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2.2 Factorisation of Quadratic Expression

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2.2 Factorisation of Quadratic Expression
 
(A) Factorisation quadratic expressions of the form ax2 + bx + c, b = 0 or c = 0
1. Factorisation of quadratic expressions is a process of finding two linear expressions whose product is the same as the quadratic expression.
2. Quadratic expressions ax2 + c and ax2 + bx that consist of two terms can be factorised by finding the common factors for both terms.
 
Example 1:
Factorise each of the following:
(a) 2x2+ 6
(b) 7p2– 3p
(c) 6x2– 9x
 
Solution:
(a) 2x2+ 6 = 2 (x2 + 3) ← (2 is common factor)
(b) 7p2– 3p = p (7p – 3) ← (p is common factor)
(c) 6x2– 9x = 3x (2x – 3) ← (3x is common factor)


(B) Factorisation of quadratic expressions in the form ax2 – c , where a and c are perfect squares
 
Example 2:
(a) 9p2– 16
(b) 25x2– 1
(c) 1 4 − 1 25 x 2

Solution: 
(a) 9p2– 16 = (3p)2 – 42= (3p – 4) (3p + 4)
(b) 25x2– 1 = (5x)2 – 12= (5x – 1) (5x + 1)
(c)
1 4 − 1 25 x 2 = ( 1 2 ) 2 − ( 1 5 x ) 2 = ( 1 2 − 1 5 x ) ( 1 2 + 1 5 x )


(C) Factorisation quadratic expressions in the form ax2 + bx + c, where a ≠ 0, b ≠ 0 and c ≠ 0
 
Example 3:
Factorise each of the following
(a) 3y2+ 2y – 8
(b) 4x2– 12x + 9
 
Solution: 
(a)
Factorise using the Cross Method
 

3y2+ 2y – 8 = (3y – 4) (y + 2)

(b)

 
 4x2– 12x + 9 = (2x – 3) (2x – 3)
 

2.1 Quadratic Expression

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(A) Identifying quadratic expression
1. A quadratic expression is an algebraic expression of the form ax2 + bx + c, where a, b and c are constants, a ≠ 0 and x is an unknown.
(a) The highest power of x is 2.
(b) For example, 5x2– 6x + 3 is a quadratic expression.

Example 1
State whether each of the following is a quadratic expression in one unknown.
(a) x2 – 5x + 3
(b) 8p2 + 10
(c) 5x + 6
(d) 2x2 + 4y + 14
(e)  2 p + 1 p + 6
(f) y3 – 3y + 1

Solution:
(a) Yes. A quadratic expression in one unknown.

(b) Yes. A quadratic expression in one unknown.

(c) Not a quadratic expression in one unknown. The highest power of the unknown x is not 2.

(d) Not a quadratic expression in one unknown. There are 2 unknowns, x and y in the quadratic expression.

(e) Not a quadratic expression in one unknown. The highest power of the unknown x is not 2. 
1 p = p − 1

(f) Not a quadratic expression in one unknown. The highest power of the unknown x is not 2.

2. A quadratic expression can be formed by multiplying two linear expressions.
 (2x + 3)(x  - 3) = 2x2 – 3x – 9


Example 2
Multiply the following pairs of linear expressions.
(a) (4x + 3)(x – 2)
(b) (y – 6)2
(c) 2x (x – 5)

Solution:
(a) (4x + 3)(x – 2)
= (4x)(x) + (4x)(-2) +(3)(x) + (3)(-2)
= 4x2 – 8x + 3x – 6
= 4x2 – 5x – 6

(b) (y – 6)2 
= (y – 6)(y – 6)
= (y)(y) + (y)(-6) + (-6)(y) + (-6)(-6)
= y2 -6y – 6y + 36
= y2 - 12y + 36

(c) 2x (x – 5) 
= 2x(x) + 2x(-5)
= 2x2 – 10x

Quadratic Equations Long Questions (Question 1 – 4)

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Question 1:
Solve the quadratic equation, (y + 3)(y – 4) = 30

Solution:
(y + 3)(y – 4) = 30
y2 – 4y + 3y– 12 = 30
y2 – y – 12 – 30 = 0
y2 – y – 42 = 0
(y + 6)(y – 7) = 0
y + 6 = 0, y = –6
Or
y – 7 = 0
y = 7


Question 2:
Solve the quadratic equation, 5x2 = 3( x – 2) + 8

Solution:
5x2 = 3x – 6 + 8
5x2 – 3x – 2 = 0
(5x + 2)(x – 1) = 0
5x + 2 = 0, x − 2 5
Or
x – 1 = 0
x = 1


Question 3:
Solve the quadratic equation
2 p 2 − 15 p = 7

Solution:
2 p 2 − 15 p = 7
2p2 – 15 = 7p
2p2 –7p – 15 = 0
(2p + 3)(p – 5) = 0
2p + 3 = 0, p =  − 3 2
Or 
p – 5 = 0
p = 5


Question 4:
Solve the quadratic equation  y ( y − 9 2 ) = 5 2

Solution:
y ( y − 9 2 ) = 5 2 y 2 − 9 y 2 = 5 2

( ×2), 2y2 – 9y= 5
2y2 – 9y – 5 = 0
(2y + 1)(y – 5) = 0
2y + 1 = 0, y = – ½
Or
y – 5 = 0
y = 5

1.1 Significant Figures

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1.1.2 Significant Figures (Part 2)
1. Perform combined operations (addition, subtraction, multiplication and division) involving numbers, the final answer is rounded off to specific significant figures. 

Example:
Find the value of each of the following and give your answer correct to 3 significant figures.
(a) 261.9 + 75.6 × 0.7
(b) 0.062 × 30.12 + 1.268
(c)  8.608 ÷ 0.08 − 28.35
(d) 0.846 ÷ 0.4 - 0.153 × 2

Solution:
(a) 261.9 + 75.6 × 0.7
= 261.9 + 52.92
= 314.82
= 315 (3 s. f.)

(b) 0.062 × 30.12 + 1.268
= 1.86744 + 1.268
= 3.13544
= 3.14 (3 s. f.)

(c)  8.608 ÷ 0.08 − 28.35
= 107.6 – 28.35
= 79.25
= 79.3 (3 s. f.)

(d)  0.846 ÷ 0.4 − 0.153 × 2
= 2.115 – 0.306
= 1.809
= 1.81 (3 s. f.)



Example 1:
Calculate the value of 5.33 + 0.33 × 17 and give your answer correct to three significant figures.

Solution:
5.33 + 0.33 × 17
= 5.33 + 5.61
= 10.94
= 10.9 (3 s. f.)



Example 2:
Calculate the value of 49.3567 + 16.73 ÷ 0.5 and give your answer correct to four significant figures.
Solution:
49.3567 + 16.73 ÷ 0.5
= 49.3567 + 33.46
= 82.8167
= 82.82 (4 s. f.)



Example 3:
Calculate the value of 3.42 ÷ 12 × 3.7 and give your answer correct to four significant figures.

Solution:
3.42 ÷ 12 × 3.7
= 1.0545
= 1.055 (4 s. f.)

1.3 SPM Practice (Short Questions)

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Question 1:
Round off each of the following numbers to the number of significant figures indicated in brackets.
(a) 80616 (3 s. f.)
(b) 60932 (3 s. f.)
(c) 0.4783 (2 s. f.)
(d) 3.047 (3 s. f.)
(e) 0.00567 (2 s. f.)
(f) 0.05086 (3 s. f.)

Solution:
(a) 80600
(b) 60900
(c) 0.48
(d) 3.05
(e) 0.0057
(f) 0.0509


Question 2:
Express each of the following as a single numbers.
(a) 8.565 × 10-5
(b) 1.304 × 105
(c) 6.754 × 10-6
(d) 1.0352 × 104

Solution:
(a) 0.00008565
(b) 130400
(c) 0.000006754
(d) 10352


Question 3:
Express each of the following in standard form.
(a) 376510
(b) 47865400
(c) 0.000507
(d) 0.00006408

Solution:
(a) 3.7651 × 105
(b) 4.78654 × 107
(c) 5.07 × 10-4
(d) 6.408 × 10-5


Question 4:
1.3 × 1015 + 3.2 × 1014

Solution:
1.3 × 1015 + 3.2 × 1014
= 1.3 × 1015 + 0.32 × 1015
= (1.3 +0. 32) × 1015
= 1.62 × 1015


Question 5:
0.0000036 – 2.1 × 10-7

Solution:
0.0000036 – 2.1 × 10-7
= 3.6 × 10-6 – 2.1 × 10-7
= 3.6 × 10-6 – 0.21 × 10-6
= (3.6 - 0.21) × 10-6
= 3.39 × 10-6

1.3 SPM Practice (Short Questions)

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1.3.2 Standard Form, SPM Practice (Paper 1)
Question 6:
(a)  Round off 0.05079 correct to three significant figures.
(b)  Find the value of 5 × 107 + 7.2 × 105 and state its answer in standard form.

Solution:
(a)  0.05079 = 0.0508 (correct to three significant figures)
(b)  5 × 107 + 7.2 × 105 = 500 × 105 + 7.2 × 105
= (500 + 7.2) × 105
= 507.2 × 105
= 5.072 × 107


Question 7:
(a)  Calculate the value of 70.2 – 3.22 × 8.4 and round off its answer correct to three significant figures.
(b)  Express 840 0.000021  as a number in standard form.

Solution:
(a)  70.2 – 3.22 × 8.4 = 70.2 – 27.048
= 43.152 = 43.2 (correct to three significant figures)

(b)   
840 0.000021 = 8.4 × 10 2 2.1 × 10 − 5 = 8.4 2.1 × 10 2 − ( − 5 ) = 4 × 10 7



Question 8:
Calculate the value of 7 × (2 × 10-2)3– 4.3 × 10-5 and state its answer in standard form.

Solution:
7 × (2 × 10-2)3 – 4.3 × 10-5= 7 × 23 × (10-2)3 – 4.3 × 10-5
= 56 × 10-6 – 4.3 × 10-5
= 5.6 × 10-5 – 4.3 × 10-5
= 1.3 × 10-5

1.2 Standard Form (Sample Questions)

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Example 1:
Find the value of 7.3 × 103 + 3.2 × 104 and express your answer in standard form.

Solution:
7.3 × 103 + 3.2 × 104
= 7.3 × 103 + 3.2 × 101 × 103
= [7.3 + (3.2 × 101)] × 103
= 39.3 × 103
= 3.93 × 104


Example 2:
Find the value of 3.3 × 105 + 6400 and express your answer in standard form.

Solution:
= 3.3 × 105 + 6.4 × 103
= 3.3 × 101 × 101 × 103+ 6.4 × 103
= 330 × 103 + 6.4 × 103
= (330 + 6.4) × 103
= 336.4 × 103
= 3.364 × 105

1.2 Standard Forms

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1.2.1 Standard Form (Part 1)
1. Standard form is a way of writing very large numbers or very small numbers in the form of A × 10n, where  1 ≤ A < 10  and n is a positive or a negative integer.

Example 1:
Express each of the following numbers in standard form.
(a) 7244
(b) 32567
(c) 750000
(d) 0.65
(e) 0.0428
(f) 0.000369

Solution:
(a) 7244 = 7.244 × 1000 = 7.244 × 103

(b) 32567 = 3.2567 × 10000 = 3.2567 × 104

(c) 750000 = 7.5 × 100000 = 7.5 × 105

(d) 0.65
= 6.5 × 1 10
= 6.5 × 10-1

(e) 0.0428
= 4.28 × 1 100
= 4.28 × 10-2

(f) 0.000369
= 3.69 × 1 10000
= 3.69 × 10-4


Example 2:
Express each of the following numbers in standard form.
(a) 63.4
(b) 2738
(c) 23000
(d) 428000000
(e) 0.0063
(f) 0.000000038

Solution:
(a) 63.4 = 6.34 × 10

(b) 2738 = 2.738 × 1000 = 2.738 × 10³

(c) 23000 = 2.3 × 10000 = 2.3 × 104

(d) 428000000 = 4.28 × 100000000 = 4.28 × 108

(e) 0.0063
= 6.3 × 1 1000
= 6.3 × 10-3
 
(f) 0.000000038
= 3.8 × 1 100000000
= 3.8 × 10-8
 

1.1 Significant Figures

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1.1.1 Significant Figures (Part 1)
1. Significant figures are the relevant digits in an integer or a decimal number which has been rounded up to a value according to a degree of accuracy.

2. In rounding off positive numbers to a given number of significant figures, the significance of zero is shown as below.

(a) All non-zero digits in a number are significant figures (s. f.).
Example:
(i) 568 (3 s. f.)
(ii) 36.97 (4 s. f.)

(b) All zeroes between non-zero digits are significant.
Example:
(i) 7001 (4 s. f.)
(ii) 3.04 (3 s. f.)
(iii) 22.054 (5 s. f.)

(c) All zeroes that lie on the right of a non- zero digit in a decimal are significant.
Example:
(i) 0.70 (2 s. f.)
(ii) 4.500 (4 s. f.)
(iii) 3.00 (3 s. f.)

(d) Zeroes that lie on the left of a non-zero digit in a decimal are not significant.
Example:
(i) 0.05 (1 s. f.)
(ii) 0.0040 (2 s. f.)
(iii) 0.07040 (4 s. f.)

(e) Zeroes at the end of a whole number are to be considered as non significant unless stated otherwise.
Example
(i) 40 (1 s. f.)
(ii) 3670 (3 s. f.)
(iii) 704200 (4 s. f.)


Example 1:
Round off the following numbers correct to three significant figures.
(a) 246 = 246 (3 s. f.)
(b) 2463 = 2460 (3 s. f.)
(c) 24632 = 24600 (3 s. f.)
(d) 0.00745 = 0.00745 (3 s. f.)
(e) 0.007453 = 0.00745 (3 s. f.)
(f) 0.007455 = 0.00746 (3 s. f.)
(g) 0.007403 = 0.00740 (3 s. f.)

Example 2:
Round off each of the following numbers to the number of significant figures indicated in brackets.
(a) 3548 (2 s. f.)
(b) 0.5089 (3 s. f.)
(c) 33.028 (1 s. f.)
(d) 0.40055 (3 s. f.)
(e) 0.681 (2 s. f.)
(f) 38.97 (3 s. f.)

Solution:
(a) 3500 (2 s. f.)
(b) 0.509 (3 s. f.)
(c) 30 (1 s. f.)
(d) 0.401 (3 s. f.)
(e) 0.68 (2 s. f.)
(f) 39.0 (3 s. f.)

1.3 SPM Practice (Short Questions)

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Question 9:
3.17× 10 −8 −1.20× 10 −9 =

Solution:
3.17× 10 −8 −1.20× 10 −9 =3.17× 10 −8 −( 0.120× 10 1 × 10 −9 ) =3.17× 10 −8 −0.120× 10 −8 =( 3.17−0.120 )× 10 −8 =3.05× 10 −8


Question 10:
Express  0.096 ( 2× 10 3 ) 3  in standard form.

Solution:
0.096 ( 2× 10 3 ) 3 = 0.096 8× 10 9 =1.2× 10 −11


Question 11:
Express 0.000064−3.5× 10 −6  in standard form.

Solution:
0.000064−3.5× 10 −6 =6.4× 10 −5 −3.5× 10 −6 =6.4× 10 −5 −0.35× 10 −5 =( 6.4−0.35 )× 10 −5 =6.05× 10 −5


Question 12:
200.7× 10 11  is written as 2.007× 10 b  in the standard form. State the value of b.

Solution:
200.7× 10 11 =2.007× 10 2 × 10 11                   =2.007× 10 13                 b=13


Question 13:
Find the product of 0.1985 and 0.5.
Round off the answer correct to two significant figures.

Solution:
0.1985 × 0.5
= 0.09925
= 0.10 (two significant figures)