Chapter 03 More About Polynomials

Certificate Mathematics in Action Full Solutions 4A

3 More about Polynomials • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Activity

4.

∵

Activity 3.1 (p. 143) ∴ 3

1.

2.

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

2

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

(b)

∵

Yes. Since f (3) = 0 , i.e. when f(x) is divided by x − 3 , the remainder is 0, it means that f(x) is divisible by x − 3 , so x − 3 is a factor of x3 − 2 x 2 − 5 x + 6 .

∴ ∵

f (−2) = ( −2)3 − 2(−2) 2 − 5(−2) + 6 = −8 − 8 + 10 + 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 , therefore x + 1 is

∴ ∵

∴

not a factor of x 3 − 2 x 2 − 5 x + 6 . ∵

Activity 3.2 (p. 149) 1.

∴

f ( x ) = ( x + a )( x 2 + bx + c) = x 2 + ax 2 + bx 2 + abx + cx + ac

∵

= x 3 + ( a + b) x 2 + ( ab + c) x + ac By comparing the coefficients, ac = 6 ∴ The product of a and c is 6. 2.

∴ ∵

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

∴ ∴

3. √

x −1

√

x +1

√

x−2

√

x+2

√

x−3

√

x+3

√

59

x−4

x+4

x−5

x+5

x−6

√

f (1) = 13 − 2(1) 2 − 5(1) + 6 = 1− 2 − 5 + 6 =0 x − 1 is a factor of f(x).

x+6

5.

∵ ∴

= −1 − 2 + 5 + 6 =8 ≠0 x + 1 is not a factor of f(x).

f (2) = 2 3 − 2( 2) 2 − 5( 2) + 6 = 8 − 8 − 10 + 6 = −4 ≠0 x − 2 is not a factor of f(x). f (−2) = (−2) 3 − 2(−2) 2 − 5( −2) + 6 = −8 − 8 + 10 + 6 =0 x + 2 is a factor of f(x). f (3) = 33 − 2(3) 2 − 5(3) + 6 = 27 − 18 − 15 + 6 =0 x − 3 is a factor of f(x). f (−3) = ( −3)3 − 2( −3) 2 − 5( −3) + 6 = −27 − 18 + 15 + 6 = −24 ≠0 x + 3 is not a factor of f(x). f (6) = 63 − 2(6) 2 − 5(6) + 6 = 216 − 72 − 30 + 6 = 120 ≠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 ≠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 x + 2 . f ( x ) = ( x − 3)( x − 1)( x + 2)

4 p. 132

Follow-up Exercise

(5 x 3 + 4 x ) ÷ x

p.130 (5 − x − 4 x 2 ) + (7 x 3 − x + 1) 1.

More about Polynomials

1.

(a)

= 5 − x − 4x + 7 x − x + 1 2

3

= 7 x3 − 4 x 2 − x − x + 5 + 1 = 7 x3 − 4 x 2 − 2 x + 6

5x3 + 4 x x 5x3 4 x = + x x = 5x2 + 4 =

(8 x 3 + 4 x 2 + 4 x) ÷ 2 x

Alternative Solution − 4x2 − x + 5

(b)

+) 7 x3 + 0 x 2 − x + 1 7 x3 − 4 x 2 − 2 x + 6

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

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

3x − 2

= 3x3 + 7 x 2 − 2 x + 1 − x3 + 2 x + 5 = 3x − x + 7 x − 2 x + 2 x + 1 + 5 3

3

2

2.

= 2x + 7 x + 6 3

2

(a)

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

x 3x 2 − 2 x 3x − 2x − 2x 3x 2 + 4 x + 3

−) x 3 + 0 x 2 − 2 x − 5

2 x 6 x3 + 8 x 2 + 6 x − 1

2 x3 + 7 x 2 + 0 x + 6 ∴ The answer is 2x3 + 7x2 + 6.

6 x3 8x 2

(b)

( 2 x 3 + 4 x − 1)(2 − 3 x)

8x 2

= (2 x 3 − 4 x − 1)(−3 x + 2) 3.

6x 6x

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

−1

= −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 4

3

p.135

2

= − 6 x + 4 x − 12 x + 11x − 2

x−1 x − 1 x2 − 2x + 3

Alternative Solution 2x3 + 0x 2 + 4x −1 − 3x + 2

×) 4

3

2

+ 3x

3

2

+ 8x − 2

− 6 x + 0 x − 12 x +)

4x + 0x

x2 − x

1.

−x+3 − x +1 2 ∴

− 6 x 4 + 4 x 3 − 12 x 2 + 11x − 2

The quotient is x − 1 and the remainder is 2. 3x + 2

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

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

2x + 1 6x 2 + 7 x − 5

3

6 x 2 + 3x 4x − 5 4x + 2

2.

= ( 4 x − x 2 ) + ( x 2 + x)( x) + ( x 2 + x)(−1) − ( x 3 − 2) = ( 4 x − x 2 ) + x 3 + x 2 − x 2 − x − ( x 3 − 2) = 4 x − x 2 + x3 − x − x3 + 2 = x3 − x3 − x 2 + 4 x − x + 2

−7 ∴

The quotient is 3 x + 2 and the remainder is -7.

= − x 2 + 3x + 2

60

Certificate Mathematics in Action Full Solutions 4A x 2 − 7 x + 23 (c)

x + 3 x3 − 4 x 2 + 2 x + 1

1 = f  3

x3 + 3x 2 − 7 x2 + 2x

3.

3

− 7 x − 21x 23 x + 1 23 x + 69 − 68 The quotient is x 2 − 7 x + 23 and the remainder is −68.

(d) Let f ( x ) = x 4 + x 2 + 3 .  1 = f −   2

2x − 5

4

4 x3 + 2 x 2 + 2 x − 10 x 2 − 2 x + 7 − 10 x 2 − 5 x − 5 ∴

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

3.

p.141 1.

2.

Remainder = f (1)

(b)

Remainder = f (−2)

(c)

1 Remainder = f   5

(d)

 Remainder = f  − 

(e)

Remainder = f (0)

(f)

Remainder = f (2)

(a)

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

When f(x) is divided by

x−2 f ( 2) = 28

,

3( 2)3 + 4(2) 2 + a (2) + b = 28 2a + b = −12 …… (2) (2) − (1), 3a = −15 a = −5

3  2

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

p.146

= −3

1.

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

3 2 Remainder = 2(−2) − 3(−2) + (−2) + 1 = −16 − 12 − 2 + 1 = − 29

61

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

(a)

3

2

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

2 x + x + 1 4 x3 − 8 x 2 + 0 x + 7 2

4.

2

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

2

∴

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

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

f (1) = 13 − 4(1) 2 − 7(1) + 10 = 1 − 4 − 7 + 10 =0 ∴ x − 1 is a factor of x 3 − 4 x 2 − 7 x + 10 .

f (2) = 23 − 4( 2) 2 − 7(2) + 10 (b) = 8 − 16 − 14 + 10 = −12 ≠0 ∴ x − 2 is not a factor of x 3 − 4 x 2 − 7 x + 10 .

4

f (−1) = ( −1) − 4(−1) − 7( −1) + 10 = −1 − 4 + 7 + 10 = 12 ≠0 ∴ x + 1 is not a factor of x 3 − 4 x 2 − 7 x + 10 . 3

(c)

(d)

2.

2

(a)

∵

∴ (b)

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 .

f (3) = 33 − 5(3) 2 + 8(3) − 6 = 27 − 45 + 24 − 6 =0 x − 3 is a factor of f(x).

By long division, x2 − 2x + 2 x − 3 x3 − 5x 2 + 8 x − 6 x3 − 3x 2 − 2x2 + 8x − 2x2 + 6x 2x − 6 2x − 6

Let f ( x ) = 2 x3 − 5 x 2 − x + 1 . (a)

(b)

f (3) = 2(3) 3 − 5(3) 2 − 3 + 1 = 54 − 45 − 3 + 1 =7 ≠0 ∴ x − 3 is not a factor of 2 x 3 − 5 x 2 − x + 1 . f (−4) = 2(−4)3 − 5( −4) 2 − (−4) + 1 = −128 − 80 + 4 + 1 = −203 ≠0 ∴ x + 4 is not a factor of 2 x 3 − 5 x 2 − x + 1 . 3

(c)

(d)

2 Hence, f ( x ) = ( x − 3)( x − 2 x + 2)

p.152 1.

∴

x − 1 x3 + 2 x 2 − x − 2 x3 − x 2 3x 2 − x 3x 2 − 3 x 2x − 2 2x − 2

2

∴

2.

∵ ∴

f (1) = 13 + 2(1) 2 − 15(1) − 36 = −48 ≠ 0 f ( −1) = (−1) 3 + 2(−1) 2 − 15( −1) − 36 = −20 ≠ 0 ∵

f (3) = 33 + 2(3) 2 − 15(3) − 36 = −36 ≠ 0

2

3

2

f (2) = 2 3 + 2(2) 2 − 15(2) − 36 = −50 ≠ 0 f (−2) = (−2) 3 + 2(−2) 2 − 15(−2) − 36 = −6 ≠ 0

2

4 x − kx + 9 x − 2 is divisible by 4 x − 1 . By the converse of the factor theorem, 1 f =0 2 3

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

Let f ( x ) = x 3 + 2 x 2 − 15 x − 36 .

Let f ( x ) = 4 x − kx + 9 x − 2 . 3

f (1) = 13 + 2(1) 2 − 1 − 2 = 0 x − 1 is a factor of f(x).

By long division, x 2 + 3x + 2

2

1 1 1 1 f   = 2  − 5  −   + 1 2  2 2 2 1 5 1 = − − +1 4 4 2 1 =− 2 ≠0 ∴ 2 x − 1 is a factor of 2 x 3 − 5 x 2 − x + 1 .

Let f ( x ) = x 3 + 2 x 2 − x − 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 2 x 3 − 5 x 2 − x + 1 . 3

3.

4.

More about Polynomials

∴

f ( −3) = (−3) 3 + 2(−3) 2 − 15( −3) − 36 = 0 x + 3 is a factor of f(x).

By long division,

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

62

Certificate Mathematics in Action Full Solutions 4A f (1) = 13 + 12 − 8(1) − 12 = −18 ≠ 0

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

∵

f (2) = 23 + 2 2 − 8( 2) − 12 = −16 ≠ 0

x3 + 3 x 2 − x 2 − 15 x − x2

∴

− 3x − 12 x − 36 − 12 x − 36

∴

∴

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

6.

∴

f (1) = 13 − 7(1) + 6 = 0 x − 1 is a factor of f(x).

x3 − x 2 x −7x

f (1) = 2(1) 3 + 7(1) 2 − 44(1) + 35 = 0 x − 1 is a factor of f(x).

By long division, 2 x 3 + 7 x 2 − 44 x + 35 = ( x − 1)(2 x 2 + 9 x − 35) = ( x − 1)(2 x − 5)( x + 7)

By long division, x2 + x − 6 x − 1 x3 + 0 x 2 − 7 x + 6

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

Let f ( x ) = 2 x3 + 7 x 2 − 44 x + 35 . ∴

Let f ( x ) = x 3 − 7 x + 6 . ∵

f (−2) = ( −2) 3 + ( −2) 2 − 8( −2) − 12 = 0 x + 2 is a factor of f(x).

By long division, x 3 + x 2 − 8 x − 12 = ( x + 2)( x 2 − x − 6) ∴ x 3 + x 2 − 8 x − 12 = ( x − 3)( x + 2) 2

∴ 3.

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

Exercise

2

Exercise 3A (p.130)

x2 − x − 6x + 6 − 6x + 6 ∴

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

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

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

4.

Let f ( x) = x3 + 3 x 2 − x − 3 . ∵ ∴

f (1) = 13 + 3(1) 2 − 1 − 3 = 0 x − 1 is a factor of f(x).

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

By long division, x2 + 4x + 3 x − 1 x3 + 3x 2 x −x 3

4x −

− x−3

(2 x − 3 x 2 + 1) − (3 x 2 − 4 x + 1)

x

3.

4x − 4x

= − 6x 2 + 6x

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

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

2

∴

= 6 x 3 − 2 x3 − 4 x 2 − 3 x + 2 x + 2 − 1 = 4 x3 − 4 x 2 − x + 1

2 2

= 6 x3 − 3x + 2 − 2 x3 − 4 x 2 + 2 x − 1

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

= 3x 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 = − 2x3 − x 2 − x + 1

5.

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

63

4

5.

More about Polynomials

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

( x 2 + 2 x − 1)(3 x 2 − 5 x − 2)

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

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

= 4 x3 + 3x 2 + x + 4 x 2 + 3x + 1

12.

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

2

2

= 4 x3 + 7 x 2 + 4 x + 1

( x 2 + 2 x − 1)(−2) = 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

( x + 3)( x − 2 x ) 2

6.

= ( x + 3)( x 2 ) + ( x + 3)(−2 x )

5 P − ( x − 1)Q

= x + 3x − 2 x − 6 x

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

3

2

2

= x3 + x 2 − 6 x

7.

13.

= 5 x 2 − 15 x − 25 − 3 x3 + 3 x 2 + 5 x − 5

(3 x − 2)(2 x 2 − 4 x − 1)

= −3 x3 + 5 x 2 + 3 x 2 − 15 x + 5 x − 25 − 5

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

= − 3 x 3 + 8 x 2 − 10 x − 30

= 6 x − 4 x − 12 x + 8 x − 3 x + 2 3

2

2

(3 − x )(Q − P )

= 6 x 3 − 16 x 2 + 5 x + 2

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

( 2 x + 1)( x 2 − x + 1) + (2 x + 1) 2 = ( 2 x + 1)( x 2 ) + ( 2 x + 1)(− x) + ( 2 x + 1)(1) + 8.

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

14.

(4 x 2 + 4 x + 1)

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

= 2 x3 + x 2 − 2 x 2 − x + 2 x + 1 + 4 x 2 + 4 x + 1

= ( − x + 3)(2 x 2 ) + (− x + 3)(3 x)

= 2 x3 + x 2 − 2 x 2 + 4 x 2 − x + 2 x + 4 x + 1 + 1

= −2 x 3 + 6 x 2 − 3 x 2 + 9 x

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

= − 2 x3 + 3x 2 + 9 x Q 2 − PQ = (3 x 2 − 5) 2 − ( x 2 − 3 x − 5)(3 x 2 − 5)

Level 2

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

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

9.

3

2

2

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

3

2

2

15.

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

3

2

2

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

= 2 x3 − 2 x 2 + 9 x − 2

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

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

= 6 x 4 + 9 x 3 − 10 x 2 − 15 x

( x − 2 x + 2 x − 5) 3

10.

2

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

Exercise 3B (p.136) Level 1 x+ 3

= 11x3 − 9 x 2 − 4 x + 11 1.

( 2 x 2 − 3 x − 5)( x 2 − 2) 11.

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

3

2

2

= 2 x 4 − 3 x 3 − 9 x 2 + 6 x + 10

7 x 7 x 2 + 21x 7 x2 21x 21x ∴ The quotient is x + 3 and the remainder is 0.

64

Certificate Mathematics in Action Full Solutions 4A 2d − 1

5x 2 − 3 x 2.

3 x 15 x 3 − 9 x 2

8.

15 x 3

2d − 1 4d 2 − 4d + 1 4d 2 − 2d − 2d + 1 − 2d + 1

− 9x2 − 9x2

∴ The quotient is 2d − 1 and the remainder is 0. ∴ The quotient is 5 x 2 − 3x and the remainder is 0.

3x + 8

a+ 3

2x − 3 6x2 + 7 x + 2

a a 2 + 3a + 2 3.

a

9.

2

16 x + 2 16 x − 24

3a 3a

26 2

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

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

4a − 1

x− 2

4a + 1 16a 2 + 0a − 4

2x 2x − 4x + 3 2

4.

2x

10.

2

16a 2 + 4a − 4a − 4 − 4a − 1

− 4x − 4x

−3 3

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

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

s 2 − 4 s + 11

p+ 2

s + 2 s 3 − 2 s 2 + 3s + 2

p + 1 p + 3p + 5 2

5.

6x2 − 9x

p + p 2

s 3 + 2s 2 11.

2p + 5 2p + 2

− 4 s 2 + 3s − 4 s 2 − 8s 11s + 2 11s + 22

3

− 20

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

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

z−8 z + 3 z 2 − 5z + 6 6.

2h 2 + 3h + 10

z 2 + 3z − 8z + 6 − 8 z − 24 30 ∴ The quotient is z − 8 and the remainder is 30. 2r

7.

r − 2 2 r 2 − 4r + 1 2 r − 4r

h − 3 2h3 − 3h 2 + h − 5 2 h 3 − 6h 2 12.

3h 2 + h 3h 2 − 9h 10h − 5 10h − 30 25

2

1 ∴ The quotient is 2r and the remainder is 1.

65

∴ The quotient is 2h 2 + 3h + 10 and the remainder is 25.

4 2x2 − 4x + 2

x−2

2x + 1 4x − 6x + 0x − 7

x + x + 3 x3 − x 2 + 3 x − 3

3

2

2

4 x3 + 2 x 2 13.

x3 + x 2 + 3 x

17.

− 8x2 + 0 x

− 2x2 + 0x − 3

− 8x2 − 4 x

− 2x2 − 2x − 6 2x + 3

4x − 7 4x + 2

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

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

p 18.

p + 2 p − 1 p3 + 2 p2 − 5 p + 1 2

p3 + 2 p2 − p

3m + 2m + 1 2

− 4 p +1

3m + 2 9m + 12m + 7 m − 5 3

2

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

9 m3 + 6m 2 6m 2 + 7 m

14.

2z + 3

6m + 4m 2

3m − 5 3m + 2

z 2 − 3z − 2 2 z 3 − 3z 2 − z + 4 2z3 − 6z 2 − 4z

19.

3z 2 + 3z + 4

−7

3z 2 − 9 z − 6 12 z + 10

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

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

3w − 2 w + 1 2

2d + 5

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

15.

2

d − 3d − 4 2d 3 − d 2 − 24d − 18 2

− 8w − 2w 2

2d 3 − 6d 2 − 8d

20.

− 8w − 6 w 2

5d 2 − 16d − 18

4w + 3 4w + 3

5d 2 − 15d − 20 −d + 2

∴ The quotient is 3w2 − 2 w + 1 and the remainder is 0.

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

3q − 4

Level 2

2q + 5q − 2 6q 3 + 7q 2 − 23q + 2 2

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

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x − x + 2x 3

2

6x − 4x + 3 2

6 x 2 − 6 x + 12 2x − 9

21.

6q 3 + 15q 2 − 6q − 8q 2 − 17 q + 2 − 8q 2 − 20q + 8 3q − 6 ∴ The quotient is 3q − 4 and the remainder is 3q − 6 .

∴ The quotient is x + 6 and the remainder is 2 x − 9 .

66

Certificate Mathematics in Action Full Solutions 4A 3s + 5

x+3

3s − 2s − 1 9 s + 9 s − 16 s − 4 2

3

x − 3x − 2 x 3 + 0 x 2 + 2 x − 5

2

2

9 s 3 − 6 s 2 − 3s

22.

x3 − 3x 2 − 2 x

26.

15s 2 − 13s − 4

3x 2 + 4 x − 5

15s − 10s − 5

3x 2 − 9 x − 6

− 3s + 1

13 x + 1

2

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

∴

x −1

2x2 + 2x − 1

x + x + 1 x3 + 0 x 2 + 0 x + 1 2

2 x 2 − x + 2 4 x 4 + 2 x3 + 0 x 2 + 6 x − 3 4 x 4 − 2 x3 + 4 x 2 23.

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

x3 + x 2 + x

27.

4 x3 − 4 x 2 + 6 x

− x2 − x + 1

4 x3 − 2 x 2 + 4 x

− x2 −

x −1

− 2x + 2x − 3 2

2

− 2x + x − 2 2

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

x −1

x 2 − 5 x + 20

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

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

2w − 3 4 w2 + w − 2 8w3 − 10 w2 − 9 w + 1

− 5 x3 + 5 x 2 − 3x

28.

− 5 x 3 − 15 x 2 + 5 x

8 w3 + 2 w 2 − 4 w

24.

20 x 2 − 8 x − 5

− 12 w2 − 5w + 1

20 x 2 + 60 x − 20

− 12 w2 − 3w + 6

− 68 x + 15

− 2w − 5 ∴ The quotient is 2 w − 3 and the remainder is −2 w − 5 . 5 3e − 2 2e 2 + 1 6e3 − 5e 2 + 7e + 1 25.

6e 3

+ 3e − 5e 2 + 4e + 1 5 − 5e 2 − 2 4e +

7 2

5 ∴ The quotient is 3e − and the remainder is 2 7 4e + . 2

∴ The quotient is x 2 − 5 x + 20 and the remainder is −68 x + 15 . [(3x 3 + 2 x − 3) − ( x 3 − 4 x 2 + 4)] ÷ ( x 2 − 3 x + 5) 29.

= (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) = ( 2 x 3 + 4 x 2 + 2 x − 7) ÷ ( x 2 − 3 x + 5) By long division, 2 x + 10 x − 3x + 5 2 x 3 + 4 x 2 + 2 x − 7 2

2 x 3 − 6 x 2 + 10 x 10 x 2 − 8 x − 7 10 x 2 − 30 x + 50 22 x − 57 ∴ The quotient is 2 x + 10 and the remainder is 22 x − 57 .

67

4

30.

[(2t − 1)(t 2 + 1)] ÷ (t 2 + 2t + 1)

6.

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

By long division,

t + 2t + 1 2t − t + 2t − 1 2

2

2

7.

− 5t 2 + 0t − 1

3

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

− 5t − 10t − 5 2

10t + 4

8.

3

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

9.

2

3

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

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

3.

Let f ( x ) = 2 x3 − x 2 + 7 x + 1 . = f (−1)

10. Let f ( x ) = 2 x3 + 2 x 2 − 4 x + 7 . 3 = f  2

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

4.

2

5.

3

Let f ( x ) = 2 x3 − 3 x 2 − 4 x + 5 . = f (−3) 3 2 Remainder = 2(−3) − 3(−3) − 4(−3) + 5 = −54 − 27 + 12 − 5 = − 64

2

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

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

Let f ( x ) = 8 x 3 − 2 x + 1 . 1 = f  2

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

2

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

Level 1

2.

Let f ( x ) = 4 x3 − x 2 + 8 x − 11 . 1 = f  4

Exercise 3C (p.142)

1.

Let f ( x ) = 27 x 3 − 6 x + 2 .  1 = f −   3

2t 3 + 4t 2 + 2t

∴ The quotient is 2t − 5 and the remainder is 10t + 4 .

Let f ( x ) = x 3 + 4 x 2 − x − 3 . = f (5) 3 2 Remainder = 5 + 4(5) − 5 − 3 = 125 + 100 − 5 − 3 = 217

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

2t − 5

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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

68

Certificate Mathematics in Action Full Solutions 4A 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 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

x 3 + x 2 + 2ax − 10 . (or any other reasonable answers) 19. By the remainder theorem, we have f ( −1) = 3

k2 −k2 −2 = k k = −2 14. Let f ( x ) = kx 4 + 30 x 3 + 18 x + 20 . By the remainder theorem, we have  1 f  −  = 13  3 4

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

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

3

 1  1  1 4 −  + 30 −  + 18 −  + 20 = 13 3 3      3 k 10 − − 6 + 20 = 13 81 9 k 1 = 81 9 k =9

Level 2 20. Let f ( x ) = x 2 + 2 x + 3 . By the remainder theorem, we have k f   = 11 2 2

15. Let f ( x ) = 2 x3 − 5 x 2 + 4kx − 7 . By the remainder theorem, we have 1 f   = −6 2 3

2

1 1 1 2  − 5  + 4k   − 7 = −6 2 2 2 1 5 − + 2k − 7 = −6 4 4 2k = 2 k =1 16. Let f ( x ) = x1997 − 1 . = f (−1) 1997 Remainder = ( −1) − 1 = −1 − 1 = −2

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

69

k k   + 2   + 3 = 11 2 2 2 k + k + 3 = 11 4 k 2 + 4k − 32 = 0 (k + 8)(k − 4) = 0 k +8 = 0 or k − 4 = 0 k = − 8 or 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 + 2 k − 1 = k 2

k +1 = 0 k = −1

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

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

4 f (− k ) = −6

f (n) = 11

( −k ) 2 − 4( −k ) − 3 = −6

n − mn + 3 = 11

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

n 2 − mn − 8 = 0

2

By substituting (2) into (1), we have ( 2m) 2 − m( 2m) − 8 = 0

2

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

2m 2 − 8 = 0 m = ±2 When m = 2, n = 4 When m = −2, n = −4

k 2 − 2k + 1 = 0 (k − 1) 2 = 0 k =1

∴ The values of m and n are m = 2 m = −2 or  .  n=4  n = −4

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

27. (a)

When f(x) is divided by x + 2 , f ( −2) = −31 2(−2)3 − ( −2) 2 + p ( −2) + q = −31 − 2 p + q = −11 3 p = 9 (1) – (2), p=3

……(2)

By substituting p = 3 into (1), we have 3 + q = −2 q = −5

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

……(1)

……(2) 3a = 24 (1) – (2), a =8 By substituting a = 8 into (1), we have 8 + b = −4 b = −12 ∴ The remainder when P(x) is divided by ( x − 1)( x + 2) is 8 x − 12 .

When f(x) is divided by x − 3 , f (3) = −4 33 − p (3) 2 + 2(3) + q = −4 − 9 p + q = −37 (1) – (2), 5 p = 20 p=4

∵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.

(b) Let Q(x) and ax + b be the quotient and the remainder respectively when P(x) is divided by ( x − 1)( x + 2) . ∴ P ( x) = ( x − 1)( x + 2)Q ( x ) + ax + b By the remainder theorem, we have P(1) = −4 (1 − 1)(1 + 2)Q(1) + a (1) + b = −4 ……(1) a + b = −4

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 g(x) is divided by x − n , g ( n) = 0 2m − n = 0 ……(2) n = 2m

23. Let f ( x ) = 4 x 2 − 2 x + 1 . By the remainder theorem, we have  k f  −  = 3k  2

2(1) 3 − 12 + p(1) + q = −1 p + q = −2

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……(2)

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 ,

28. (a)

By the remainder theorem, we have f ( −1) = 1 ( −1)99 + k = 1 −1 + k = 1 k=2

70

Certificate Mathematics in Action Full Solutions 4A (b)

Let Q(x) be the quotient when f(x) is divided by x +1 . f ( x) = ( x + 1)Q ( x ) + 1 ∴ 99 x + 2 = ( x + 1)Q ( x ) + 1 x

99

9 99 = (9 + 1)Q (9) − 1 = 10Q(9) − 1 = 10[Q (9) − 1] + 9 ∴ The remainder when 999 is divided by 10 is 9.

Exercise 3D (p. 147) Level 1

2.

f (1) = 13 − 12 − 3(1) + 3 = 1 −1 − 3 + 3 =0 ∴ x − 1 is a factor of f(x). f (−1) = (−1)3 − (−1) 2 − 3( −1) + 3 = −1 − 1 + 3 + 3 (b) =4 ≠0 ∴ x + 1 is not a factor of f(x).

f (−4) = (−4)3 + 5(−4) 2 + 3(−4) + 15 = −64 + 80 − 12 + 15 (a) = 19 ≠0 ∴ x + 4 is not a factor of f(x). (b)

f (−5) = ( −5) + 5( −5) + 3( −5) + 15 = −125 + 125 − 15 + 15 =0 ∴ x + 5 is a factor of f(x).

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

71

2

f (4) = 2( 4)3 − 9(4) 2 + 5( 4) − 4 = 128 − 144 + 20 − 4 =0

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 2(−2)3 + k (−2) 2 − (−2) − 6 = 0 − 16 + 4k + 2 − 6 = 0 4k = 20 k =5 Let f ( x ) = x3 − 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 + 10a + 15 = 0 10a = 110 a = 11

8.

f (2) = 2 3 + 2(2) 2 − 5(2) − 6 = 8 + 8 − 10 − 6 =0 ∴ x - 2 is a factor of f(x).

3

4.

6.

(a)

(a)

Let f ( x ) = x3 − a 3 . 3 3 ∵ f (a ) = a − a  =0 ∴ x3 – a3 is divisible by x – a.

7.

f (3) = 33 + 2(3) 2 − 5(3) − 6 (b) = 27 + 18 − 15 − 6 = 24 ≠0 ∴ x - 3 is not a factor of f(x).

3.

5.

= ( x + 1)Q ( x ) − 1 ……(1)

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

1.

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

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 + 32 − 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 P ( x) = ( x 2 + x − 2)(2 x + 3) = 2 x 3 + 3x 2 + 2 x 2 + 3x − 4 x − 6 = 2x3 + 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,

4

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f (1) = 0 3

2

1 + m(1) + n(1) + 5 = 0 1+ m + n + 5 = 0 n = −6 − m ∴ m = –3, n = –3 or m = –1, n = –5 or m = 2, n = –8. (or any other reasonable answers)

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)3 + a (−2) 2 − ( −2) − b = 0 − 8 + 4a + 2 − b = 0 b = 4a − 6 ∴ 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. g (−5) = ( −5) + 12(−5) + 41(−5) + 30 = −125 + 300 − 205 + 30 =0 ∴ x + 5 is a factor of g(x). 3

3

 f −  15. (a) ∵

(b) By long division, x2 + 7 x + 6 x + 5 x3 + 12 x 2 + 41x + 30 x3 + 5 x 2 7 x 2 + 41x 7 x 2 + 35 x 6 x + 30 6 x + 30

3 +9 2

(b) By long division, 2x2 − 7x + 3 2 x + 3 4 x 3 − 8 x 2 − 15 x + 9 4x3 +

6x 2

− 14 x 2 − 15 x − 14 x 2 − 21x 6x + 9 6x + 9 Hence, f(x) = (2x + 3)(2x2 – 7x + 3) By the cross method, 2x2 – 7x + 3 = (2x – 1)(x – 3) ∴ f ( x ) = (2 x + 3)(2 x − 1)( x − 3) 3

2

 4  4  4  4 h −  = 3 −  + 4 −  − 75 −  − 100 3 3 3       3 16. (a) ∵  64 64 =− + + 100 − 100 9 9 =0 ∴ 3x + 4 is a factor of h(x).

(b) By long division, x2 − 25 3 x + 4 3x 3 + 4 x 2 − 75 x − 100 3x 3 + 4 x 2 − 75 x − 100 − 75 x − 100

2

14. (a) ∵

2

3  3  3   = 4 −  − 8 −  − 15 − 2 2 2      27 45 =− − 18 + +9 2 2 =0 ∴ 2x + 3 is a factor of f(x).

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. ∵x3 – 4x2 + kx + 6 is divisible by x – 3. ∴ By the converse of the factor theorem, f (3) = 0 33 − 4(3) 2 + k (3) + 6 = 0 27 − 36 + 3k + 6 = 0 3k = 3 k =1 (b) By long division,

2

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

72

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

x2 − x − 2 x − 3 x3 − 4 x 2 + x + 6 x3 − 3x 2

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

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

20. (a) ∵When f(x) is divided by x + 1, the remainder is 10. ∴ By the remainder theorem, we have f (−1) = 10 2(−1) 3 − (−1) 2 + a ( −1) + b = 10 − 2 − 1 − a + b = 10 …… (1) a − b = −13 ∵x – 1 is a factor of f(x). ∴ By the converse of the factor theorem, f (1) = 0

18. (a) Let f(x) = 8x3 + mx2 – 25x + 6. ∵8x3 + mx2 – 25x + 6 is divisible by 4x – 1. ∴ By the converse of the factor theorem, 1 f =0 4 3

2(1) 3 − 12 + a (1) + b = 0 a + b = −1 2 a = − 14 (1) + (2), a = −7

2

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

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

(b) By long division, 2x 2 + x − 6

x − 1 2x3 − x 2 − 7 x + 6 2x3 − 2x 2

4 x − 1 8 x 3 + 2 x 2 − 25 x + 6

x2 − 7x

8x3 − 2 x 2

x2 − x − 6x + 6 − 6x + 6

4 x 2 − 25 x 4x 2 −

x

− 24 x + 6 − 24 x + 6 Hence, f(x) = (4x – 1)(2x2 + x – 6) By the cross method, 2x2 + x – 6 = (2x – 3)(x + 2) 3 2 ∴ 8 x + 2 x − 25 x + 6 = ( 4 x − 1)(2 x − 3)( x + 2) 3

2

3

2

19. Let f(x) = x + px + qx + 2 and g(x) = x + qx + px – 6. ∵x – 2 is a common factor of x3 + px2 + qx + 2 and x3 + qx2 + px – 6. f ( 2) = 0 ∴ 3 2 2 + p (2) + q (2) + 2 = 0 2 p + q = −5 …… (1) g ( 2) = 0

3

2

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

73

…… (2)

…… (2)

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

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

s (−1) 3 + 9(−1) 2 − t (−1) − 5 = 8

t = s + 4 …… (1)

∵x + 5 is a factor of g(x). ∴ By the converse of the factor theorem, g (−5) = 0 s (−5) 3 + 9(−5) 2 − t ( −5) − 5 = 0 − 25s + t = −44 …… (2) By substituting (1) into (2), we have

4 −25s + s + 4 = −44 24 s = 48 s=2

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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

By substituting s = 2 into (1), we have t = 2+4 =6

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 + 21x + 18 = ( x + 2)( x + 3)

(b) By long division, 2x 2 − x − 1

5.

x + 5 2x + 9x − 6x − 5 3

∵

2 x 3 + 10 x 2 − x 2 − 6x − x−5 − x−5 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)

6.

7.

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

f (2) = 23 + 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.

3.

f (3) = 33 + 3(3) 2 − 25(3) − 75 = −96

4.

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)

2

Let f(x) = x + x – 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) 3 2 ∴ x + x − 10 x + 8 = ( x − 2)( x − 1)( x + 4)

Let f(x) = x3 + 3x2 – 25x – 75. f (1) = 13 + 3(1) 2 − 25(1) − 75 = −96 3 2 ∵ f (−1) = ( −1) + 3( −1) − 25(−1) − 75 = −48

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

Let f(x) = x3 + x2 – 8x – 12. f (1) = 13 + 12 − 8(1) − 12 = −18 3 2 ∵ f (−1) = ( −1) + ( −1) − 8( −1) − 12 = −4

Let f(x) = x3 – 7x – 6. 3 ∵ f (1) = 1 − 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)

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)

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

2.

f (−1) = ( −1) 3 + 3(−1) 2 − 4( −1) − 12 = −6

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

Exercise 3E (p. 152) Level 1

f (1) = 13 + 3(1) 2 − 4(1) − 12 = −12

f ( 2) = 23 + 3( 2) 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 2 − 5x

1.

Let f(x) = x3 + 3x2 – 4x – 12.

2

3 2 ∴ x + 3 x − 25 x − 75 = ( x − 5)( x + 3)( x + 5)

9.

Let f(x) = x3 + 4x2 – 11x – 30.

Let f(x) = x3 + 8x2 + 21x + 18.

74

Certificate Mathematics in Action Full Solutions 4A f (1) = 13 + 4(1) 2 − 11(1) − 30 = −36

f (1) = 13 + 10(1) 2 + 33(1) + 36 = 80

f (−1) = ( −1) 3 + 4( −1) 2 − 11(1) − 30 = −16

∵ 

f (2) = 23 + 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) 3 2 ∴ x + 4 x − 11x − 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 + 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) = ( −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)

15. Let f(x) = x3 – 12x2 + 47x – 60. f (1) = 13 − 12(1) 2 + 47(1) − 60 = −24

3 2 2 ∴ 3 x − 6 x − 12 x + 24 = 3( x − 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) 3 2 ∴ 5 x + 20 x + 5 x − 30 = 5( x − 1)( x + 2)( x + 3)

12. 5x3 – 65x – 60 = 5(x3 – 13x – 12) Let f(x) = x3 – 13x – 12. 3 ∵ f (1) = 1 − 13(1) − 12 = −24 3

f (−1) = ( −1) − 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) 3 ∴ 5 x − 65 x − 60 = 5( x − 4)( x + 1)( x + 3)

∵ 

f (2) = 2 3 − 12( 2) 2 + 47(2) − 60 = −6 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) 3 2 ∴ x − 12 x + 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 3 2 ∵ f (−1) = ( −1) + 16(−1) + 52( −1) + 48 = 11

f (2) = 23 + 16(2) 2 + 52( 2) + 48 = 224 f (−2) = ( −2) 3 + 16(−2) 2 + 52( −2) + 48 = 0 ∴ x + 2 is a factor of f(x). By long division, x3 + 16x2 + 52x + 48 = (x + 2)(x2 + 14x + 24) 3 2 2 ∴ x + 16 x + 52 x + 48 = ( x + 2) ( x + 12) 17. Let f(x) = x3 + x2 – 24x + 36.

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

f (2) = 2 3 − 8(2) 2 + 4( 2) + 48 = 32 f (−2) = ( −2) 3 − 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) 3 2 ∴ x − 8 x + 4 x + 48 = ( x − 6)( x − 4)( x + 2 )

∵

14. Let f(x) = x3 + 10x2 + 33x + 36.

f (1) = 13 + 12 − 24(1) + 36 = 14 f (−1) = ( −1) 3 + (−1) 2 − 24( −1) + 36 = 60

f ( 2) = 23 + 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) 3 2 ∴ x + x − 24 x + 36 = ( x − 3)( x − 2)( x + 6)

18. Let f(x) = x3 – 4x2 – 20x + 48. ∵

75

f (−1) = ( −1) 3 − 12( −1) 2 + 47(−1) − 60 = −120

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

4 ∴ x – 2 is a factor of f(x). By long division, x3 – 4x2 – 20x + 48 = (x – 2)(x2 – 2x – 24)

3

19. Let f(x) = x3 + 11x2 + 4x – 60. f (1) = 13 + 11(1) 2 + 4(1) − 60 = −44 f (−1) = ( −1) 3 + 11(−1) 2 + 4( −1) − 60 = −54

f (2) = 23 + 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) 3 2 ∴ x + 11x + 4 x − 60 = ( x − 2)( x + 3)( x + 10)

2

 2  2  2  2 f  −  = 3 −  + 8 −  − 68 −  − 48 3  3 3    3 ∵ 8 32 136 =− + + − 48 9 9 3 =0 ∴ 3x + 2 is a factor of f(x). By long division, 3x3 + 8x2 – 68x – 48 = (3x +2)(x2 + 2x – 24)

3 2 ∴ x − 4 x − 20 x + 48 = ( x − 6)( x − 2)( x + 4)

∵

More about Polynomials

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

24. Let f(x) = 5x3 – 6x2 – 29x + 6. 3

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).

20. Let f(x) = x3 – 27x – 54. f (1) = 13 − 27(1) − 54 = −80

By long division, 5x3 – 6x2 – 29x + 6 = (5x – 1)(x2 – x – 6) 3 2 ∴ 5 x − 6 x − 29 x + 6 = (5 x − 1)( x − 3)( x + 2)

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, x3 – 27x – 54 = (x + 3)(x2 – 3x – 18) ∴ x − 27 x − 54 = ( x − 6) ( x + 3) 3

2

Revision Exercise 3 (p. 154) Level 1 ( x 2 − x 4 + 2 x 3 − 4 x + 5) + (−4 x 3 + x + 2 x 2 + 3 x 4 − 8) 1.

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

3

2

21. Let f(x) = x – 15x + 72x – 108. f (1) = 13 − 15(1) 2 + 72(1) − 108 = −50 3

∵

f (−1) = ( −1) − 15( −1) + 72( −1) − 108 = −196 3

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

2

2

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

2.

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

2

f (−2) = ( −2) − 15(2) + 72( −2) − 108 = −320 3

= 3x + 5 x 2 − 8 − 4 x 3 − 3x 2 + 2 x − 2 x 3 + 2 = − 6 x3 + 2x 2 + 5x − 6

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)

( 2 x 2 − x + 2)( x + 3) 3.

= 2 x 3 − x 2 + 2 x + 6 x 2 − 3x + 6

3 2 2 ∴ x − 15 x + 72 x − 108 = ( x − 6) ( x − 3)

= 2 x 3 + 5x 2 − x + 6

22. Let f(x) = 2x3 – 25x2 + 67x + 40. 3

( x 3 + 3 x 2 − 2 x + 1)(2 x − 1)

2

 1  1  1  1 f  −  = 2 −  − 25 −  + 67 −  + 40 2 2 2        2 ∵ 1 25 67 =− − − + 40 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 2 − x + 2)( x) + (2 x 2 − x + 2)(3)

4.

= ( x 3 + 3 x 2 − 2 x + 1)(2 x) + ( x 3 + 3 x 2 − 2 x + 1)(−1) = 2 x 4 + 6 x 3 − 4 x 2 + 2 x − x 3 − 3x 2 + 2 x − 1 = 2x 4 + 5x3 − 7 x 2 + 4 x − 1

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

23. Let f(x) = 3x3 + 8x2 – 68x – 48.

76

Certificate Mathematics in Action Full Solutions 4A x(3 − 2 x)(4 − 3 x) + (5 − 2 x 2 + 4 x 3 ) − (8 − 5 x + 3 x 2 ) = x( −2 x + 3)(−3x + 4) + (4 x 3 − 2 x 2 + 5) − (3 x 2 − 5 x + 8) 5.

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

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

12. Let f(x) = 16x3 – 2x + 1.  1 = f −   2

3y 6.

11. Let f(x) = x3 – 2x2 + 7x + 1.

2

4 y 12 y 3 + 0 y 2 + 8 y

Remainder

12 y 3 8y

3

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

∴ The quotient is 3y and the remainder is 8y. 13. Let f(x) = (x – 699)699 – (x – 701)701.  1400  = f   2  = f (700) Remainder = (700 − 699) 699 − (700 − 701) 701

z 2 − 2z + 6 z + 2 z 3 + 0z 2 + 2z − 3 z 3 + 2z 2 7.

− 2z 2 + 2z − 2z 2 − 4z 6z − 3 6 z − 12

= 1699 − ( −1) 701 = 1+1 =2

− 15 ∴ The quotient z2 – 2z + 6 and the remainder is –15. p3 − p2 p − 2 p4 − 3 p3 + 2 p2 + 0 p − 3 8.

p 4 − 2 p3 − p3 + 2 p2

f (1) = 14 + 29(1) + 6 14. (a) = 1 + 29 + 6 = 36 ≠0 ∴ x – 1 is not a factor of f(x). (b)

− p 3 + 2p 2 −3 ∴ The quotient is p3 – p2 and the remainder is –3. 2r 2 + 2r − 2 2r + 1 4 r 3 + 6 r 2 − 2r + 1 4r 3 + 2r 2 9.

∴ 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

77

f (−2) = (−2) 3 − 11( −2) 2 + 32( −2) − 28 15. (a) = −8 − 44 − 64 − 28 = −144 ≠0 ∴ x + 2 is not a factor of f(x).

4r 2 − 2 r 4r 2 + 2r − 4r + 1 − 4r − 2 3

f (−3) = ( −3) 4 + 29( −3) + 6 = 81 − 87 + 6 =0 ∴ x + 3 is a factor of f(x).

(b)

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

4 x−4

f (2) = 0 3

2

2 + k (2) − 5( 2) + 6 = 0 8 + 4k − 10 + 6 = 0 k = −1

2

x − 2 x − 1 x 2 − 6 x 2 + 3x − 4 x3 − 2x 2 − x

22.

− 4x 2 + 4x − 4 − 4x 2 + 8x + 4 − 4x − 8

17. Let f(x) = kx3 – 3x2 – 8kx + 9. By the remainder theorem, we have f (−3) = 3

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

k (−3) 3 − 3( −3) 2 − 8k ( −3) + 9 = 3 − 27 k − 27 + 24k + 9 = 3 3k = −21 k = −7 18. Let f(x) = x3 – 2x2 – 7x – 4. 3 2 ∵ f (1) = 1 − 2(1) − 7(1) − 4 = −12

2x2 − x + 4 x 2 − 2 2x4 − x3 + 0x 2 + x − 4 2x4

− 4x2

− x3 + 4x 2 + x

23.

− x3

f (−1) = ( −1) 3 − 2( −1) 2 − 7(−1) − 4 = 0 ∴ x + 1 is a factor of f(x). By long division, x3 – 2x2 – 7x – 4 = (x + 1)(x2 – 3x – 4)

+ 2x 2

4x − x − 4 4x 2

2

3

f ( −1) = (−1) 3 + 7( −1) 2 + 16(−1) + 12 = 2

− x 2 − 4 x + 5 3 x 3 + 0 x 2 − 15 x + 8 3 x 3 + 12 x 2 − 15 x

24.

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

− 12 x 2 + 0 x + 8 − 12 x 2 − 48 x + 60 48 x − 52

f (−2) = (−2) 3 + 7( −2) 2 + 16( −2) + 12 = 0 ∴ 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)

∴ The quotient is –3x + 12 and the remainder is 48x – 52.

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

20. Let f(x) = x3 – kx2 + x + c. ∵ x + k is factor of x3 – kx2 + x + c. ∵ By the converse of the factor theorem, f (−k ) = 0

2x − 4 2x 2 + x −1 4x3 − 6x 2 + 2x − 5 4x3 + 2x 2 − 2x

25.

− 8x 2 + 4 x − 5 − 8x 2 − 4 x + 4 8x − 9

(−k ) 3 − k (−k ) 2 + (−k ) + c = 0 c = 2k 3 + k ∴ k = 2, c = 18 or k = 1, c = 3 or k = −1, c = −3. (or any other reasonable answers) 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

x+ 4

∴ The quotient is 2x2 – x + 4 and the remainder is –x + 4. − 3 x + 12

2

19. 2x + 14x + 32x + 24 = 2(x + 7x + 16x + 12) Let f(x) = x3 + 7x2 + 16x + 12. f (1) = 13 + 7(1) 2 + 16(1) + 12 = 36 ∵

−8 −

3 2 2 ∴ x − 2 x − 7 x − 4 = ( x − 4 ) ( x + 1) 3

More about Polynomials

∴ The quotient is 2x – 4 and the remainder is 8x – 9. 26. Let f(x) = 3x3 + 16x2 + x + c. By the remainder theorem, we have  1 f  −  = −5  3 3

2

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

78

Certificate Mathematics in Action Full Solutions 4A 27. By the remainder theorem, we have f (1) = g (1) 13 − 2(1) 2 + p(1) − 4 = 3(1) 3 − 12 + 1 + 1 1− 2 + p − 4 = 3 −1 + 1+ 1 p =9 28. (a) Let f(x) = x(x – 1)(x – 2) – (x – 3)(x + 20). f (4) = 4(4 − 1)(4 − 2) − (4 − 3)(4 + 20) ∵ = 4(3)(2) − 1(24) =0 ∴ x – 4 is a factor of x(x – 1)(x – 2) – (x – 3)(x + 20). (b) x(x –1)(x – 2) – (x – 3)(x + 20) = ( x 2 − x )( x − 2) − ( x 2 + 17 x − 60) = x 3 − 3 x 2 + 2 x − x 2 − 17 x + 60 = x 3 − 4 x 2 − 15 x + 60 By long division, x3 – 4x2 – 15x + 60 = (x – 4)(x2 – 15) 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 f (−1) = ( −1) 3 + 5(−1) 2 − 2( −1) − 24 = −18

f (2) = 23 + 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 2 ∴ x + 5 x − 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

k=

or

1 2

32. Let f(x) = x3 – 2x2 + ax + b When f(x) is divided by x + 1, f (−1) = −8 ( −1) 3 − 2(−1) 2 + a ( −1) + b = −8 a − b = 5 …… (1) When f(x) is divided by x – 2, f (2) = 4 2 3 − 2( 2) 2 + a (2) + b = 4 2a + b = 4 (1) + (2), 3a = 9 a=3

…… (2)

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 3

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 (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)

f (2) = 2 3 − 7(2) 2 − 6( 2) + 72 = 40

∵

f (−2) = (−2) 3 − 7( −2) 2 − 6( −2) + 72 = 48 f (3) = 3 3 − 7(3) 2 − 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 − 7 x − 6 x + 72 = ( x − 6)( x − 4)( x + 3) 3

2

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 2

k k 8  + 2  − 1 = 0 2 2 2k 2 + k − 1 = 0 ( k + 1)(2k − 1) = 0 k+1=0 or 2k – 1 = 0

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k = −1

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 2(1) 3 − 12 − a(1) + b = 0 a − b = 1 …… (1) g (1) = 0 b(1) 3 − 2(1) 2 − 1 + a = 0 a + b = 3 …… (2) (1) + (2), 2a = 4 a=2 By substituting a = 2 into (1), we have 2 −b =1 b =1 35. (a) By the remainder theorem, we have

4

h ( − 2) = 0

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

(b)

f ( x ) − g ( x) = 4 x 3 + 3x 2 + 5( −10) x + 15 − (− x 3 − 10 x 2 + 46 x − 21) = 5 x 3 + 13 x 2 − 96 x + 36 Let h(x) = 5x3 + 13x2 – 96x + 36. 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) ∴

36. (a)

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

f ( x + 1) = ( x + 1) 3 + k = ( x + 1)( x + 1) 2 + k

f (2 x) = (2 x) 3 + 1 = 8x 3 + 1 Let g(x) = 8x3 + 1. By the remainder theorem, = g (2) remainder = 8( 2) 3 + 1 = 65

37. (a)

1− a + b = 0 b = a −1 (c) (1) + (2),

…… (2)

2b = 2 b =1

By substituting b = 1 into (1), we have 1= 3−a a=2 38. Let f(x) = x99 + 1. (a) By the remainder theorem, = f (−1) 99 remainder = ( −1) + 1 = −1 + 1 =0

x99 + 1 = ( x + 1)Q( x) 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. 39. (a) By the remainder theorem, remainder = f ( −1)

= ( −1) n + 2( −1) + 3 = −1 − 2 + 3 ( n is a positive odd integer ) =0

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

(b)

(−2) + ( a + 2)(−2) + 1 + a + b = 0 2

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

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

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

(b) By the remainder theorem, when f(x) is divided by x – 1, = f (1) remainder n = 1 + 2(1) + 3 =6 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) By substituting x = –1 into (1), we have f (−1) = (−1 + 1)(−1 − 1)Q(−1) + [ a(−1) + b] ……(3) 0 = −a + b

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Certificate Mathematics in Action Full Solutions 4A ∴ By the factor theorem, x – 1 and 3x + 4 are factors of f(x). (x – 1)(3x + 4) = 3x2 + x – 4 By long division,

(2) + (3), 2b = 6 b=3 By substituting b = 3 into (3), we have 0 = −a + 3

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. 40. (a) Let f(x) = 2x2 + px + q and g(x) = 2x2 + qx + p. ∵ x – r is a common factor of f(x) and g(x). ∴ By the converse of the factor theorem, f ( r ) = g (r )

∴ 5.

2r 2 + pr + q = 2r 2 + qr + p ( p − q) r = p − q r = 1 ( p and p are distinct real

f ( x) = ( x − 1)( x + 2)(3 x + 4)

Answer: C By the remainder theorem, we have f (−k ) = k ( −k + 2)(−k − 3) + 2 = k k2 + k −6+ 2 = k k 2 = 4 or –2 (rejected) k =2

numbers) 6. (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)

Multiple Choice Questions (p. 156) 1.

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

7.

8.

3 2 remainder = 2 − 2 + 2(2) + 1 = 8 − 4 + 4 −1 =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) = 3x 2 + 4 x + 1 2 ∴ f (1) = 3(1) + 4(1) + 1 = 3 + 4 +1 =8

4.

Answer: A  4 ∵ f(1) = 0 and f  −  = 0  3

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

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) 4 = 2 b (2) + (3), b=2 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. 10. Answer: D ∵ f(x) is divisible by x + 2 and 2x – 1.

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∴ By the converse of the factor theorem, f (−2) = 0 2(−2) 3 + a( −2) 2 + b( −2) + 4 = 0 2a − b = 6 1 f =0 2 3

…… (1)

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

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