MATHEMATICS FORM 2 – ALGEBRA

– BINARY OPERATIONS

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

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

– Bi means two

Example1.

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

Example2

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

From p * q = 5q – p

3 * 2 = 5q – p

= 10 – 3

= 7

Then, 6 * 7 = 5q – p

6 * 7 = 5(7) – p

35 – 6 = 29

6 *(3 * 2) = 29

35 – 6 = 29

6 * (3 * 2) = 29

BRACKETS IN COMPUTATION

– In expression where there are a mixture of operations, the order of performing the operation is BODMAS

(ii) B = BRACKET

O = OPEN

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

– Is the equation which are true for all values of the variable

Example

Determine which of the following are identity.,

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

Solution:

3y + 1 = 2(y + 1)

Test y = 3

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

9 + 1 = 3(3)

10 = 9

Now, 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

Now, 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 x 3

= 3(8) + 6

= 24 + 6 = 30

Then

(3 * 2) = 3(3)³ + 2(2)

a * b = 30 * 85

30 * 85 = 3(30)³ + 2(85)

= 3(27000) + 170

= 81000 + 170

(2 * 3) * (3 * 2) = 81170

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

Solution:

Consider (3 * 4)

From x * y = 3x + 6y

3 * 4 = 3(3) + 6(4)

= 9 + 24

= 33

Then 2 * 33 = 3x + 6y

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

= 6 + 198 = 204

2 * (3 * 4) = 204

3. If m*n = 4m² – n

Find y if 3 * y = 34

Solution:

= m * n = 4m² – n

= 3 * y = 34

= 3 * y = 4(3)² – y = 34

= 4(3²) – y = 34

= 4(9) – y = 34

36 – y = 34

y = 2

4. Determine which of the following is identities

2y + 1 = 2(y + 1)

Solution:

2y + 1 = 2(y + 1)

Test y = 7

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

14 + 1 = 2(8)

15 = 16

Now, LHS RHS (The equation is not an identity).

QUADRATIC EXPRESSION
Is an expression of the form of ax²+ bx + c.

– Is an expression whose highest power is 2.

– General form of quadratic expression is ax² + bx + c where a, b, and c are real numbers and a≠ 0.

Note

(i) a≠ o

bx – middle term

y = mx² + cx – linear equation

y = ax + b

y= mx² + 2 – quadratic equation

y = mx²+ c

example

(i) 2x² + 3x + 6 (a =2, b =3, c =6)

ii) 3x² – x (a =3, b = -1, c = 0)

iii) 1/2x² – 1/yx – 5 (a = ½, b = -1/4, 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 + x and width x – 5 find its area

Solution:

From, A = l x w where A is area, l is length and w is width

= (2x + 3) (x – 5) Alternative way:

= 2x(x – 5) + 3(x – 5) (2x + 3) X (x-5)

= 2x² – 10x + 3x – 15 2x² -10N + 3x-15

2x² – 7x – 15unit area 2x² – 7x-15 Unit area

EXPANSION

Example 1

Expand i) (x + 2) (x + 1)

Solution:

(x + 2) (x + 1) Alternative way:

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

= x² + x + 2x + 2x² +x+2x+2

= x² + 3x + 2x²+3x+2

ii) (x – 3) (x + 4) Alternative way:

x (x + 4) – 3(x + 4) (x-3) (x+4)

x² + 4x – 3x – 12 x²+4x-3x-12

= x² + x – 12x²+x-12

iii) (3x + 5) (x – 4) Alternative way:

3x(x -4) + 5 (x – 4) (3x+5) (x-4)

= 3x² – 12x + 5x – 203x²-12x+5x-20

= 3x² – 7 – 203x²-7x-20

iv) (2x + 5) (2x – 5) Alternative way:

2x (2x – 5) + 5(2x – 5) (2x+5) (2x-5)

4x² – 10x + 10x – 25 4x²-10x+10x-25

= 4x² – 25 4x²-25

EXERCISE

I. Expand the following

(x + 3) (x + 3) Alternative way:

x(x + 3) + 3x + 9 (x+3) (x+3)

= x² + 3x + 3x + 9x²+3x +3x+9

= x² + 6x + 9 x²+6x+9

iii) (2x – 1) (2x – 1)

Solution:

2x(2x – 1) – 1 (2x – 1)
=(2x-1) ( 2x-1)

= 4x² – 2x – 2x + 1

= 4x²– 4x +1

iii) (3x – 2) (x +2)

Solution:

3x(x + 2) – 2(x + 2) Alternative way:

= 3×2 + 6x – 2x – 4(3x-2) (x+2)

= 3x² + 4x – 43x²+6x-2x-4

3x²+4x-4

2) Expand the following

i) (a + b) (a + b)

Solution:

a(a + b) + b(a + b)
=(a+b) (a+b)

= a² + ab + ba + b²

= a² + 2ab + b²

ii) (a + b) (a –b)

Solution:

a(a + b) – b(a + b)
= (a+b) (a-b)

= a²– ab + ab -b²

= a² – b²

iii) (p + q) (p – q)

Solution:

p(p – q) + q(p – q) Alternative way:

= p² – pq + qp – q^{2 (p+q) (p-q)}

= p² – q^{2p2-pq+pq-q2
p2– q2}

iv) (m – n) (m + n)

Solution:

m(m + n) – n(m + n) Alternative way:

= m² +mn – nm + n^{2 (m-n) (m+n)}

= m² – n^{2 m2+ mn -nm – n2
m2- n2}

v) (x – y) (x – y)

Solution:

x(x – y) – y(x – y)
= (x-y) (x-y)

= x² – xy – yx + y²

= x² – 2xy + y²

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 i) x² + 6x + 8

Solution:
Find the number such that

i) Sum = 6; coefficient of x

ii) Product = 1 x 8; Product of coefficient of x² and constant term

= 8 = 1 x 8

= 2 x 4

Now

x² + 2x + 4x + 8

(x² + 2x) + (4x + 8)

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

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

ii) 2x² + 7x + 6

Solution:

Sum = 7

Product, = 2 x 6 = 12

– 12 = 1 x 12

= 2 x 6

= 3 x 4

Now,

2x² + 3x + 4x + 6

(2x² + 3x) + (4x + 6)

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

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

iii) 3x² – 10x + 3

Solution:

Sum = -10

Product = 3 x 3 = 9

9 = 1 x 9

= 3 x 3

Now,

3x² – x – 9x + 3

(3x² – x) – (9x + 3)

x(3x – 1) – 3(3x + 1)

(x – 3) (3x – 1)

iv) x² + 3x – 10

Solution:

Sum = 3

Product = 1 x -10 = -10

= -2 x 5

Now,

X² – 2x + 5x – 10

(x² – 2x) + (5x – 10)

x (x – 2) + 5(x – 2)

= (x + 5) (x – 2)

EXERCISE

i) Factorize the following

4x² + 20x + 25

Solution:

Sum = 20

Product = 4 x 25 = 100

100 = 1 x 100

= 2 x 50

= 4 x 25

= 5 x 20

= 10 x 10

= 4x² + 10x + 10x + 25

(4x² + 10x) + (10x + 25)

2x(2x + 5) + 5 (2x + 5)

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

ii) 2x² + 5x – 3

Solution:

Sum = 5

Product = -6

number = (- 1,6)

= 2x² – x + 6x – 3
= 2x²+ 5x – 3

(2x² – x) + (6x – 3)

x (2x – 1) + 3(2x – 1)

= (x + 3) (2x – 1)

iii) x² – 11x + 24

Solution:

Sum = -11

Product = 1 x 24 = 24

24 = 1 x 24

= 1 x 24

= 2 x 12

= 3 x 8 = -3 x -8

= 4 x 6

x² – 3x – 8x + 24

(x² – 3x) – (8x – 24)

x(x – 3) – 8(x – 3)

= (x – 8) (x – 3)

iv) x² – 3x – 28

Solution:

Sum = -3

Product = 1 x -28 = -28

28 = 1 x 28

= 2 x 14

= 4 x7

= x² + 4x – 7x – 28

(x² + 4x) – (7 + 28)

x(x +4) – 7(x +4)

(x – 7) (x + 4)

BY INSPECTION

Example

Factorize

i) x² + 7x + 10

Solution:

(x + 2) (x + 5)

ii) x² + 3x – 40

Solution:

(x – 5) (x + 8)

iii) x² + 6x + 7

Solution:

Has no factor.

DIFFERENT OF TWO SQUARE

Consider a square with length ‘’a’’ unit

1^st case, At = (a x a) – (b x b)

= a² – b²

2^nd case

A₁ = a (a – b) …….(i)

A₂ = b (a – b)…….(ii)

Now, 1^st case = 2^nd case

A_T = A₁ + A₂

a² – b² = a (a – b) + b(a – b)

= (a + b) (a – b)

Generally a² – b² = (a + b) (a – b)

Example 1

Factorize i) x² – 9

ii) 4x² – 25

iii) 2x² – 3

Solution:

i) x² – 9 = x² – 3²

= (x + 3) (x – 3)

ii) 4x² – 25 = 2²x² – 5²

= (2x)² – 5²

iii)2x² – 3 =()² x² – ( )²

= (x)² – ()²

=(x + )(x – )

EXERCISE

I. Factorize by inspection

i) x² + 11x – 26

Solution:

(x + 13) (x -2)

ii) x² – 3x – 28

Solution:

(x – 7) (x + 4)

2. Factorization by difference of two square

i) x² – 1

Solution:

X² – 1 = ()² – ()²

= (x)² – 1

= (x + 1) (x – 1)

ii) 64 – x²

Solution:

64 – x² = 8² – x²

= (8 + x) (8 – x)

iii) (x + 1)² – 169

solution:

(x + 1)² – 169

(x + 1)² – 13²

= (x + 1 – 13) (x + 1 + 13)

= (x – 12) (x + 14)

iv) 3x² – 5

Solution:

3x² – 5 = (x)² – ()²

= (x – )(x + )

APPLICATION OF DIFFERENCES OF TWO SQUARE

Example 1

Find the value of i) 755² – 245²

ii) 5001² – 4999²

Solution:

i) 755² – 745²

From a² – b² = (a + b) (a – b)

755² – 245² = (755 – 245)(755 + 245)

= (510) (1000)

= 510, 000

ii) 5001² – 4999²

5001² – 4999² = (5001 – 4999) (5001 + 4999)
5001²– 4999²=(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

Sum = 6

Product = 9 x 1 = 9

= 9 = 1 x9

= 3 x 3

x² + 3x + 3x + 9

(x² + 3x) + (3x + 9)

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

= (x + 3)²

ii) 2x² + 8x + 8

Sum = 8

Product = 2 x 8 = 16

16 = 1 x 16

= 2 x 8

= 4 x4

2x² + 4x + 4x + 8

(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:

ax² + 8x + 4

a = a, b = 8, c = 4

From,

4ac = b²

4(a) (4) = 8²

16a/16 = 64/16

a = 4

Example2

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

Solution:

2x² + kx + 18

a = 2, b = kx, c = 18

from

4ac = b²

4(2)(18) = k²

From

4ac = b²

4(2) (18) = k²

=

K =

K = 12

– Other example

Factorize i) 2x² – 12x

Solution:

2x(x – 6)

ii) x² + 10x

= x(x + 10)