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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)
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).
RHS (The equation is not an identity).QUADRATIC EXPRESSION
Is an expression of the form of ax2 + bx + c.
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 + 2 x2 +x+2x+2
= x2 + 3x + 2 x2+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 – 12 x2+x-12
iii) (3x + 5) (x – 4) Alternative way:
3x(x -4) + 5 (x – 4) (3x+5) (x-4)
= 3x2 – 12x + 5x – 20 3x2-12x+5x-20
= 3x2 – 7 – 20 3x2-7x-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 + 9 x2+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)
=(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 – 4 3x2+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)
=(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)
= (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 – q2 (p+q) (p-q)
= p2 – q2 p2-pq+pq- q2
p2 – q2
p2 – q2
iv) (m – n) (m + n)
Solution:
m(m + n) – n(m + n) Alternative way:
= m2 +mn – nm + n2 (m-n) (m+n)
= m2 – n2 m2 + mn -nm – n2
m2 – n2
m2 – n2
v) (x – y) (x – y)
Solution:
x(x – y) – y(x – y)
= (x-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
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)
(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 + 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 x 7
7 = 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
1st case, At = (a x a) – (b x 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 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
)2 x2 – ( 
)2 = ( x)2 – ()2
x)2 – ()2 =( x + ) ( x – )
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
)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)2 – (
)2= ( x – ) ( x + )
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)
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)
