MATHEMATICS O LEVEL(FORM TWO) NOTES – ALGEBRA

ALGEBRA

BINARY OPERATIONS

This is the operation in which two numbers are combined according to the instruction.

The instruction may be explained in words or by symbols, e.g., x, *.

Bi means two.

Example 1

Evaluate:

(i) 5 x 123

Solution:

5 x 123 = 5(100 + 20 + 3)

= 500 + 100 + 15

= 615

(ii) (8 x 89) – (8 x 79)

= 8(89 – 79)

= 8(10)

= 80

Example 2

If a * b = 4a – 2b, find 3 * 4.

Solution:

a * b = 4a – 2b

3 * 4 = 4(3) – 2(4)

= 12 – 8

3 * 4 = 4

Example 3

If p * q = 5q – p, find 6 * (3 * 2).

Solution:

Consider 3 * 2:

3 * 2 = 5q – p = 5(2) – 3 = 10 – 3 = 7

Then, 6 * 7 = 5q – p = 5(7) – 6 = 35 – 6 = 29

Therefore, 6 * (3 * 2) = 29

BRACKETS IN COMPUTATION

In expressions where there are a mixture of operations, the order of performing the operations is BODMAS:

• B = Bracket
• O = Order (powers and roots)
• D = Division
• M = Multiplication
• A = Addition
• S = Subtraction

Example

Simplify the following expression:

(i) 10x – 4(2y + 3y)

Solution:

10x – 4(2y + 3y) = 10x – 4(5y) = 10x – 20y

IDENTITY

An identity is an equation which is true for all values of the variable.

Example

Determine which of the following are identities:

(i) 3y + 1 = 2(y + 1)

Solution:

Test y = 3:

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

9 + 1 = 2(4)

10 = 8

Since LHS ≠ RHS, the equation is not an identity.

(ii) 2(p – 1) + 3 = 2p + 1

Test p = 4:

2(4 – 1) + 3 = 2(4) + 1

2(3) + 3 = 8 + 1

6 + 3 = 9

9 = 9

Since LHS = RHS, the equation is an identity.

EXERCISE

1. If a * b = 3a³ + 2b, find (2 * 3) * (3 * 2).

Solution:

a * b = 3a³ + 2b

(2 * 3) = 3(2)³ + 2(3) = 3(8) + 6 = 24 + 6 = 30

(3 * 2) = 3(3)³ + 2(2) = 3(27) + 4 = 81 + 4 = 85

Then, 30 * 85 = 3(30)³ + 2(85) = 3(27000) + 170 = 81000 + 170 = 81170

2. If x * y = 3x + 6y, find 2 * (3 * 4).

Solution:

3 * 4 = 3(3) + 6(4) = 9 + 24 = 33

2 * 33 = 3(2) + 6(33) = 6 + 198 = 204

Therefore, 2 * (3 * 4) = 204

3. If m * n = 4m² – n, find y if 3 * y = 34.

Solution:

3 * y = 4(3)² – y = 4(9) – y = 36 – y = 34

y = 2

4. Determine which of the following are identities:

2y + 1 = 2(y + 1)

Solution:

Test y = 7:

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

14 + 1 = 2(8)

15 = 16

Since LHS ≠ RHS, the equation is not an identity.

QUADRATIC EXPRESSION

A quadratic expression is an expression of the form ax² + bx + c.

• It is an expression whose highest power is 2.
• The general form is ax² + bx + c where a, b, and c are real numbers and a ≠ 0.

Note:

• a ≠ 0
• bx is the middle term
• y = mx² + cx is a linear equation
• y = ax + b is a linear equation
• y = mx² + 2 is a quadratic equation
• y = mx² + c is a quadratic equation

Examples

(i) 2x² + 3x + 6 (a=2, b=3, c=6)

(ii) 3x² – x (a=3, b=-1, c=0)

(iii) ½x² – ¼x – 5 (a=½, b=-¼, c=-5)

(iv) –x² – x – 1 (a=-1, b=-1, c=-1)

(v) x² – 4 (a=1, b=0, c=-4)

(vi) x² (a=1, b=0, c=0)

Example

If a rectangle has length 2x + 3 and width x – 5, find its area.

Solution:

Area = length × width = (2x + 3)(x – 5)

= 2x(x – 5) + 3(x – 5)

= 2x² – 10x + 3x – 15

= 2x² – 7x – 15 unit area

EXPANSION

Example 1

Expand (x + 2)(x + 1)

Solution:

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

= x² + x + 2x + 2

= x² + 3x + 2

Example 2

Expand (x – 3)(x + 4)

Solution:

(x – 3)(x + 4) = x(x + 4) – 3(x + 4)

= x² + 4x – 3x – 12

= x² + x – 12

Example 3

Expand (3x + 5)(x – 4)

Solution:

(3x + 5)(x – 4) = 3x(x – 4) + 5(x – 4)

= 3x² – 12x + 5x – 20

= 3x² – 7x – 20

Example 4

Expand (2x + 5)(2x – 5)

Solution:

(2x + 5)(2x – 5) = 2x(2x – 5) + 5(2x – 5)

= 4x² – 10x + 10x – 25

= 4x² – 25

EXERCISE

I. Expand the following:

(x + 3)(x + 3)

Solution:

x(x + 3) + 3(x + 3) = x² + 3x + 3x + 9 = x² + 6x + 9

iii) (2x – 1)(2x – 1)

Solution:

2x(2x – 1) – 1(2x – 1) = 4x² – 2x – 2x + 1 = 4x² – 4x + 1

iii) (3x – 2)(x + 2)

Solution:

3x(x + 2) – 2(x + 2) = 3x² + 6x – 2x – 4 = 3x² + 4x – 4

Example 2

Expand (a + b)(a + b)

Solution:

a(a + b) + b(a + b) = a² + ab + ba + b² = a² + 2ab + b²

ii) (a + b)(a – b)

Solution:

a(a – b) – b(a – b) = a² – ab + ab – b² = a² – b²

iii) (p + q)(p – q)

Solution:

p(p – q) + q(p – q) = p² – pq + qp – q² = p² – q²

iv) (m – n)(m + n)

Solution:

m(m + n) – n(m + n) = m² + mn – nm – n² = m² – n²

v) (x – y)(x – y)

Solution:

x(x – y) – y(x – y) = x² – xy – yx + y² = x² – 2xy + y²

FACTORIZATION

Factorization is the process of writing an expression as a product of its factors.

(i) By Splitting the Middle Term

In quadratic form ax² + bx + c:

• Sum = b
• Product = ac

Example

Factorize x² + 6x + 8

Solution:

Sum = 6 (coefficient of x)

Product = 1 × 8 = 8 (product of coefficient of x² and constant term)

Factors of 8 that add to 6 are 2 and 4.

Rewrite as x² + 2x + 4x + 8

= (x² + 2x) + (4x + 8)

= x(x + 2) + 4(x + 2)

= (x + 4)(x + 2)

Example

Factorize 2x² + 7x + 6

Solution:

Sum = 7

Product = 2 × 6 = 12

Factors of 12 that add to 7 are 3 and 4.

Rewrite as 2x² + 3x + 4x + 6

= (2x² + 3x) + (4x + 6)

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

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

DIFFERENCE OF TWO SQUARES

Consider a square with length ‘a’ units.

1st case: A = (a × a) – (b × b) = a² – b²

2nd case:

A₁ = a(a – b) … (i)

A₂ = b(a – b) … (ii)

Now, 1st case = 2nd case

AT = A₁ + A₂

a² – b² = a(a – b) + b(a – b) = (a + b)(a – b)

Generally, a² – b² = (a + b)(a – b)

Example 1

Factorize x² – 9

Solution:

x² – 9 = x² – 3² = (x + 3)(x – 3)

Example 2

Factorize 64 – x²

Solution:

64 – x² = 8² – x² = (8 + x)(8 – x)

Example 3

Factorize (x + 1)² – 169

Solution:

(x + 1)² – 169 = (x + 1)² – 13² = (x + 1 – 13)(x + 1 + 13) = (x – 12)(x + 14)

Example 4

Factorize 3x² – 5

Solution:

3x² – 5 = (√3 x)² – (√5)² = (√3 x – √5)(√3 x + √5)

APPLICATION OF DIFFERENCE OF TWO SQUARES

Example 1

Find the value of:

(i) 755² – 245²

(ii) 5001² – 4999²

Solution:

(i) 755² – 245² = (755 – 245)(755 + 245) = 510 × 1000 = 510,000

(ii) 5001² – 4999² = (5001 – 4999)(5001 + 4999) = 2 × 10000 = 20,000

PERFECT SQUARE

Note:

(a + b)² = (a + b)(a + b)

(a – b)² = (a – b)(a – b)

Example

Factorize:

(i) x² + 6x + 9

Solution:

Sum = 6

Product = 9 × 1 = 9 = 3 × 3

x² + 3x + 3x + 9 = x(x + 3) + 3(x + 3) = (x + 3)²

(ii) 2x² + 8x + 8

Solution:

Sum = 8

Product = 2 × 8 = 16 = 4 × 4

2x² + 4x + 4x + 8 = 2x(x + 2) + 4(x + 2) = (x + 2)(2x + 4)

For a perfect square ax² + bx + c, then 4ac = b².

Example 1

If ax² + 8x + 4 is a perfect square, find the value of a.

Solution:

a = a, b = 8, c = 4

From 4ac = b²:

4(a)(4) = 8²

16a = 64

a = 4

Example 2

If 2x² + kx + 18 is a perfect square, find k.

Solution:

a = 2, b = k, c = 18

From 4ac = b²:

4(2)(18) = k²

144 = k²

k = 12

Other examples

Factorize:

(i) 2x² – 12x

Solution:

2x(x – 6)

(ii) x² + 10x

Solution:

x(x + 10)