) is a linear factor of P(x).
Given ( ) = ( − 1)( − 2) ( ) + +

(1) = 0 + (1) +
3 = + − − − − − −(1)

(2) = 0 + (2) +
1 = 2 + − − − − − −(2)

Solve Eq(2) – Eq(1) : = −2
From Eq(1) : = 3 − (−2) = 5
Therefore, the remainder of P(x) is −2 + 5

Lim Hwee Cheng pg. 28

Chapter 6
Polynomials

(b) Let ( ) = +

( ) = ( − 1)( − 2)( + ) − 2 + 5
(3) = 7
(3) = (3 − 1)(3 − 2)(3 + ) − 2(3) + 5
7 = (2)(1)(3 + ) − 1
3 + = 4
= 1

Given the coefficient of x3 is 1 , ∴ ( ) = ( + 1)
Therefore, ( ) = ( − 1)( − 2)( + 1) − 2 + 5

To solve for ( ) = 7 − 3
( − 1)( − 2)( + 1) − 2 + 5 = 7 − 3
( − 1)( − 2)( + 1) + ( − 2) = 0
( − 2) [( − 1)( + 1) + 1)] = 0
( − 2)( 2 − 1 + 1) = 0
2 ( − 2) = 0
∴ = 0 , = 2

Q23 : 2014 / 2015

Given that ( − ) is a factor of a polynomial ( ) = − + − where and
are real numbers. If ( ) is divided by ( + ) the remainder is −24, find the values of
and . Hence, find the remainder when ( ) is divided by ( + ).

( ) = 3 − 10 2 + − 2
(2) = 0
(2)3 − 10(2)2 + (2) − 2 = 0

8 + 2 = 42
4 + = 21 ---------- (1)

Lim Hwee Cheng pg. 29

Chapter 6
Polynomials

(−1) = −24
(−1)3 − 10(−1)2 + (−1) − 2 = −24

− − 10 − − 2 = −24
− − = −12 − − − − − − − (2)

Eq(1) + Eq(2) : 3 = 9
From Eq(2) : = 3

= 12 − 3 = 9

( ) = 3 3 − 10 2 + 9 − 2

1 13 −1 2 −1
(− 2) = 3 (− 2) − 10 ( 2 ) + 9 ( 2 ) − 2

= −3 − 10 − 9 − 2

8 42

= − 75

8

Q24 : 2014 / 2015

Expand ( + )( + ) , and are real numbers with > . Hence, find the values of

and if ( + )( + ) = − − . Express − + − in the form of partial
− −

fractions.

( + )( + )2 = ( + )( 2 + 2 + 2)
= 3 + 2 2 + 2 + 2 + 2 + 2
= 3 + (2 + ) 2 + ( 2 + 2 ) + 2

Compare 3 − 3 − 2 = 3 + (2 + ) 2 + ( 2 + 2 ) + 2

x2 : 2 + = 0
= −2 − − − − − − − − − −(1)

Lim Hwee Cheng pg. 30

Chapter 6
Polynomials

xo : 2 = −2

−2 = (−2 )( 2)

3 = 1

= 1

4−4 2+5 −1 is improper faction.
3−3 −2

4 − 4 2 + 5 − 1 (− 2 + 7 − 1)
= + 3 − 3 − 2
3 − 3 − 2

= + (− 2+7 −1)
( −2)( +1)2

(− 2 + 7 − 1)
( − 2)( + 1)2 = − 2 + + 1 + ( + 1)2

= ( +1)2+ ( −2)( +1)+ ( −2)
( −2)( +1)2

− 2 + 7 − 1 = ( + 1)2 + ( − 2)( + 1) + ( − 2)

When = 2 ; −(2)2 + 7(2) − 1 = (2 + 1)2

9 = 9

= 1

When = −1 ; −(−1)2 + 7(−1) − 1 = (−1 − 2)

−9 = −3

= 3

When = 0 ; −(0)2 + 7(0) − 1 = − 2 − 2

−1 = 1 − 2 − 6

= −2

Thus, − + − = + 1 − 2 + 3
− − −2 +1 ( +1)2

Lim Hwee Cheng pg. 31

Chapter 6
Polynomials

Q25 : 2015/2016

Express + + in the form of partial fractions.
( − )( + )

5 2 + 4 + 4 5 2 + 4 + 4
( 2 − 4)( + 2) = ( − 2)( + 2)( + 2)

5 2 + 4 + 4
= ( − 2)( + 2)2

Thus

5 2 + 4 + 4
( − 2)( + 2)2 = − 2 + ( + 2) + ( + 2)2

= ( +2)2+ ( −2)( +2)+ ( −2)
( −2)( +2)2

5 2 + 4 + 4 = ( + 2)2 + ( − 2)( + 2) + ( − 2)

When = 2 : 5(2)2 + 4(2) + 4 = (2 + 2)2 + (0) + (0)

32 = 16

= 2

When = −2 : 5(−2)2 + 4(−2) + 4 = (0) + (0) + (−2 − 2)

16 = −4

= −4

When = 0 : 5(0)2 + 4(0) + 4 = 4 − 4 − 2

4 = 4(2) − 4 − 2(−4)

4 = 12

= 3

Therefore,

5 2 + 4 + 4 23 4
( − 2)( + 2)2 = − 2 + ( + 2) − ( + 2)2

Lim Hwee Cheng pg. 32

Chapter 6
Polynomials

Q26: 2015/2016
A polynomial ( ) = + + − + where , and are constants, has
factors ( + ) and ( − ). When ( ) is divided by ( + ), the remainder is 8. Find the
values of , and . Hence, factorise ( ) completely and state its zeroes.

Given ( + 2) is a factor, (−2) = 0
2(−4)4 + (−2)3 + (−2)2 − 17(−2) + = 0
−8 + 4 + = −66 − − − − − − − −(1)

Given ( − 1) is a factor, (1) = 0
2(1)4 + (1)3 + (1)2 − 17(1) + = 0
+ + = 15 − − − − − − − −(2)

When divided by ( + 1), Remainder is 8 , (−1) = 8
2(−1)4 + (−1)3 + (−1)2 − 17(−1) + = 8
− + + = −11 − − − − − − − − − (3)

Eq(2) − Eq(3) : 2 = 26
= 13

Eq(2) − Eq(1) : 9 − 3 = 81
9(13) − 3 = 81
3 = 36
= 12

From Eq(2) : 13 + 12 + = 15

= −10

Therefore, ( ) = 2 4 + 13 3 + 12 2 − 17 − 10

Lim Hwee Cheng pg. 33

Chapter 6
Polynomials

Long division:

2 + − 2 2 2 + 11 + 5

2 4 + 13 3 + 12 2 − 17 − 10
(2 4 + 2 3 − 4 2 )

11 3 + 16 2 − 17
(11 3 + 11 2 − 22 )

5 2 + 5 − 10
5 2 + 5 − 10

Thus, P(x) = (2 2 + 11 + 5) ( 2 + − 2)
= (2 + 1)( + 5)( + 2)( − 1)

When P(x) = 0 , (2 + 1)( + 5)( + 2)( − 1) = 0

= −1 , = −5 , = −2 & =1

2

The zeros of P(x) are : −5, −2, − 1 , 1

2

Q27: 2016 / 2017
Express − − in partial fractions form.

2 is a improper fraction, use LONG DIVISION to write in the standard form.
2−2 −3

2 − 2 − 3 1
̅̅̅̅ ̅2̅̅+̅̅̅0̅̅ ̅ ̅+̅̅̅0̅̅̅

2 − 2 − 3

2̅̅ ̅ ̅̅+̅̅̅3̅

2 2 + 3
2 − 2 − 3 = 1 + 2 − 2 − 3

= 1 + 2 +3
( −3)( +1) Express this in partial
fraction

Lim Hwee Cheng pg. 34

Chapter 6
Polynomials

2 + 3
( − 3)( + 1) = − 3 + + 1

= ( +1)+ ( −3)

( −3)( +1)

2 + 3 = ( + 1) + ( − 3)

When = −1 : 2(−1) + 3 = (0) + (−1 − 3)
1 = −4
= − 1

4

When = 3 ∶ 2(3) + 3 = (3 + 1) + (0)
9 = 4
= 9

4

Therefore, 2 = 1+ 9 − 1
2−2 −3
4 ( −3) 4( +1)

Q28: 2016 / 2017

(a) Polynomial ( ) has a remainder of 3 when divided by ( + ). Find the remainder of
( ) + when divided by ( + ).

(b) Polynomial ( ) = + − − has a factor ( − ) and remainder when
divided by ( + ), while a polynomial ( ) = − + + has a remainder
when divided by ( − ). Find the value of the constants a and b if + = .
Hence, obtain the zeroes for ( )

(a) ( ) = ( ) ( ) + ( )
= ( )( + 3) + 3

( ) + 2 = ( )( + 3) + 3 + 2
= ( )( + 3) + 5

Lim Hwee Cheng pg. 35

Chapter 6
Polynomials

(b) ( ) = + − −
Given 1(1) = 0 , (1)3 + (1)2 − 5 (1) − 7 = 0
− 5 = 6 − − − − − −(1)

Given 2(−1) = 1 , (−1)3 + (−1)2 − 5 (−1) − 7 = 1
+ 5 − 8 = 1 − − − − − −(2)

( ) = − + +
Given 2(1) = 2 , (1)3 − (1)2 + (1) + 6 = 2

− + + 7 = 2 − − − − − − − −(3)

Eq(2) + Eq(3) : 6 − 1 = 1 + 2 , Given + = .
6 − 1 = 5
6 = 6
= 1

From Eq(1) : = 6 + 5(1) = 11

Thus, ( ) = + − −

− + +

̅̅ ̅ ̅ ̅̅+̅̅̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅̅−̅̅̅ ̅ ̅ ̅ ̅̅−̅̅̅ ̅
−

−
−

−
−

Lim Hwee Cheng pg. 36

Chapter 6
Polynomials

1( ) = ( − 1)( 2 + 12 + 7)
Use quadratic formula for 2 + 12 + 7

= −12±√122 −4(1)(7) = −12 ±√116 = −6 ± √29
2(1) 2

When 1( ) = 0
= 1 , = −6 + √29 , = −6 − √29 { roots }

Thus the zeros of 1( ) ∶ 1, −6 − √29 , − 6 + √29

Q29: 2017 / 2018

Express ( − ) ( − + )in partial fractions

− +
( − )( + ) = − + +

( + ) + ( + )( − )
= ( − )( + )
3 2 − 5 = ( 2 + 2 ) + ( + )( − 3)

When = 3 ∶ 3(3)2 − 5 = (32 + 2) + 0
22 = 11
= 2

When = 0 ∶ 3(0)2 − 5 = 2 − 3
−5 = 2(2) − 3
= 3

Lim Hwee Cheng pg. 37

Chapter 6
Polynomials

When = 1 ∶ 3(1)2 − 5 = (3) + ( + )(−2)
−2 = 3 − 2 − 2
−2 = 3(2) − 2 − 2(3)
= 1

Thus , − = + +
( − )( + ) +
−

Q30 : 2017/2018
Given the polynomial ( ) = − and ( ) = + + + + .

(a) Find all zeroes of ( )
(b) When ( ) is divided by ( ) , the remainder is + . Use the

remainder theorem to find the value of and .
(c) Using the values of and obtained from part (b), find the remainder when

( ) + is divided by ( ).

(a) ( ) = −
= ( − 2 )( + 2)

The zeroes of ( ) are −2 and 2

(b) ( ) = ( − 2)( + 2) ( ) + (14 + 52)

(2) = 14(2) + 52
(2)4 + (2)3 + 2(2)2 + (2) + 28 = 80
16 + 2 = 36

8 + = 18 − − − − − −(1)

(−2) = 14(−2) + 52
(−2)4 + (−2)3 + 2(−2)2 + (−2) + 28 = 80

Lim Hwee Cheng pg. 38

Chapter 6
Polynomials

16 − 2 = −4
8 − = −2 − − − − − − − (2)

Eq(1) + Eq(2) : 16 = 16
= 1

From Eq(1) : 8 + = 18
= 10

(c)
( ) = 4 + 3 + 2 2 + 10 + 28.
2 ( ) + = 2 4 + 2 3 + 4 2 + 20 + 56 +
= 2 4 + 2 3 + 4 2 + 21 + 56

2 − 4 2 2 + 2 + 12

2 4 + 2 3 + 4 2 + 21 + 56
2 4 + 0 3 − 8 2

2 3 + 12 2 + 21
2 3 + 0 2 − 8

12 2 + 29 + 56

12 2 − 48

29 + 104

The remainder is 29 + 104

Lim Hwee Cheng pg. 39

Chapter 6
Polynomials

Q31 : 2018 / 2019

The polynomial ( ) = 4 + 3 − 7 2 − 4 + has a factor ( + 3) and the remainder 60
when divided by ( − 3). Find the values of a and b. Hence, factorise ( ) completely.

Given ( ) = 4 + 3 − 7 2 − 4 +
(−3) = 0
(−3)4 + (−3)3 − 7(−3)2 − 4 (−3) + = 0
81 − 27 − 63 + 12 + = 0
15 − = 18 − − − − − −(1)

(3) = 60
(3)4 + (3)3 − 7(3)2 − 4 (3) + = 0

81 + 27 − 63 − 12 + = 60
15 + = 42 − − − − − −(1)

(2) − (1) ∶ 2 = 24
= 12

From Eq(1) : 15 − 12 = 18
= 2

(c)

Lim Hwee Cheng ( ) = ( + 3)( 3 − 2 − 4 + 4)
= ( + 3)( 2 ( − 1) − 4( − 1))
= ( + 3)( − 1)( 2 − 4)
= ( + 3)( − 1)( − 2)( + 2)

pg. 40

Chapter 6
Polynomials

Q32 : 2019 /2020
Polynomial ( ) = − − + is divisible by − + . Find the values of
p and q. Hence, factorise ( ) completely.

Divisible

Lim Hwee Cheng pg. 41