4ach03(more About Polynomials)

Certificate Mathematics in Action Full Solutions 4A

More about Polynomials

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Activity

4.

∵

Activity 3.1 (p. 143)

2.

=0

∴ ∵

f (3) =33 −2(3) 2 −5(3) +6 =27 −18 −15 +6 =0

1.

(a)

= −1 −2 +5 +6

Yes. Since f (3) = 0 , i.e. when f(x) is

=8 ≠0

∴

factor of x 3 − 2 x 2 − 5 x + 6 . ∵

≠0

∴ ∵

f ( −2) = ( −1)3 −2( −1) 2 −5( −1) +6 = −1 −2 +5 +6 =8 ≠0 f ( −2) = 0 , therefore x + 2 is a factor of x 3 − 2 x 2 − 5 x + 6 ; f ( −1) = 8 ,

=0

∴

x + 2 is a factor of f(x).

∵

f (3) =33 −2(3) 2 −5(3) +6 = 27 −18 −15 +6 =0

x3 − 2 x 2 − 5 x + 6 .

∴ ∵

Activity 3.2 (p. 149)

= −24

= x 2 + ax 2 + bx 2 + abx + cx + ac = x 3 + ( a + b) x 2 + (ab + c) x + ac

∴

≠0 x +3 is not a factor of f(x). f (6) = 6 3 −2(6) 2 −5(6) +6

By comparing the coefficients,

ac = 6

∵

∴ The product of a and c is 6.

∴ ∵

3. √

x −1

√

x +1

√

x −2

√

x +2

√

x −3

√

x +3

x −4

x +4

x −5

x +5

= 216 −72 −30 +6 =120

The possible values of a are ± 1, ± 2, ± 3, ± 6. a is a factor of the constant term of f(x).

√

x −3 is a factor of f(x).

f ( −3) = ( −3)3 −2( −3) 2 −5( −3) +6 = −27 −18 +15 +6

f ( x) = ( x + a)( x 2 + bx + c )

x −6

x − 2 is not a factor of f(x).

f ( −2) =( −2) 3 −2( −2) 2 −5( −2) +6 =−8 −8 +10 +6

therefore x +1 is not a factor of

≠0 x −6 is not a factor of f(x).

f ( −6) = ( −6)3 −2(−6) 2 −5(−6) +6 = −216 −72 +30 +6 = −252

x +6 5.

59

= 8 −8 −10 +6 = −4

f ( −2) = ( −2)3 −2( −2) 2 −5( −2) +6 = −8 −8 +10 +6 =0

√

x +1 is not a factor of f(x). f ( 2) = 2 3 −2( 2) 2 −5( 2) +6

(b)

2.

x −1 is a factor of f(x).

f ( −1) =( −1) 3 −2( −1) 2 −5( −1) +6

divided by x −3 , the remainder is 0, it means that f(x) is divisible by x −3 , so x −3 is a

1.

f (1) =13 −2(1) 2 −5(1) +6 =1 −2 −5 +6

∴ ∴

≠0 x + 6 is not a factor of f(x). One of the factors of f(x) is x −3 , x −1 or x +2 .

∵

The factor of f(x) are x −3 , x −1 and

More about Polynomials x +2 . ∴

f ( x ) =( x − 3)( x − 1)( x +2)

60

Certificate Mathematics in Action Full Solutions 4A

Follow-up Exercise

( 4 x − x 2 ) + x ( x +1)( x −1) −( x 3 − 2) = ( 4 x − x 2 ) +[ x( x ) + x (1)]( x −1) −( x 3 − 2)

p.130

1.

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

(5 − x −4 x 2 ) +(7 x 3 − x +1)

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

= 5 − x −4 x 2 +7 x 3 − x +1

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

= 7 x 3 −4 x 2 − x − x +5 +1

= 4 x − x 2 + x3 − x − x3 + 2

= 7 x 3 − 4 x 2 −2 x + 6

= x3 − x3 − x 2 + 4 x − x + 2 = − x 2 +3 x + 2

Alternative Solution −4 x 2 −x +5

p. 132

+) 7 x 3 +0 x 2 −x +1

(5 x 3 + 4 x ) ÷ x

7 x −4 x −2 x +6 3

2

(3 x 3 +7 x 2 − 2 x +1) −( x 3 −2 x −5) 2.

1.

(a)

= 3 x 3 +7 x 2 −2 x +1 − x 3 + 2 x +5 = 3 x 3 − x 3 +7 x 2 − 2 x + 2 x +1 +5

5 x3 + 4 x x 3 5x 4x = + x x 2 = 5x + 4 =

= 2 x 3 +7 x 2 + 6

(8 x 3 + 4 x 2 + 4 x) ÷ 2 x

Alternative Solution 3 x 3 +7 x 2 −2 x +1 −)

(b)

x 3 +0 x 2 −2 x −5 2 x 3 +7 x 2 +0 x +6

∴ The answer is 2x3 + 7x2 + 6.

8 x3 + 4 x 2 + 4 x 2x 3 8x 4x2 4x = + + 2x 2x 2x 2 = 4x + 2x + 2 =

3.

3x − 2

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

2.

(a)

=−6 x 4 −12 x 2 +3 x +4 x 3 +8 x −2 =−6 x 4 +4 x 3 −12 x 2 +3 x +8 x −2

x 3x 2 − 2 x 3x − 2x − 2x

=−6 x 4 +4 x 3 −12 x 2 +11 x −2

3x 2 + 4 x + 3

Alternative Solution 2x3 ×) −6 x

4

+0 x

3

−12 x

2

4x3 + 0x 2

+) −6 x

4

+4 x

3

2 x 6 x3 + 8 x 2 + 6 x −1

+0 x 2 +4 x −1 −3 x +2

−12 x

2

+3 x +8 x −2

6 x3 (b)

8x2 8x2

+11 x −2

6x 6x

4.

−1 p.135

61

3

x −1 x −1 x 2 − 2 x + 3 x2

1.

2 ∴



The quotient is x −1 and the remainder is 2.

2.

Remainder = f  −

(e)

Remainder = f (0)



(f)

Remainder = f ( 2)

(a)

Let f ( x ) = x 3 − 4 . = f (1) Remainder =13 −4

3x + 2

=−3

2 x +1 6 x 2 + 7 x − 5 6 x 2 + 3x 4x − 5 4x + 2

2.

(b) Let f ( x ) = 2 x 3 −3 x 2 + x +1 . Remainder = f ( −2) = 2( −2) 3 −3( −2) 2 +( −2) +1

−7 ∴ -7.

3  2

(d)

−x − x +3 − x +1

More about Polynomials

The quotient is 3 x + 2 and the remainder is

= −16 −12 −2 +1 = −29

x 2 − 7 x + 23 x + 3 x3 − 4 x 2 + 2 x + 1 x3 + 3x 2 − 7 x2 + 2x

3.

(c)

− 7 x 2 − 21 x 23 x + 1 23 x + 69 − 68 ∴

The quotient is x − 7 x + 23 and the remainder is −68. 2

2x − 5

1  = f  3 3

2

1  1  1  = 27   −18   + 3  − 5 3 3     3 = 1 − 2 +1 − 5 = −5

2 x 2 + x + 1 4 x3 − 8 x 2 + 0 x + 7

(d) Let f ( x ) = x 4 + x 2 + 3 .

4 x3 + 2 x 2 + 2 x

4.

Let f ( x ) = 27 x 3 −18 x 2 +3 x −5 . Remainder

 1 = f −   2

− 10 x 2 − 2 x + 7 − 10 x 2 − 5 x − 5

4

3 x + 12 ∴

The quotient is 2 x − 5 and the remainder is 3 x +12 .

p.141 1.

(a)

Remainder = f (1) (b) (c)

Remainder = f (−2)

1   5 

Remainder = f 

2

 1  1 = −  + −  + 3 Remainder  2  2 1 1 = + +3 16 4 53 = 16 3.

Let f ( x ) = 3 x 3 + 4 x 2 + ax + b . When f(x) is divided by x +1 ,

f ( −1) = 4 3( −1) + 4( −1) + a ( −1) +b = 4 …… − a +b = 3 3

2

62

Certificate Mathematics in Action Full Solutions 4A f (3) = 2(3) 3 −5(3) 2 −3 +1

(1)

=54 −45 −3 +1

(a)

When f(x) is divided by

=7

,

x −2 f ( 2) = 28

≠0

∴

x −3 is not a factor of

2 x − 5 x 2 − x +1 . 3

3( 2) + 4( 2) + a ( 2) + b = 28 …… (2) 2a + b = −12 3

2

(b)

(2) − (1), 3a = −15 a = −5

f ( −4) = 2( −4)3 −5( −4) 2 −( −4) +1 = −128 −80 +4 +1

By substituting a = −5 into (1), we have −( −5) +b = 3 . b = −2

= −203 ∴

≠0 x + 4 is not a factor of

2 x 3 − 5 x 2 − x +1 .

p.146

(c) Let f ( x ) = x − 4 x − 7 x +10 . 3

1. (a)

2

3

=0

∴

x −1 is a factor of

x 3 − 4 x 2 − 7 x +10 . f ( 2) = 23 −4( 2) 2 −7( 2) +10

2 x 3 − 5 x 2 − x +1 .

=8 −16 −14 +10

(b)

3

≠0

x − 2 is not a factor of x 3 − 4 x 2 − 7 x +10 .

(c)

f ( −1) = ( −1)3 −4( −1) 2 −7( −1) +10 = −1 −4 +7 +10

2 x 3 − 5 x 2 − x +1 .

=12 ∴

≠0 x +1 is not a factor of

x 3 − 4 x 2 − 7 x +10 . (d)

3.

Let f ( x ) = 4 x 3 − kx 2 + 9 x − 2 .

∵

4 x 3 − kx 2 + 9 x − 2 is divisible by

4 x −1 . ∴

By the converse of the factor theorem,

1 f =0 2

f ( −2) = ( −2)3 −4( −2) 2 −7( −2) +10 = −8 −16 +14 +10 =0

∴

x + 2 is a factor of x 3 − 4 x 2 − 7 x +10 .

2.

63

2

1 1 1 1 f   = 2  − 5  −   + 1 2 2 2 2 1 5 1 (d) = − − +1 4 4 2 1 =− 2 ≠0 ∴ 2 x −1 is a factor of

= −12 ∴

2

 1  1  1  1 f  −  = 2 −  − 5 −  −  −  + 1 2 2      2  2 1 5 1 = − − + +1 4 4 2 =0 ∴ 2 x +1 is a factor of

f (1) =13 −4(1) 2 −7(1) +10 =1 −4 −7 +10

Let f ( x ) = 2 x 3 −5 x 2 − x +1 .

3

2

1 1 1 4  − k   + 9  − 2 = 0 4 4 4 . 1 k 9 − + −2 = 0 16 16 4 k 5 = 16 16 k =5

3

4.

(a)

f (3) = 33 −5(3) 2 +8(3) −6

∵

= 27 −45 + 24 −6 =0

(b)

x 2 − x − 12 x + 3 x 3 + 2 x 2 − 15 x − 36 x3 + 3 x 2

x −3 is a factor of f(x).

∴

More about Polynomials

− x 2 − 15 x

By long division,

− x2

x2 − 2x + 2

− 3x − 12 x − 36

x − 3 x3 − 5 x 2 + 8 x − 6

− 12 x − 36

x3 − 3x 2 − 2x2 + 8x − 2x2 + 6x 2x − 6 2x − 6 Hence,

f ( x ) =( x −3)( x 2 −2 x +2)

∴ x 3 +2 x 2 −15 x −36 = ( x +3)( x 2 −x −12 ) = ( x −4)( x +3) 2

∴ x 3 +2 x 2 −15 x −36 = ( x +3)( x 2 −x −12 ) = ( x −4)( x +3) 2

p.152 1.

3. Let f ( x) = x 3 + 2 x 2 − x − 2 .

∵

f (1) =13 +2(1) 2 −1 −2 =0

∴

x −1 is a factor of f(x).

Let f ( x) = x 3 − 7 x + 6 .

∵

f (1) =13 −7(1) +6 =0

∴

x −1 is a factor of f(x).

By long division,

x2 + x − 6

By long division,

x 2 + 3x + 2

x −1 x 3 + 0 x 2 − 7 x + 6

x −1 x 3 + 2 x 2 − x − 2

x3 − x 2

x3 − x 2

x2 − 7 x x2 −

3x 2 − x 3x − 3x 2

− 6x + 6 − 6x + 6

2x − 2 2x − 2 ∴

∴

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

2. ∵

Let f ( x ) = x 3 + 2 x 2 −15 x −36 .

f (1) =13 +2(1) 2 −15 (1) −36 =−48 ≠0 3

f ( −1) =( −1) +2( −1) 3

f ( 2) =2 +2( 2)

2

2

−15 ( −1) −36 =−20 ≠0

−15 ( 2) −36 =−50 ≠0

4.

x 3 −7 x 2 +6 =( x −1)( x 2 + x −6) =( x −2)( x −1)( x +3)

Let f ( x ) = x 3 + 3 x 2 − x −3 .

∵

f (1) =13 +3(1) 2 −1 −3 =0

∴

x −1 is a factor of f(x).

By long division,

x2 + 4x + 3 x −1 x 3 + 3 x 2

− x −3

x3 − x 2

f ( −2) =( −2) 3 +2( −2) 2 −15 ( −2) −36 =−6 ≠0

4x2 −

f (3) =33 +2(3) 2 −15 (3) −36 =−36 ≠0

x

4x2 − 4x

f ( −3) =( −3) 3 +2( −3) 2 −15 ( −3) −36 =0 ∴

x

x + 3 is a factor of f(x).

3x − 3 3x − 3

By long division, ∴

64

Certificate Mathematics in Action Full Solutions 4A ( 2 x −3 x 2 +1) −(3 x 2 −4 x +1)

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

3.

= −3 x 2 −3 x 2 +2 x +4 x +1 −1

5. 2 x 3 + 2 x 2 −16 x − 24 = 2( x 3 + x 2 −8 x −12 )

= −6 x 2 +6 x

Let f ( x ) = x 3 + x 2 −8 x −12 . ∵ f (1) =13 +12 −8(1) −12 =−18 ≠0

(3 x 2 −2 x 3 −7 x −4) −( 4 x 2 −6 x −5)

4.

= −2 x 3 − x 2 − x +1

f ( 2) =2 3 +2 2 −8( 2) −12 =−16 ≠0

( 4 x 2 +3 x +1)( x +1)

f ( −2) =( −2) 3 +( −2) 2 −8( −2) −12 =0

x + 2 is a factor of f(x).

= ( 4 x 2 +3 x +1)( x ) +( 4 x 2 +3 x +1)(1) 5.

By long division,

= 4 x 3 +7 x 2 +4 x +1

x 3 + x 2 −8 x −12 = ( x −3)( x + 2) 2

( x +3)( x 2 −2 x )

∴ 2 x 3 +2 x 2 − 16 x −24 =2( x −3)( x +2) 2

6. ∴ ∴

6.

= ( x +3)( x 2 ) +( x +3)( −2 x ) = x 3 +3 x 2 −2 x 2 −6 x = x 3 + x 2 −6 x

Let f ( x ) = 2 x 3 + 7 x 2 − 44 x + 35 .

f (1) =2(1) 3 +7(1) 2 −44 (1) +35 =0

= 4 x 3 +3 x 2 + x +4 x 2 +3 x +1 = 4 x 3 +3 x 2 +4 x 2 + x +3 x +1

x 3 + x 2 −8 x −12 = ( x + 2)( x 2 − x − 6) ∴

= 3 x 2 −2 x 3 −7 x −4 −4 x 2 +6 x +5 = −2 x 3 +3 x 2 −4 x 2 −7 x +6 x −4 +5

f ( −1) =( −1) 3 +( −1) 2 −8( −1) −12 =−4 ≠0

∴

= 2 x −3 x 2 +1 −3 x 2 +4 x −1

7.

x −1 is a factor of f(x).

(3 x −2)( 2 x 2 −4 x −1)

By long division,

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

2 x 3 +7 x 2 −44 x +35 =( x −1)( 2 x 2 +9 x −35 ) =( x −1)( 2 x −5)( x +7)

= 6 x 3 −4 x 2 −12 x 2 +8 x −3 x +2 = 6 x 3 −16 x 2 +5 x +2

8.

Exercise

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

Exercise 3A (p.130)

( 4 x 2 +4 x +1)

Level 1

= 2 x 3 + x 2 −2 x 2 − x +2 x +1 +4 x 2 +4 x +1 = 2 x 3 + x 2 −2 x 2 +4 x 2 − x +2 x +4 x +1 +1

( x 3 −5 x + 4) + ( x 3 + 2 x 2 −5) 1.

= x 3 + x 3 + 2 x 2 − 5 x + 4 −5 = 2 x 3 + 2 x 2 −5 x −1 (6 x 3 −3 x + 2) + ( −2 x 3 − 4 x 2 + 2 x −1)

2.

= 2 x 3 +3 x 2 +5 x +2

= x 3 − 5 x + 4 + x 3 + 2 x 2 −5

Level 2 9.

= 6 x 3 −3 x + 2 − 2 x 3 − 4 x 2 + 2 x −1

( 4 x 3 −3 x −7) −( 2 x 3 − 4 x 2 −7 x ) −(6 x 2 −5 x −5)

= 6 x 3 − 2 x 3 − 4 x 2 −3 x + 2 x + 2 −1

= 4 x 3 −3 x − 7 − 2 x 3 + 4 x 2 + 7 x − 6 x 2 + 5 x + 5

= 4 x 3 − 4 x 2 − x +1

= 4 x 3 − 2 x 3 + 4 x 2 −6 x 2 −3 x + 7 x +5 x −7 +5 = 2 x 3 − 2 x 2 +9 x − 2 10.

65

3

More about Polynomials

(5 x 3 − 6 x 2 − x + 3) + (7 x 3 − 5 x 2 − x + 3) −

Q 2 − PQ

( x 3 − 2 x 2 + 2 x − 5)

= (3 x 2 −5) 2 −( x 2 −3 x −5)( 3 x 2 −5)

= 5 x3 − 6 x 2 − x + 3 + 7 x3 −5 x 2 − x + 3 −

= 9 x 4 −30 x 2 + 25 − ( x 2 −3 x −5)( 3 x 2 ) −

x3 + 2 x 2 − 2 x + 5

( x 2 −3 x −5)( −5)

= 5 x3 + 7 x3 − x3 − 6 x 2 − 5 x 2 + 2 x 2 − x − x −

= 9 x 4 −30 x 2 + 25 −3 x 4 + 9 x 3 +15 x 2 + 5 x 2 −

2x +3 +3 +5

15 x − 25

= 11 x − 9 x − 4 x +11 3

= 9 x 4 −3 x 4 + 9 x 3 −30 x 2 +15 x 2 + 5 x 2 −15 x + 25 − 25

2

11.

= 6 x 4 + 9 x 3 −10 x 2 −15 x ( 2 x 2 −3 x −5)( x 2 −2) = ( 2 x 2 −3 x −5)( x 2 ) +( 2 x 2 −3 x −5)( −2) = 2 x 4 −3 x 3 −5 x 2 −4 x 2 +6 x +10 = 2 x 4 −3 x 3 −9 x 2 +6 x +10

Exercise 3B (p.136) Level 1

x +3

12.

( x 2 + 2 x −1)( 3 x 2 − 5 x − 2)

1.

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

7 x 7 x 2 + 21 x 7 x2 21 x

( x 2 + 2 x −1)( −2)

21 x

= 3 x 4 + 6 x 3 − 3 x 2 − 5 x 3 −10 x 2 + 5 x − 2 x 2 − 4 x + 2 = 3 x 4 + 6 x 3 − 5 x 3 − 3 x 2 −10 x 2 − 2 x 2 + 5 x − 4 x + 2 = 3 x 4 + x 3 −15 x 2 + x + 2

5x

13.

5 P −( x −1)Q

2.

= 5( x 2 −3 x −5) −( x −1)( 3 x 2 −5)

− 3x

3 x 15 x 3 − 9 x 2 15 x 3 − 9x2

= 5 x 2 −15 x −25 −3 x 3 +3 x 2 +5 x −5

− 9x2 ∴ The quotient is 5 x 2 −3 x and the remainder is 0.

= −3 x 3 +8 x 2 −10 x −30

a +3

(3 − x )( Q − P ) = (3 − x )[( 3 x −5) −( x −3 x −5)] 2

2

a a 2 + 3a + 2

= ( −x +3)( 3 x −5 − x +3 x +5) 2

2

= ( −x +3)( 3 x 2 − x 2 +3 x −5 +5) = ( −x +3)( 2 x 2 +3 x ) = ( −x +3)( 2 x 2 ) + ( −x +3)( 3 x) = −2 x 3 + 6 x 2 −3 x 2 +9 x = − 2 x +3 x +9 x 3

15.

2

= 5 x 2 −15 x −25 −( x −1)( 3 x 2 ) −( x −1)( −5) = −3 x 3 +5 x 2 +3 x 2 −15 x +5 x −25 −5

14.

∴ The quotient is x + 3 and the remainder is 0.

3.

a2 3a 3a 2

2

∴ The quotient is a + 3 and the remainder is 2.

66

Certificate Mathematics in Action Full Solutions 4A

x−2

3x + 8

2x 2x2 − 4x + 3 4.

2x2

2x − 3 6x2 + 7 x + 2 9.

6x2 − 9x

− 4x − 4x

16 x + 2 16 x − 24 3

26

∴ The quotient is x − 2 and the remainder is 3.

∴ The quotient is 3 x +8 and the remainder is 26.

4 a −1

p +2 p +1 p 2 + 3 p + 5 p2 + p

5.

4a +1 16 a 2 + 0a − 4 10.

2 p +5 2 p +2

− 4a − 4 − 4a − 1

3

−3 ∴ The quotient is 4a −1 and the remainder is −3.

∴ The quotient is p + 2 and the remainder is 3.

z −8

s 2 − 4 s +11

z + 3 z 2 − 5z + 6 6.

16 a 2 + 4a

s + 2 s 3 − 2 s 2 + 3s + 2

z 2 + 3z

s3 + 2s 2

−8z + 6 − 8 z − 24

− 4 s 2 + 3s

11.

− 4 s 2 − 8s

30

11 s + 2 11 s + 22

∴ The quotient is z −8 and the remainder is 30.

− 20

2r 7.

∴ The quotient is s 2 −4 s +11 and the remainder is −20.

r − 2 2r − 4r + 1 2

2r 2 − 4r

2h 2 + 3h +10

1 ∴ The quotient is

2d − 1 4d − 4d + 1 2

4d 2 − 2d − 2d + 1 − 2d + 1 ∴ The quotient is 2d −1 and the remainder is 0.

67

2 h − 6h 3

2r and the remainder is 1.

2d − 1 8.

h − 3 2h3 − 3h 2 + h − 12.

5

2

3h 2 + h 3h 2 − 9h 10 h − 5 10 h − 30 25 ∴ The quotient is 2h 2 + 3h +10 and the remainder is 25.

3 2 x −9 .

2x2 − 4x + 2 2 x +1 4 x 3 − 6 x 2 + 0 x − 7 4x + 2x 3

x −2

2

x 2 + x + 3 x3 − x 2 + 3x − 3

−8 x + 0 x 2

13.

More about Polynomials

x3 + x 2 + 3x

17.

−8 x − 4 x 2

− 2x2 + 0x − 3

4 x −7

− 2x2 − 2x − 6

4x +2

2x + 3

−9

∴ The quotient is x − 2 and the remainder is 2x +3 .

∴ The quotient is 2 x 2 − 4 x + 2 and the remainder is −9.

p

3m 2 + 2m +1 3m + 2 9m +12 m + 7 m − 5 3

2

18.

p 2 + 2 p − 1 p3 + 2 p 2 − 5 p + 1 p3 + 2 p 2 − p

9m 3 + 6m 2

− 4p + 1

6m 2 + 7 m

14.

6m 2 + 4m

∴ The quotient is p and the remainder is −4 p +1 .

3m − 5 3m + 2

2z + 3

−7 ∴ The quotient is 3m 2 + 2m +1 and the remainder is −7.

z 2 − 3z − 2 2 z 3 − 3z 2 − 19.

3w − 2 w + 1

3z 2 − 9 z − 6 12 z + 10

4 w + 3 12 w3 + w2 − 2 w + 3 12 w + 9 w

15.

2

3z 2 + 3z + 4

2

3

z+4

2z − 6z − 4z 3

2

∴ The quotient is 2 z + 3 and the remainder is 12 z +10 .

− 8w2 − 2 w − 8w2 − 6 w

2d + 5

4w + 3 4w + 3 ∴ The quotient is 3w 2 − 2 w +1 and the remainder is 0.

d 2 − 3d − 4 2d 3 − d 2 − 24d − 18 20.

2 d 3 − 6 d 2 − 8d 5d 2 − 16d − 18 5d 2 − 15d − 20 −d+ 2

Level 2

x +6 x2 − x + 2 x3 + 5 x 2 − 2 x + 3 16.

∴ The quotient is 2d + 5 and the remainder is −d + 2 .

x3 − x 2 + 2 x 6x2 − 4x + 3 6 x 2 − 6 x + 12 2x − 9 ∴ The quotient is x + 6 and the remainder is

68

Certificate Mathematics in Action Full Solutions 4A

3q − 4

3e −

2q 2 + 5q − 2 6q 3 + 7 q 2 − 23q + 2

2e 2 + 1 6e3 − 5e 2 + 7e + 1

6q 3 + 15q 2 − 6q

21.

5 2

− 8q 2 − 17q + 2

25.

− 8q − 20q + 8 2

+ 3e

6e3

− 5e + 4e + 1 2

3q − 6 ∴ The quotient is 3q −4 and the remainder is 3q −6 .

3s + 5

∴ The quotient is 3e −

3s − 2 s − 1 9 s + 9 s − 16 s − 4 2

3

2

9 s 3 − 6 s 2 − 3s

22.

−

5 2

4e +

7 2

− 5e 2

4e +

15s 2 − 13s − 4

5 and the remainder is 2

7 . 2

x +3

15s 2 − 10 s − 5 − 3s + 1

x 2 − 3x − 2 x3 + 0 x 2 + 2 x − 5

∴ The quotient is 3s + 5 and the remainder is −3s +1 .

x3 − 3x 2 − 2 x

26.

3x 2 + 4 x − 5 3x 2 − 9 x − 6

2 x + 2 x −1 2

13 x + 1

2 x − x + 2 4 x4 + 2 x3 + 0 x 2 + 6 x − 3 2

∴

4 x 4 − 2 x3 + 4 x 2 4 x3 − 4 x 2 + 6 x

23.

x −1

4 x3 − 2 x 2 + 4 x − 2x2 + 2x − 3 − 2x2 + x − 2 x −1

x 2 + x +1 x 3 + 0 x 2 + 0 x +1 27.

∴ The quotient is 2 x 2 + 2 x −1 and the remainder is x −1 .

− x2 −

x +1

− x2 −

x −1

∴ The quotient is x −1 and the remainder is 2.

4w2 + w − 2 8w3 − 10w2 − 9 w + 1 24.

x3 + x 2 + x

2

2w − 3

x 2 − 5 x + 20

8w + 2 w − 4 w 3

The quotient is x + 3 and the remainder is 13 x +1 .

2

x 2 + 3 x −1 x 4 − 2 x3 + 4 x 2 − 3 x − 5

− 12w − 5w + 1 2

− 12w − 3w + 6

x 4 + 3x3 −

2

− 2w − 5 ∴ The quotient is 2 w −3 and the remainder is −2 w −5 .

28.

x2

− 5 x3 + 5 x 2 − 3x − 5 x 3 − 15 x 2 + 5 x 20 x 2 − 8 x − 5 20 x 2 + 60 x − 20 − 68 x + 15 ∴ The quotient is x 2 − 5 x + 20 and the

69

3 = f (1)

remainder is −68 x +15 .

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

29.

[( 3 x 3 + 2 x − 3) − ( x 3 − 4 x 2 + 4)] ÷ ( x 2 − 3 x + 5) = (3 x 3 + 2 x − 3 − x 3 + 4 x 2 − 4) ÷ ( x 2 − 3 x + 5) = (3 x 3 − x 3 + 4 x 2 + 2 x − 3 − 4) ÷ ( x 2 − 3 x + 5)

3.

= ( 2 x 3 + 4 x 2 + 2 x − 7) ÷ ( x 2 − 3 x + 5)

2 x + 10 4.

2 x 3 − 6 x 2 + 10 x

Let f ( x ) = x 3 + 7 x 2 −5 x +1 . = f ( 2) 3 2 Remainder = 2 +7( 2) −5( 2) +1 =8 +28 −10 +1 = 27

10 x − 8 x − 7 2

10 x 2 − 30 x + 50 22 x − 57

Let f ( x ) = 2 x 3 − x 2 + 7 x +1 . = f ( −1) 3 2 Remainder =2( −1) −( −1) +7( −1) +1 =−2 −1 −7 +1 =−9

By long division,

x 2 − 3x + 5 2 x3 + 4 x 2 + 2 x − 7

More about Polynomials

5.

∴ The quotient is 2 x +10 and the remainder is 22 x −57 .

Let f ( x ) = 2 x 3 −3 x 2 − 4 x + 5 . Remainder = f ( −3) = 2( −3) 3 −3( −3) 2 −4( −3) +5

30.

= −54 −27 +12 −5 = −64

[( 2t −1)( t 2 +1)] ÷(t 2 +2t +1) =[( 2t −1)( t 2 ) +( 2t −1)(1)] ÷(t 2 +2t +1) =( 2t 3 −t 2 +2t −1) ÷(t 2 +2t +1)

6.

By long division,

3 2 Remainder = 5 + 4(5) −5 −3 =125 +100 −5 −3 = 217

2t − 5 t 2 + 2t + 1 2t 2 − t 2 + 2t − 1 2t 3 + 4t 2 + 2t

Let f ( x ) = x 3 + 4 x 2 − x −3 . = f (5)

7.

− 5t + 0t − 1 2

Let f ( x ) =27 x 3 −6 x +2 .

 1 = f −   3

− 5t − 10t − 5 2

10t + 4

Remainder

∴ The quotient is 2t − 5 and the remainder is 10 t + 4 . Exercise 3C (p.142) Level 1 1.

Let f ( x ) = x 3 −8 x −7 . = f (3) 3 Remainder =3 −8(3) −7 =27 −24 −7 =−4

2.

8.

3

 1  1 = 27  −  − 6 −  + 2  3  3 = −1 + 2 + 2 =3

Let f ( x ) = 4 x 3 − x 2 +8 x −11 .

1 = f  4 3

2

1 1 1  −   + 8  −11 4 4 4 1 1 = − + 2 −11 16 16 = −9

Remainder = 4

Let f ( x) = x 3 + x 2 + 2 x +1 .

70

Certificate Mathematics in Action Full Solutions 4A 9.

Let f ( x ) =8 x 3 −2 x +1 .

 1 f  −  = 13  3

1  = f  2

4

3

1 1  Remainder = 8    − 2  +1 2 = 1 −1 +1

2

=1 10. Let f ( x ) = 2 x 3 + 2 x 2 − 4 x + 7 .

3 = f  2 3

2

3 3 3 = 2   + 2  − 4   + 7 Remainder 2 2 2 27 9 = + −6 +7 4 2 49 = 4 11.

Let f ( x ) =3 x 3 +7 x 2 +kx −5 . By the remainder theorem, we have f ( −3) = 4

3( −3)3 + 7( −3) 2 + k ( −3) −5 = 4 −81 + 63 −3k −5 = 4 3k = −27 k = −9

3

 1  1  1 4 −  + 30  −  + 18  −  + 20 3 3      3 k 10 − − 6 + 20 81 9 k 81 k

= 13 = 13 1 9 =9 =

15. Let f ( x ) = 2 x 3 −5 x 2 + 4kx − 7 . By the remainder theorem, we have

1  f   = −6 2 3

2

1  1  1  2  − 5  + 4k   − 7 2 2 2 1 5 − + 2k − 7 4 4 2k k

= −6 = −6 =2 =1

16. Let f ( x ) = x1997 −1 . = f ( −1) 1997 −1 Remainder =( −1) =−1 −1

=−2 12. Let f ( x ) = 2 x 3 +kx 2 +5 x +4 . By the remainder theorem, we have f ( 2) = 6

2(2) 3 + k ( 2) 2 + 5(2) + 4 = 6 16 + 4k +10 + 4 = 6 4k = −24

17. Let f ( x ) = x100 −1 . = f (1) 100 −1 Remainder =1 =1 −1 =0

k = −6 13. Let f ( x) = x 2 −kx −2 . By the remainder theorem, we have f (k ) = k k 2 −k (k ) −2 = k k 2 −k 2 −2 = k k = −2 14. Let f ( x ) = kx 4 +30 x 3 +18 x + 20 . By the remainder theorem, we have

18. Let f ( x ) = x 3 + 2ax −b . ∵When f(x) is divided by x −3 , the remainder is 1. ∴ When f(x) − 1 is divided by x −3 , the remainder is 0. ∴ When f(x) − 1 + b is divided by x −3 , the remainder is b, and when f(x) −1 + b+ x2 − 9 is divided by x − 3, the remainder is also b (since x2 − 9 = 0 when x = 3). ∴ The required polynomial is x 3 + 2ax −1 or x 3 + x 2 + 2ax −10 . (or any other reasonable answers) 19. By the remainder theorem, we have

71

3 f ( −1) = 3

 k f  −  = 3k  2

( −1) + 2( −1) + k ( −1) + c = 3 3

More about Polynomials

2

−1 + 2 − k + c = 3 k = c −2

2

∴ k =1, c =3 or k = −1, c =1 or k = −3, c = −1 (or any other reasonable answers)

 k  k 4 −  − 2 −  +1 = 3k  2  2 k 2 + k +1 = 3k k 2 − 2k +1 = 0 ( k −1) 2 = 0 k =1

Level 2

24. When f(x) is divided by x −1 , f (1) =−1

20. Let f ( x) = x 2 + 2 x + 3 . By the remainder theorem, we have

2(1) 3 −12 + p (1) +q =−1 p +q =−2

k  f   = 11 2 2

k  k    + 2   + 3 = 11 2   2 2 k + k + 3 = 11 4

When f(x) is divided by x + 2 , f ( −2) = −31

2( −2)3 −( −2) 2 + p ( −2) +q = −31 −2 p +q = −11

k 2 + 4k −32 = 0 ( k +8)( k −4) = 0 k +8 = 0 k = −8

or or

k −4 = 0 k =4

21. Let f ( x ) = x 3 + ( k + 4) x 2 − 2 x −1 . By the remainder theorem, we have f ( −k ) = k 2 ( −k ) 3 +( k +4)( −k ) 2 −2( −k ) −1 = k 2 −k 3 +k 3 +4k 2 +2k −1 = k 2 3k 2 +2k −1 = 0

k +1 = 0

or

k = −1

or

( k +1)(3k −1) = 0 3k −1 = 0 1 k = 3

22. Let f ( x) = x 2 − 4 x −3 . By the remainder theorem, we have

f ( −k ) = −6

( −k ) −4( −k ) −3 = −6 2

k 2 + 4k +3 = 0 ( k +3)( k +1) = 0 k +3 = 0 k = −3

or or

k +1 = 0 k = −1

23. Let f ( x ) = 4 x − 2 x +1 . By the remainder theorem, we have 2

……(1)

……(2) (1) – (2), 3 p =9 p =3 By substituting p = 3 into (1), we have 3 +q = −2 q = −5 25. When f(x) is divided by x − 2 , f ( 2) = −5

2 3 − p ( 2) 2 + 2( 2) + q = −5 − 4 p + q = −17

……(1)

When f(x) is divided by x −3 , f (3) = −4

33 − p (3) 2 +2(3) +q = −4 −9 p +q = −37

……(2)

(1) – (2), 5 p = 20 p =4 By substituting p = 4 into (1), we have −4( 4) +q =−17 q =−1 26. Let f ( x ) = x 2 − mx +3 and

g ( x) = 2m − x .

When f(x) is divided by x − n , f ( n) =11

n 2 −mn +3 =11

……(1)

n 2 −mn −8 = 0

When g(x) is divided by

x−n

,

72

Certificate Mathematics in Action Full Solutions 4A g ( n) = 0

f ( −1) =1

2m − n = 0

( −1)99 +k =1 −1 +k =1

n = 2m

……(2)

k =2 By substituting (2) into (1), we have

(b)

( 2 m ) 2 − m( 2 m ) − 8 = 0 2m 2 − 8 = 0 m = ±2

∴

When m = 2, n = 4 When m = −2, n = −4 ∴ The values of m and n are

+ 2 = ( x +1)Q ( x) +1

9 99 = (9 +1)Q(9) −1

∵The degree of ( x −1)( x +2) is 2, ∴ The highest possible degree of the remainder when P(x) is divided by ( x −1)( x +2) is 1.

=10 Q (9) −1 =10 [Q (9) −1] +9 ∴ The remainder when 999 is divided by 10 is 9.

Exercise 3D (p. 147) Level 1 1.

(a)

f (1) =13 −12 −3(1) +3 =1 −1 −3 +3 =0

∴ x − 1 is a factor of f(x).

By the remainder theorem, we have

P (1) = −4

(b)

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

(1 −1)(1 + 2)Q (1) +a (1) +b = −4

=4

a +b = −4

≠0 ∴ x + 1 is not a factor of f(x).

……(1)

P ( −2) = −28

2.

(a)

f ( 2) = 2 3 + 2(2) 2 −5( 2) −6 =8 +8 −10 −6 =0

( −2 −1)( −2 +2)Q ( −2) +a ( −2) +b = −28

∴ x - 2 is a factor of f(x).

−2a +b = −28

f (3) = 33 +2(3) 2 −5(3) −6

……

= 27 +18 −15 −6

(b)

3a = 24 a =8 By substituting a = 8 into (1), we have 8 + b = −4 b = −12

= 24

(1) – (2),

∴ The remainder when P(x) is divided by ( x −1)( x +2) is 8 x −12 .

≠0 ∴ x - 3 is not a factor of f(x). 3.

(a)

f ( −4) =( −4)3 +5( −4) 2 +3( −4) +15 = −64 +80 −12 +15 =19

By the remainder theorem, we have

≠0 ∴ x + 4 is not a factor of f(x). (b)

73

……

By substituting x = 9 into (1), we have

P ( x ) = ( x −1)( x + 2)Q ( x ) + ax +b

28. (a)

x

(1)

(b) Let Q(x) and ax + b be the quotient and the remainder respectively when P(x) is divided by ( x −1)( x +2) . ∴

(2)

f ( x ) = ( x +1)Q ( x) +1 99

x 99 = ( x +1)Q ( x) −1

m= 2 m= −2 .  or   n= 4  n= −4 27. (a)

Let Q(x) be the quotient when f(x) is divided by x +1 .

3 f ( −5) = ( −5) 3 +5( −5) 2 +3( −5) +15

P ( x ) = ( x 2 + x − 2)( 2 x + 3)

= −125 +125 −15 +15

= 2 x 3 + 3x 2 + 2 x 2 + 3x − 4 x − 6

=0

∴ x + 5 is a factor of f(x). 4.

Let f ( x ) = 2 x 3 −9 x 2 + 5 x − 4 .

f ( 4) = 2( 4)3 −9( 4) 2 +5( 4) − 4

∵

=128 −144 + 20 − 4

= 2 x 3 + 5x 2 − x − 6 ∴ The required polynomial is 2x3 + 3x2 – 2x – 3 or 2x3 + 5x2 – x – 6. (or any other reasonable answers) 10. ∵x3 + mx2 + nx + 5 is divisible by x – 1. ∴ By the converse of the factor theorem,

f (1) = 0

=0

13 + m(1) 2 + n(1) + 5 = 0

∴ x - 4 is a factor of 2 x 3 −9 x 2 +5 x −4 . 5.

More about Polynomials

1+ m + n + 5 = 0 n = −6 − m

Let f ( x) = x 3 − a 3 .

∴ m = –3, n = –3 or m = –1, n = –5 or m = 2, n = –8. (or any other reasonable answers)

3 3 ∵ f ( a ) = a − a 

=0

∴ x3 – a3 is divisible by x – a. 6.

Let f ( x ) = 2 x 3 + kx 2 − x − 6 .

∵ 2 x 3 + kx 2 − x − 6 is divisible by x + 2. ∴ By the converse of the factor theorem, f ( −2) = 0

11. Let f(x) = x3 + ax2 – x – b. ∵x + 2 is a factor of x3 + ax2 – x – b. ∴ By the converse of the factor theorem,

f ( −2) = 0

( −2) + a ( −2) − ( −2) − b = 0 − 8 + 4a + 2 − b = 0 b = 4a − 6 3

2( −2) +k ( −2) −( −2) −6 = 0 3

2

−16 + 4k +2 −6 = 0 4k = 20 k =5 7. Let f ( x ) = x 3 −2ax +15 . ∵x + 5 is a factor of x3 – 2ax + 15. ∴ By the converse of the factor theorem, f ( −5) = 0

( −5) 3 −2a ( −5) +15 = 0 −125 +10 a +15 = 0 10 a =110 a =11 8. Let f ( x) = 2 x 3 + x 2 − mx +12 . ∵x + 3 is a factor of 2x3 + x2 – mx + 12. ∴ By the converse of the factor theorem, f (3) = 0

2(3) 3 +3 2 − m(3) +12 = 0 54 +9 −3m +12 = 0 3m = 75 m = 25 9.

Let P(x) be the polynomial of degree 3 with 2x + 3 as one of its factors. ∵P(x) = (ax2 + bx + c)(2x + 3) where a, b and c are integers. By substituting a = 1, b = 0 and c = -1 into P(x), we have

P ( x) = ( x 2 −1)( 2 x + 3) = 2 x3 + 3x 2 − 2 x − 3

or by substituting a = 1, b = 1 and c = −2 into P(x), we have

2

∴ a = –3, b = –18 or a = –1, b = –10 or a =1, b = –2 (or any other reasonable answers)

Level 2 12. Let f(x) = 2x3 – x2 – 7x + 6. 3

2

3 3 3 3 f   = 2  −   − 7  + 6 2 2 2       2 ∵ 27 9 21 = − − +6 4 4 2 =0 ∴ 2x – 3 is a factor of 2x3 – x2 – 7x + 6.

13. Let f(x) = 8x3 – 14x2 + 7x – 1. 3

2

1 1 1 1 f   = 8  − 14   + 7  − 1 4 4 4 4 ∵ 1 7 7 = − + −1 8 8 4 =0 ∴ 8x3 – 14x2 + 7x –1 is divisible by 4x – 1.

14. (a) ∵

g ( −5) = ( −5)3 +12 ( −5) 2 + 41 ( −5) +30 = −125 +300 −205 +30 =0

∴ x + 5 is a factor of g(x). (b) By long division,

74

Certificate Mathematics in Action Full Solutions 4A x x

2

+ 7 x ∵x3+ – 4x2 + kx6 + 6 is divisible by x – 3.

∴ x By the 2 converse of the factor theorem, x + + 5 x + 1 2 + 4 1 3 0 3

x3

f (3) + 5x2

=0

2 7 x 1 x 33 −4(3) 2 + +k (3) +4 6 =0

7 x

2

+ 3 5 x

27 −36 +3k +6 = 0 6 3k = 3 6

Hence, g(x) = (x + 5)(x2 + 7x + 6) By the cross method, x2 + 7x + 6 = (x + 1)(x + 6) 1)( x ∴ g ( x ) =( x +5)( x +

x x

k =1 (b) By long division, + 6)

x x

15. (a) ∵

2

− x −

− 3 x3 − 4 x x3

3

+ 3 0 + 3 0

− 3x

− x 2

2

 3  3  3  3 f  −  = 4 −  − 8 −  − 15 −  + 9  2  2  2  2 27 45 =− − 18 + +9 2 2 =0

− x 2

Hence, f(x) = (x – 3)(x2 – x – 2) By the cross method, x2 – x – 2 = (x – 2)(x + 1) ∴

∴ 2x + 3 is a factor of f(x).

x 3 −4 x 2 +x +6 =( x −3)( x −2)( x + 1)

(b) By long division,

18. (a) Let f(x) = 8x3 + mx2 – 25x + 6. 3 2 ∵8x + mxx – 25x + 6 is divisible by 4x – 1. − 7 + 3 ∴ By the converse of the factor theorem, 3 2 + 3 4 x − 8x − 1 5 x 2 x

2 x

4 x

2

3

1 f  =0  42 

+ 6 x 2

− 1 4 x 2 − 1 5 x

− 1 42 x − 2 1 x 3 6 x 6 x

Hence, f(x) = (2x + 3)(2x2 – 7x + 3) By the cross method, 2x2 – 7x + 3 = (2x – 1)(x – 3) 1)( ∴ f ( x ) =( 2 x +3)( 2 x −

1  1 1 8  + m  − 25  + 6 = 0 4 4 4 1 m 25 + − +6 =0 8 16 4 m 1 = 16 8 m =2

x − 3)

16. (a) ∵ 3

2

 4  4  4  4 h −  = 3 −  + 4 −  − 75 −  −100 3 3 3        3 64 64 =− + + 100 −100 9 9 =0

(b) By long division, 2 x 4 x

2

+

− 1 8x 3 + 2 8x

3

− 2

4 x

4 x

∴ 3x + 4 is a factor of h(x). (b) By long division, x2

3x + 4 3x3 + 4x2 3x3 + 4x2

2 Hence, h( x ) =(3 x +4)( x −25 )

=(3 x +4)( x +5)( x −5)

17. (a) Let f(x) = x3 – 4x2 + kx + 6.

75

Hence, f(x) = (4x – 1)(2x2 + x – 6) By the cross method, 2x2 + x – 6 = (2x – 3)(x + 2) − 7 5 x − 1 0 0 ∴

− 2 5

8 x 3 +2 x 2 −25 x +6 =( 4 x − 1)( 2 x −3)( x +2)

− 7 5 x − 1 0 0 f(x)1 =0 x3 + px2 + qx + 2 and g(x) = x3 + qx2 + px – − 7 5 19.xLet− 0 6. ∵x – 2 is a common factor of x3 + px2 + qx + 2 and x3 + qx2 + px – 6.

3

∴

More about Polynomials

f ( 2) = 0 3

g ( −1) = 8

2

2 + p ( 2) + q ( 2) + 2 = 0 2 p + q = −5 g ( 2) = 0

23 + q ( 2) 2 + p ( 2) − 6 = 0 p = −2q −1 By substituting (2) into (1), we have 2( −2q −1) +q =−5

s ( −1) + 9( −1) − t ( −1) − 5 = 8 t = s +4 3

…… (1)

2

…… (1) ∵x + 5 is a factor of g(x). ∴ By the converse of the factor theorem,

…… (2)

g (−5) = 0

s (−5) + 9( −5) −t ( −5) −5 = 0 − 25 s + t = −44 3

3q =3 q =1 By substituting q = 1 into (2), we have p =−2(1) −1

=−3 20. (a) ∵When f(x) is divided by x + 1, the remainder is 10. ∴ By the remainder theorem, we have

2

…… (2) By substituting (1) into (2), we have −25 s +s +4 =−44 24 s =48 s =2 By substituting s = 2 into (1), we have t =2 +4 =6

f ( −1) =10 2( −1) −( −1) + a ( −1) +b =10 3

2

− 2 −1 −a +b =10 a −b = −13

……

(1) ∵x – 1 is a factor of f(x). ∴ By the converse of the factor theorem,

(b) By long division,

f (1) = 0

x

3

+ 1 0

−

……

−

(1) + (2), 2a =−14 a =−7

Hence, g(x) = (x + 5)(2x2 – x – 1) By the cross method, 2x2 – x – 1 = (2x + 1)(x – 1) ∴ g ( x ) =( x −1)( x +5)( 2 x +1)

By substituting a = –7 into (2), we have −7 +b =−1 b =6

Exercise 3E (p. 152)

(b) By long division, 2 x x

−

+ 5 2 x 3 + 2 x

2(1) 3 −12 + a (1) + b = 0 a + b = −1 (2)

2

2 x

2

Level 1

+ x − 6

2 − 1 2 x 3 − 6 1. Letx f(x) = x3 – 3x2 –− 6x + 8. 7 x +

2 x

3

2 − 2 x ∵f(1) = 13 – 3(1)2 – 6(1) + 8 = 0

∴ xx – 1 is a factor − of f(x). 7 x 2 − x By x long division, 6 x3 – 3x2 – 6x + 8 = (x− – 1)(x2 –6 2x – 8) x + − 6 x + 6 ∴ 2

Hence, f(x) = (x – 1)(2x2 + x – 6) By the cross method, 2x2 + x – 6 = (2x – 3)(x + 2) 1)( 2 x − 3)( x +2) ∴ f ( x ) =( x − 21. (a)

∵When g(x) is divided by x + 1, the remainder is 8. ∴ By the remainder theorem, we have

x 3 −3 x 2 −6 x +8 =( x −4)( x − 1)( x +2)

Let f(x) = x3 – 7x – 6. f (1) =13 −7(1) −6 =−12 ∵ f ( −1) =( −1) 3 −7( −1) −6 =0 ∴ x + 1 is a factor of f(x). By long division, x3 – 7x – 6 = (x + 1)(x2 – x – 6) 3 ∴ x −7 x −6 =( x −3)( x +1)( x +2)

2.

76

Certificate Mathematics in Action Full Solutions 4A 3. Let f(x) = x3 + x2 – 10x + 8. ∵f(1) = 13 + 12 – 10(1) + 8 = 0 ∴ x – 1 is a factor of f(x). By long division, x3 + x2 – 10x + 8 = (x – 1)(x2 + 2x – 8) ∴ x

3

+x

2

− 10 x + 8 =( x −2)( x − 1)( x +4)

4. Let f(x) = x3 + 8x2 + 21x + 18. ∵  f (1) =13 +8(1) 2 +21 (1) +18 =48 f ( −1) =( −1) 3 +8(−1) 2 +21 ( −1) +18 =4 f ( 2) =2 3 +8( 2) 2 +21 ( 2) +18 =100 f ( −2) =( −2) 3 +8(−2) 2 +21 (−2) +18 =0 ∴ x + 2 is a factor of f(x). By long division, x3 + 8x2 + 21x + 18 = (x + 2)(x2 + 6x + 9) 3 2 2 ∴ x +8 x +21 x +18 =( x +2)( x +3)

f (1) =13 +3(1) 2 −25 (1) −75 =−96 f (−1) =( −1) 3 +3( −1) 2 −25 ( −1) −75 =−48 f (3) =33 +3(3) 2 −25 (3) −75 =−96 f ( −3) =( −3) 3 +3( −3) 2 −25 ( −3) −75 =0 ∴ x + 3 is a factor of f(x). By long division, x3 + 3x2 – 25x – 75 = (x + 3)(x2 – 25) ∴ x 3 +3 x 2 −25 x −75 =( x −5)( x +3)( x +5)

9.

Let f(x) = x3 + 4x2 – 11x – 30. ∵  f (1) =13 +4(1) 2 −11 (1) −30 =−36

f ( −1) =( −1) 3 +4( −1) 2 −11 (1) −30 =−16 f ( 2) =2 3 +4( 2) 2 −11 ( 2) −30 =−28 f ( −2) =( −2) 3 +4( −2) 2 −11 ( −2) −30 =0 ∴ x + 2 is a factor of f(x). By long division, x3 + 4x2 – 11x – 30 = (x + 2)(x2 + 2x – 15) ∴

5.

Let f(x) = x3 + 3x2 – 4x – 12. ∵ f (1) =13 +3(1) 2 −4(1) −12 =−12 f ( −1) =( −1) 3 +3( −1) 2 −4( −1) −12 =−6 3

2

f ( 2) =2 +3( 2) −4( 2) −12 =0 ∴ x – 2 is a factor of f(x). By long division, x3 + 3x2 – 4x – 12 = (x – 2)(x2 + 5x + 6) ∴ x 3 +3 x 2 −4 x − 12 =( x −2)( x +2)( x +3)

6.

7.

Let f(x) = x3 – 4x2 + 5x – 2. ∵f(1) = 13 – 4(1)2 + 5(1) – 2 = 0 ∴ x – 1 is a factor of f(x). By long division, x3 – 4x2 + 5x – 2 = (x –1)(x2 – 3x + 2) 3 3 2 ∴ x −4 x +5 x −2 =(x −2 )( x −1) Let f(x) = x3 + x2 – 8x – 12. ∵ 3

2

x 3 +4 x 2 −11 x −30 =( x −3)( x +2)( x +5)

10. 3x3 – 6x2 – 12x + 24 = 3(x3 – 2x2 – 4x + 8) Let f(x) = x3 – 2x2 – 4x + 8. ∵ f (1) =13 −2(1) 2 −4(1) +8 =3 f ( −1) =( −1) 3 −2( −1) 2 −4( −1) +8 =9 f ( 2) =2 3 −2( 2) 2 −4( 2) +8 =0 ∴ x – 2 is a factor of f(x). By long division, x3 – 2x2 – 4x + 8 = (x – 2)(x2 – 4) ∴ x3 – 2x2 – 4x + 8 = (x – 2)2(x + 2) ∴ 3 x 3 −6 x 2 −12 x +24 =3( x −2) 2 ( x +2)

11. 5x3 + 20x2 + 5x – 30 = 5(x3 + 4x2 + x – 6) Let f(x) = x3 + 4x2 + x – 6. ∵f(1) = 13 + 4(1)2 + (1) – 6 = 0 ∴ x – 1 is a factor of f(x). By long division, x3 + 4x2 + x – 6 = (x – 1)(x2 + 5x + 6) ∴ x3 + 4x2 + x – 6 = (x – 1)(x + 2)(x + 3) ∴ 5 x 3 +20 x 2 +5 x −30 =5( x − 1)( x +2)( x +3)

f (1) =1 +1 −8(1) −12 =−18 f ( −1) =( −1) 3 +( −1) 2 −8( −1) −12 =−4 f ( 2) =2 3 +2 2 −8( 2) −12 =−16 f ( −2) =( −2) 3 +( −2) 2 −8( −2) −12 =0 ∴ x + 2 is a factor of f(x). By long division, x3 + x2 – 8x – 12 = (x + 2)(x2 – x – 6) 3 2 2 ∴ x +x −8 x −12 =(x −3)( x +2) 8.

Let f(x) = x3 + 3x2 – 25x – 75. ∵

12. 5x3 – 65x – 60 = 5(x3 – 13x – 12) Let f(x) = x3 – 13x – 12. ∵ f (1) =13 −13 (1) −12 =−24 f ( −1) =( −1) 3 −13 ( −1) −12 =0 ∴ x + 1 is a factor of f(x). By long division, x3 – 13x – 12 = (x + 1)(x2 – x – 12) ∴ x3 – 13x – 12 = (x + 1)(x – 4)(x + 3) ∴ 5 x 3 −65 x −60 =5( x −4)( x + 1)( x +3)

Level 2 77

3

∴ x + 2 is a factor of f(x). By long division, x3 + 16x2 + 52x + 48 = (x + 2)(x2 + 14x + 24) ∴

13. Let f(x) = x3 – 8x2 + 4x + 48. ∵ f (1) =13 −8(1) 2 +4(1) +48 =45 f ( −1) =( −1) 3 −8( −1) 2 +4( −1) +48 =35 f ( 2) = 2

3

−8( 2)

2

More about Polynomials

+4( 2) +48 =32

f ( −2) =( −2) −8( −2) 2 +4( −2) +48 =0 ∴ x + 2 is a factor of f(x). By long division, x3 – 8x2 + 4x + 48 = (x + 2)(x2 – 10x + 24) ∴

x 3 +16 x 2 +52 x +48 =( x +2) 2 ( x +12 )

17. Let f(x) = x3 + x2 – 24x + 36. ∵

3

x 3 −8 x 2 +4 x +48 =( x −6)( x −4)(x +2 )

14. Let f(x) = x3 + 10x2 + 33x + 36. ∵  f (1) =13 +10 (1) 2 +33 (1) +36 =80 f ( −1) =( −1) 3 +10 ( −1) 2 +33 ( −1) +36 =12 f ( 2) =2 3 +10 ( 2) 2 +33 ( 2) +36 =150 f ( −2) =( −2) 3 +10 ( −2) 2 +33 ( −2) +36 =2 f (3) =33 +10 (3) 2 +33 (3) +36 =252 f ( −3) =( −3) 3 +10 ( −3) 2 +33 ( −3) +36 =0 ∴ x + 3 is a factor of f(x). By long division, x3 + 10x2 + 33x + 36 = (x + 3)(x2 + 7x + 12) 3 2 2 ∴ x +10 x +33 x +36 =( x +3) ( x +4)

f (1) =13 +12 −24 (1) +36 =14 f ( −1) =( −1) 3 +( −1) 2 −24 ( −1) +36 =60 f ( 2) =2 3 +2 2 −24 ( 2) +36 =0 ∴ x – 2 is a factor of f(x). By long division, x3 + x2 – 24x + 36 = (x – 2)(x2 + 3x – 18) ∴ x 3 +x 2 −24 x +36 =( x −3)( x −2)( x +6)

18. Let f(x) = x3 – 4x2 – 20x + 48. ∵ f (1) =13 −4(1) 2 −20 (1) +48 =25 f ( −1) =( −1) 3 −4( −1) 2 −20 ( −1) +48 =63 f ( 2) =2 3 −4( 2) 2 −20 ( 2) +48 =0 ∴ x – 2 is a factor of f(x). By long division, x3 – 4x2 – 20x + 48 = (x – 2)(x2 – 2x – 24) ∴ x 3 −4 x 2 −20 x +48 =( x −6)( x −2)( x +4)

19. Let f(x) = x3 + 11x2 + 4x – 60. ∵ f (1) =13 +11 (1) 2 +4(1) −60 =−44 15. Let f(x) = x3 – 12x2 + 47x – 60. ∵  f (1) =13 −12 (1) 2 +47 (1) −60 =−24 f ( −1) =( −1) 3 −12 ( −1) 2 +47 ( −1) −60 =−120 f ( 2) =2 3 −12 ( 2) 2 +47 ( 2) −60 =−6

f ( −1) =( −1) 3 +11 ( −1) 2 +4( −1) −60 =−54 f ( 2) =2 3 +11 ( 2) 2 +4( 2) −60 =0 ∴ x – 2 is a factor of f(x). By long division, x3 + 11x2 +4x – 60 = (x – 2)(x2 + 13x + 30) ∴ x3 + 11 x 2 +4 x −60 =( x −2)( x +3)( x + 10 )

f ( −2) =( −2) 3 −12 ( −2) 2 +47 ( −2) −60 =−210 f (3) =33 −12 (3) 2 +47 (3) −60 =0 ∴ x – 3 is a factor of f(x). By long division, x3 – 12x2 + 47x – 60 = (x – 3)(x2 – 9x + 20) ∴ x3 − 12 x 2 +47 x −60 =( x − 5)( x −4)( x − 3)

16. Let f(x) = x3 + 16x2 + 52x + 48. ∵ f (1) =13 +16 (1) 2 +52 (1) +48 =117 f ( −1) =( −1) 3 +16 ( −1) 2 +52 ( −1) +48 =11 f ( 2) =2 3 +16 ( 2) 2 +52 ( 2) +48 =224 f ( −2) =( −2) 3 +16 ( −2) 2 +52 ( −2) +48 =0

20. Let f(x) = x3 – 27x – 54. ∵  f (1) =13 −27 (1) −54 =−80 f ( −1) =( −1) 3 −27 ( −1) −54 =−28 f ( 2) =2 3 −27 ( 2) −54 =−100 f ( −2) =( −2) 3 −27 ( −2) −54 =−8 f (3) =33 −27 (3) −54 =−108 f ( −3) =( −3) 3 −27 ( −3) −54 =0 ∴ x + 3 is a factor of f(x). By long division,

78

Certificate Mathematics in Action Full Solutions 4A x3 – 27x – 54 = (x + 3)(x2 – 3x – 18) 3 2 ∴ x −27 x −54 =(x −6 )( x +3) 21. Let f(x) = x3 – 15x2 + 72x – 108. ∵

5 x 3 −6 x 2 −29 x +6 =(5 x −1)( x −3)( x +2)

Revision Exercise 3 (p. 154)

f (1) =13 −15 (1) 2 +72 (1) −108 =−50

Level 1

f ( −1) =( −1) 3 −15 ( −1) 2 +72 ( −1) −108 =−1961. f ( 2) =2 3 −15 ( 2) 2 +72 ( 2) −108 =−16 f ( −2) =( −2) 3

3

−15 ( 2)

2

( x 2 − x 4 + 2 x 3 − 4 x + 5) + ( −4 x 3 + x + 2 x 2 + 3 x 4 −8)

+72 ( −2) −108 =−320

= x 2 − x 4 + 2 x 3 − 4 x + 5 − 4 x 3 + x + 2 x 2 + 3 x 4 −8

2

f (3) =3 −15 (3) +72 (3) −108 =0 ∴ x – 3 is a factor of f(x). By long division, x3 – 15x2 + 72x – 108 = (x – 3)(x2 – 12x + 36) ∴ x3 − 15 x 2 +72 x − 108

= −x 4 + 3 x 4 + 2 x 3 − 4 x 3 + x 2 + 2 x 2 − 4 x + x + 5 −8 = 2 x 4 − 2 x 3 + 3 x 2 −3 x −3 2.

=( x −6) 2 (x −3)

(3 x +5 x 2 −8 −4 x 3 ) −(3 x 2 −2 x +2 x 3 −2) = 3 x +5 x 2 −8 −4 x 3 −3 x 2 +2 x −2 x 3 +2

22. Let f(x) = 2x3 – 25x2 + 67x + 40. ∵

= −4 x 3 −2 x 3 +5 x 2 −3 x 2 +3 x +2 x −8 +2

3

2

 1  1  1  1 f  −  = 2 −  − 25 −  + 67  −  + 40  2  2  2  2 1 25 67 =− − − + 40 3. 4 4 2 =0

∴ 2x + 1 is a factor of f(x). By long division, 2x3 – 25x2 + 67x + 40 = (2x + 1)(x2 – 13x + 40) ∴

= 2 x 3 − x 2 + 2 x + 6 x 2 −3 x + 6 = 2 x 3 +5 x 2 − x + 6

( x 3 +3 x 2 −2 x +1)( 2 x −1) =2 x 4 +6 x 3 −4 x 2 +2 x −x 3 −3 x 2 +2 x −1

3

=2 x 4 +5 x 3 −7 x 2 +4 x −1

2

 2  2  2  2 f  −  = 3 −  + 8 −  − 68 −  − 48 3 3 3        3 8 32 136 =− + + − 48 9 9 3 =0

5.

x (3 −2 x )( 4 −3 x ) +(5 −2 x 2 +4 x 3 ) −(8 −5 x +3 x 2 ) = x (−2 x +3)( −3 x +4) +( 4 x 3 −2 x 2 +5) − (3 x 2 −5 x +8)

∴ 3x + 2 is a factor of f(x). By long division, 3x3 + 8x2 – 68x – 48 = (3x +2)(x2 + 2x – 24) ∴

= ( −2 x 2 +3 x )( −3 x +4) +(4 x 3 −2 x 2 +5) − (3 x 2 −5 x +8)

3 x 3 +8 x 2 −68 x −48 =( x −4)( x +6)(3 x +2 )

= 6 x 3 −17 x 2 +12 x + 4 x 3 −2 x 2 +5 −3 x 2 +5 x −8 =10 x 3 −22 x 2 +17 x −3

24. Let f(x) = 5x3 – 6x2 – 29x + 6. 2

1 1 1 1 f   = 5  − 6  − 29   + 6 5 5 5       5 1 6 29 = − − +6 25 25 5 =0

∴ 5x – 1 is a factor of f(x).

By long division, 5x3 – 6x2 – 29x + 6 = (5x – 1)(x2 – x – 6) ∴

79

= ( 2 x 2 − x + 2)( x ) + ( 2 x 2 − x + 2)( 3)

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

23. Let f(x) = 3x3 + 8x2 – 68x – 48. ∵

∵

( 2 x 2 − x + 2)( x +3)

4.

2 x 3 −25 x 2 +67 x +40 =( x −8)( x −5)(2 x + 1)

3

= −6 x 3 +2 x 2 +5 x −6

6.

3y 4y2

1 2 y 3 +0 y 2 + 8y 12 y 3 8y

∴ The quotient is 3y and the remainder is 8y.

3

1400  =f   2  − 6 = f (700 ) 2 z +

7. z z

More about Polynomials

2

2 + 2 z 3 + 0 2 z − 699 z 701+

=+ (700 −6992 ) −(700 −701 ) z

3

z

2

699 701z 2 − 2 + 2 z

=1

−( −1)

− 2 z 2 − 4 z =1 +1 6 z − 6 z − =2 −

∴ The quotient z2 – 2z + 6 and the remainder is –15.

f (1) =14 +29 (1) +6 =1 +29 +6

14. (a)

8. p

3

p − 2 p

4

p

4

=36

− p 2

≠0 − 3 p3 + 2 p − 3 ∴ xp – 1 is2not a+ factor0 of f(x). − 2 p3

f (−2 3) − p3 + 2 p

=( −3) 4 +29 ( −3) +6

− p + 2 p

=81 −87 +6

(b)

3

2

− 3

=0

∴ x + 3 is a factor of f(x).

∴ The quotient is p3 – p2 and the remainder is –3. 15. (a)

9.

2 f ( −2 ) =( −2+ ) 3 −11 ( −2 2) 2 +32 ( −2) − 28 r − 2

2r 2r

=−8 −44 −64 −28

+ 1 4r 3 + 6r 2 − 2r 4r

3

2

+ 2 r =−1 44 2

− 2r

4r

≠0 4r ∴ x + 2 is not a factor of f(x).

∴ The quotient 2r2 + 2r – 2 and the remainder is 3. 10. Let f(x) = 4x3 – 4x2 + x – 1.

= f (1) Remainder =4(1) 3 −4(1) 2 +1 −1 =0 11. Let f(x) = x3 – 2x2 + 7x + 1.

= f ( −1) 3 2 Remainder =( −1) −2( −1) +7( −1) +1 =−1 −2 −7 +1 =−9

(b)

Remainder

=0

∴ x – 2 is a factor of f(x). 16. Let f(x) = x3 + kx2 – 5x + 6. ∵ x3 + kx2 – 5x + 6 is divisible by x – 2. ∴ By the converse of the factor theorem. f ( 2) = 0 2 3 + k ( 2) 2 −5( 2) +6 = 0 8 + 4 k −10 +6 = 0 k = −1 17. Let f(x) = kx3 – 3x2 – 8kx + 9. By the remainder theorem, we have

3

f ( −3) = 3

k ( −3) −3( −3) −8k ( −3) +9 −27 k −27 + 24 k +9 3k k 3

 1  1 =16  −  − 2 −  +1  2  2 = −2 +1 +1 =0

+ 2r

− 4r − 4r

f ( 2) = 2 3 −11 ( 2) 2 +32 ( 2) −28 =8 −44 +64 −28

12. Let f(x) = 16x3 – 2x + 1.

 1 = f −   2

2

2

=3 =3 = −21 = −7

18. Let f(x) = x3 – 2x2 – 7x – 4. ∵ f (1) =13 −2(1) 2 −7(1) −4 =−12

13. Let f(x) = (x – 699)699 – (x – 701)701. Remainder

f ( −1) =( −1) 3 −2( −1) 2 −7( −1) −4 =0 ∴ x + 1 is a factor of f(x).

80

Certificate Mathematics in Action Full Solutions 4A By long division, x3 – 2x2 – 7x – 4 = (x + 1)(x2 – 3x – 4) 3 2 2 ∴ x −2 x −7 x −4 =(x −4 )( x +1) 3

2

3

–x + 4. 24.

− 3x +

2

19. 2x + 14x + 32x + 24 = 2(x + 7x + 16x + 12) Let f(x) = x3 + 7x2 + 16x + 12. ∵

− x 2 − 4x + 5 3x

3

+

3x

3

+

−

−

f (1) =13 +7(1) 2 +16 (1) +12 =36 f ( −1) =( −1) 3 +7( −1) 2 +16 ( −1) +12 =2 ∴ The quotient is –3x + 12 and the remainder is 48x – 52.

f ( 2) =2 3 +7( 2) 2 +16 ( 2) +12 =80

f ( −2) =( −2) 3 +7( −2) 2 +16 ( −2) +12 =0 25. ∴ x + 2 is a factor of f(x). By long division, x3 + 7x2 + 16x + 12 = (x + 2)(x2 + 5x + 6) ∴ x3 + 7x2 + 16x + 12 = (x + 2)2(x + 3) 2x ∴

2x − 4 2

+ x − 1 4x3

− 6

4x3

+ 2

2 x 3 +14 x 2 +32 x +24 =2( x +2) 2 ( x +3)

− 8x

− 8x

20. Let f(x) = x3 – kx2 + x + c. ∵ x + k is factor of x3 – kx2 + x + c. ∵ By the converse of the factor theorem,

∴ The quotient is 2x – 4 and the remainder is 8x – 9.

f (−k ) = 0 ( −k ) 3 − k ( −k ) 2 + ( −k ) + c = 0

26. Let f(x) = 3x3 + 16x2 + x + c. By the remainder theorem, we have

c = 2k 3 + k

 1 f  −  = −5  3

∴ k = 2, c = 18 or k = 1, c = 3 or k = −1, c = −3. (or any other reasonable answers) 3

2

 1  1  1 3 −  +16  −  +  −  + c = −5  3  3  3 1 16 1 − + − + c = −5 9 9 3 19 c =− 3

21. (5x3 + ax + 1) + (bx2 + 7x + 3) = 5 x 3 +ax +1 +bx 2 +7 x +3

= 5 x 3 +bx 2 +( a +7) x +4 ∵ The coefficient of x3 is 5. ∴ His answer is unreasonable.

Level 2

27. By the remainder theorem, we have f (1) = g (1)

22.

13 −2(1) 2 + p (1) −4 =3(1) 3 −12 +1 +1 1 −2 + p −4 =3 −1 +1 +1 2

x − 4 x

2

− 2x − 1 x2 − 6x + 3x − 4 x3

p =9

− 2x2 − x 2 − + 4 28.4 (a) x Let f(x) = x(x – 1)(x4 – 2) x – (x – − 3)(x + 20).

− 4x 8x + 4 ∵ 2 +

− 4x − 8

∴ The quotient is x – 4 and the remainder –4x – 8.

f ( 4) = 4( 4 −1)( 4 −2) −( 4 −3)( 4 +20 ) = 4(3)( 2) −1( 24 ) =0

23. 2 x x

2

∴ x – 4 is a factor of x(x – 1)(x – 2) 2 4 – (x –− 3)(x + 20). x +

− 2 2 x 4 − x 3 + 0 x 2 2 x

− 4 x 2 (b) x(x –1)(x – 2) – (x – 3)(x + 20)

4

− x 3 2 + 4 x 2 2

= ( x − x )( x −2) −( x +17 x −60 )

− x 3 4 = x 3 −3 x 2 + 2 x − x 2 −17 x + 60

= x 3 −4 x 2 −15 x +60

∴ The quotient is 2x2 – x + 4 and the remainder is

81

By long division, x3 – 4x2 – 15x + 60 = (x – 4)(x2 – 15)

x

2

4 x

2

−

−

3 x ( x −1)( x −2) −( x −3)( x +20 )

∴

=( x −4)( x 2 −15 )

29. Let f(x) = x3 + 5x2 – 2x – 24. ∵ f (1) =13 +5(1) 2 −2(1) −24 =−20

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By substituting a = 3 into (1), we have 3 −b = 5 b = −2 33. (a) Let f(x) = 2x3 – 9x2 + px – 6. ∵ 2x – 3 is factor of 2x3 – 9x2 + px – 6. ∴ By the converse of the factor theorem,

3 f =0 2

f ( −1) =( −1) 3 +5( −1) 2 −2( −1) −24 =−18 f ( 2) =2 3 +5( 2) 2 −2( 2) −24 =0 ∴ x – 2 is a factor of f(x). By long division, x3 + 5x2 – 2x – 24 = (x – 2)(x2 + 7x + 12) ∴

3

x 3 +5 x 2 −2 x −24 =( x −2)( x +3)( x +4)

30. Let f(x) = x3 – 7x2 – 6x + 72. ∵ f (1) =13 −7(1) 2 −6(1) +72 =60 f ( −1) =( −1) 3 −7( −1) 2 −6( −1) +72 =70 f ( 2) = 2 3 −7 ( 2) 2 −6( 2) +72 = 40 f ( −2) =( −2)

3

−7 ( −2)

3

f (3) =3 −7 (3)

2

2

(b) By long division, 2x3 – 9x2 + 13x – 6 = (2x – 3)(x2 – 3x + 2) ∴

2 x 3 −9 x 2 + 13 x −6 =( x −2)( x − 1)(2 x −3)

−6( −2) +72 = 48

−6(3) +72 =18

f ( −3) =( −3) 3 −7 ( −3) 2 −6( −3) +72 =0 ∴ x + 3 is a factor of f(x). By long division, x3 – 7x2 – 6x + 72 = (x + 3)(x2 – 10x + 24) ∴ x 3 −7 x 2 −6 x +72 =( x −6)( x −4)(x +3)

31. Let f(x) = 8x2 + 2x – 1. ∵ 8x2 + 2x – 1 is divisible by 2x – k. ∴ By the converse of the factor theorem,

k  f   =0 2

34. Let f(x) = 2x3 – x2 – ax + b and g(x) = bx3 – 2x2 – x + a. ∵ x – 1 is a common factor of 2x3 – x2 – ax + b and bx3 – 2x2 – x + a. ∴ By the converse of the factor theorem,

f (1) = 0

3

2(1) −1 −a (1) +b = 0

(1) a −b =1 …… g (1) = 0

b(1) 3 −2(1) 2 −1 +a = 0

2

By substituting a = 2 into (1), we have 2 −b =1 b =1

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

or or

2k – 1 = 0 1 k = 2

35. (a) By the remainder theorem, we have

f (3) = g (3) 4(3) +3(3) +5k (3) +15 = −(3) 3 +k (3) 2 +46 (3) −21 15 k +150 =9k +90 6k = −60 k = −10 3

32. Let f(x) = x3 – 2x2 + ax + b When f(x) is divided by x + 1,

f ( −1) = −8 ( −1) −2( −1) +a ( −1) +b = −8 (1) a −b =5 …… 3

2

When f(x) is divided by x – 2,

(1) + (2), 3a =9 a =3

2

(b)

f ( 2) = 4

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

2

(2) a +b =3 …… (1) + (2), 2a = 4 a =2

k  k  8  + 2  −1 = 0 2  2 

k+1=0

2

3 3 3 2  − 9  + p  − 6 = 0 2 2 2 27 81 3 p − + −6 = 0 4 4 2 3 p 39 = 2 2 p = 13

…… (2)

f ( x ) − g ( x ) = 4 x 3 + 3 x 2 +5( −10 ) x +15 − ( −x 3 −10 x 2 + 46 x − 21 ) = 5 x 3 +13 x 2 −96 x +36

3

Let h(x) = 5x + 13x2 – 96x + 36.

82

Certificate Mathematics in Action Full Solutions 4A 1 =3 −a a =2

∵ 3

2

2 2 2 2 h  = 5  +13  − 96   + 36 = 0 5 5 5       5 ∴ 5x – 2 is a factor of h(x). By long division, 5 x 3 +13 x 2 −96 x +36

= (5 x −2)( x 2 +3 x −18 )

38. Let f(x) = x99 + 1. (a) By the remainder theorem, = f ( −1) 99 remainder =( −1) +1 =−1 +1

∴

=0

f ( x ) −g ( x ) =( x −3)(5 x −2)( x +6)

(b) Let Q(x) be the quotient when f(x) is divided by x + 1. 36. (a)

f ( x) = ( x +1)Q ( x )

f ( x + 1) = ( x + 1) 3 + k ∴ When f(x + 1) is divided by x + 1, the remainder is k. ∴ k =1

(b)

f ( 2 x ) = ( 2 x) 3 +1 =8 x 3 +1 Let g(x) = 8x3 + 1. By the remainder theorem, =g ( 2) remainder =8( 2) 3 +1

x 99 = ( x +1)Q ( x ) −1

 (1)

By substituting x = 6 into (1), we have 6 99 =(6 +1)Q (6) −1 =7Q (6) −1 ∴ If today is Monday, the day after 699 days is Sunday.

= ( −1) n + 2( −1) +3

= −1 − 2 +3 ( n is a positive

f ( x −1) = ( x −1) + a ( x −1) +b

odd integer

)

=0

= x − 2 x +1 + ax − a +b 2

= x 2 +( a −2) x +1 −a +b Let g(x) = x2 + (a – 2)x + 1 – a + b. By the remainder theorem, we have g ( 2) = 4

2 + (a − 2)( 2) +1 − a + b = 4 1 + a +b = 4 …… (1) b =3 −a

(b) By the remainder theorem, when f(x) is divided by x – 1, remainder

2

f ( x +1) = ( x +1) 2 + a ( x +1) +b

Let h(x) = x2 + (a + 2)x + 1 + a + b. ∵ x2 + (a + 2)x + 1 + a + b is divisible by x + 2. ∴ By the converse of the factor theorem, h(−2) = 0

…… (2) (c) (1) + (2),

2b =2 b =1

By substituting b = 1 into (1), we have

=1n +2(1) +3

Let Q(x) and ax + b the quotient and the remainder respectively when f(x) is divided by x2 – 1. f ( x ) = ( x 2 −1)Q ( x ) + ( ax + b)

= ( x +1)( x −1)Q ( x ) + ( ax + b) …… (1) By substituting x = 1 into (1), we have

f (1) =(1 +1)(1 −1)Q (1) +[ a (1) +b] 6 =a +b

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

= f (1) =6

= x 2 + 2 x +1 + ax + a +b = x 2 +( a + 2) x +1 + a +b

83

+1 = ( x +1)Q ( x )

= f ( −1)

2

(b)

(By (a))

39. (a) By the remainder theorem, remainder

=65

37. (a)

x

= ( x + 1)( x + 1) 2 + k

99

…… (2) By substituting x = –1 into (1), we have

f ( −1) = (−1 +1)( −1 −1)Q (−1) +[a ( −1) +b ] 0 = −a +b ……(3) (2) + (3), 2b = 6 b=3 By substituting b = 3 into (3), we have

3

0 = −a + 3

By long division, 3 x 3 +7 x 2 −2 x −8 = (3 x 2 + x −4)( x +2) = (3 x +4)( x −1)( x + 2)

a =3

∴ The remainder when f(x) is divided by x2 – 1 is 3x + 3. 2

∴

f ( x ) =( x − 1)( x +2)( 3 x +4)

2

40. (a) Let f(x) = 2x + px + q and g(x) = 2x + qx + p. ∵ x – r is a common factor of f(x) and g(x). ∴ By the converse of the factor theorem,

5. Answer: C By the remainder theorem, we have f ( −k ) = k

f ( r ) = g (r )

( −k +2)( −k −3) +2 = k

2r + pr + q = 2r + qr + p 2

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2

k 2 +k −6 +2 = k k 2 = 4 or –2 (rejected) k =2

( p − q)r = p − q r = 1 ( p and p are distinct real numbers)

6. Answer: A Let Q(x) be the quotient when P(x) is divided by 4x – 1. ∴ P ( x ) = ( 4 x −1)Q ( x ) + R

(b) By (a), put p = 3, q = –5 and r = 1. The required polynomial is 2x2 – 5x + 3 or (x – 1)(x + 5), i.e. x2 + 4x – 5. (or any other reasonable answers) 7.

Multiple Choice Questions (p. 156) 1. Answer: A Let f(x) = x3 – x2 + 2x + 1. By the remainder theorem, = f ( 2)

8.

3 2 remainder =2 −2 +2( 2) +1 =8 −4 +4 −1

= (1 − 4 x )[ −Q ( x )] + R ∴ When P(x) is divided by 1 – 4x, the remainder is R. Answer: B ∵ Q(x) is divisible by x + 1. ∴ Q(x – 1) is divisible by (x – 1) + 1 = x. Answer: A Let f(x) = 2x3 – ax2 + bx + 3. ∵ x + 3 is factor of 2x3 – ax2 + bx + 3. ∴ By the converse of the factor theorem, f ( −3) = 0

2( −3) 3 −a ( −3) 2 +b( −3) +3 = 0 −54 −9a −3b +3 = 0 −3a −b =17

=9 2.

Answer: C ∵ f(x) is divisible by x + 1. ∴ By the converse of the factor theorem, f ( −1) =0

( −1) 2005 −( −1) +k =0 −1 +1 +k =0 k =0

3.

Answer: D f ( −1) =0 ∵

3( −1) 2 − p ( −1) +1 =0 ∴ 3 + p +1 =0 p = −4 f ( x ) =3 x 2 +4 x +1 ∴

4.

f (1) =3(1) 2 +4(1) +1 =3 +4 +1 =8

Answer: B Let Q(x) and ax + b be the quotient and the remainder respectively when P(x) is divided by x2 – 1. 2 ∴ P( x ) = ( x −1)Q( x) + (ax + b) = ( x +1)( x −1)Q( x ) + ( ax + b)

…… (1) By substituting x = 1 into (1), we have

P (1) =(1 +1)(1 −1)Q (1) +[ a (1) +b] 1 = a +b …… (2) By substituting x = –1 into (1), we have

P ( −1) =( −1 +1)( −1 −1)Q ( −1) +[ a ( −1) +b] 3 =−a +b ……(3)

(2) + (3), 4 = 2b

b =2

Answer: A ∵

9.

 

f(1) = 0 and f  −

4  =0 3

∴ By the factor theorem, x – 1 and 3x + 4 are factors of f(x). (x – 1)(3x + 4) = 3x2 + x – 4

By substituting b = 2 into (2), we have

1 = a +2 a = −1

∴ The remainder when P(x) is divided by x2 – 1 is –x + 2.

84

Certificate Mathematics in Action Full Solutions 4A

10. Answer: D ∵ f(x) is divisible by x + 2 and 2x – 1. ∴ By the converse of the factor theorem,

f ( −2) = 0

2( −2) +a ( −2) +b( −2) +4 = 0 2a −b = 6 3

2

…… (1)

1 f =0  2 3

2

1 1 1 2  + a  + b  + 4 = 0 …… (2)  2  2  2 a = − 2b − 17

By substituting (2) into (1), we have 2( −2b −17 ) −b = 6 5b = −40 b = −8

By substituting b = –8 into (1), we have 2a − ( −8) = 6

a = −1

HKMO (p. 157) Let f(x) = x3 + kx2 + 3. By the remainder theorem, we have f ( −3) = f ( −1) −2

( −3) 3 +k ( −3) 2 +3 =( −1) 3 +k ( −1) 2 +3 −2 9k −24 =k 8k =24 k =3

85