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ALGEBRA
Indices (law of exponents)
Three basic rules including the Indices are:
i) am x an = am + n
ii) am ÷ an = am – n
iii) (am) n = am
Negative indices
Consider a5 ÷ a2 = a5 – 2 = a3
= a2 ÷ a5 = a2 – 5 = a -3
In general a -m = 
Fractional indices
Consider 
Similarly
Zero exponents
Consider am x ao = am + o
ao = 1Laws of logarithm
If a and b are two positive numbers there exist a third number c such that
ac = b
→c is the logarithm of b to base a
i. e 
= c
= c
Definition
Logarithm of ‘x’ to base ‘a’ is the power to which ‘a’ must be raised to give ‘x’.
If p = and q =, then 
= x and = y
and q =
, then 
= x and 
= yThus
1. 
2. 
3. 
Change of base
If y =
EXAMPLE
1. Solve for x, 
Solution
Note that:
There are two important bases of logarithms
10 and e
Series
A series is the sum of a sequentially ordered finite or infinite set of terms
Finite series
– Is the one have defined first and last term e.g. 1 + 3 + 5 + 7 + 9 + 11…… + 21 is a finite series
Infinite series
– Is the one have defined the first but not the last term e.g. 1 + 3+ 5+ 7+ 9+ 11+ …..
In both cases the first term is 1
The sigma notation
∑ stands for ‘’ sum of ‘’
e.g.
Exercise
Discuss the following and find the sum if n = 8
The sum of the first n natural numbers
The sum of squares of the first n natural numbers
The sum of the cubes of the first n natural numbers
Example
I. If an = n2 + 3n + 1 determine an expression for n
II. If an = n3 + 2n2 + 4n evaluate
a) a1 b) a4 c)
The sum of the first n natural numbers
The sum of the squares of the first n natural numbers
Proof,
Exercise
1. Evaluate 
Proof by the mathematical induction
Example
Prove that n (n2 + 5) is exactly divisible by 3 for all positive integers n
Proof: I
Let n = 1; 1(12 + 5) = 6 = 3 x 2
n = 2; 2 (22 + 5) = 18 = 3 x 6
n = 3; 3 (32 + 5) = 42 = 3 x 14
n = 4; 4 (42 + 5) = 84 = 3 x 28
n = 7; 7 (72 + 5) = 378 = 3 x 126
Proof: II
i) Let n = 1 = 1 (12 + 5) = 6 = 3 x 2
ii) Let n (n2 + 5) be divisible for n = k
i.e. k (k2 + 5) = 3p, where p is any integers
iii) When n = k + 1
(k + 1) ( (k +1)2 + 5) = (k + 1) (k2 + 2k + 1 + 5)
= (k + 1) ((k2 + 5) + (2k + 1))
= k (k2 + 5) + k (2k + 1) + (k2 + 5) + (2k +1)
= 3p + 2k2 + k + k2 + 5 + 2k +1
= 3p + 3k2 + 3k + 6
= 3 (p + k2 + k + 2)
Since p and k are positive integers
So the number in the bracket is positive
iv) Since when n = 1 the values 1 (12 + 5) is divisible by 3 then the value n (n2 + 5) will be divisible by 3 for n = 2, n = 3, n = 4…… by the above working
→n (n2 + 5) is divisible by 3 for
all n∈
+
all n∈
+Principle of proof by mathematical induction
It states if s1, s2, s3…..Sn…. is a sequence of statements and if
i) s1 is true
ii) Sn → Sn + 1, n = 1, 2, 3 … are true, then s1, s2, s3…… Sn… are true statement
Examples
1. Prove by mathematical induction that 2 + 4+ 6 +…..2n = n (n + 1)
Solution
When n = 1
L. H. S = 2, R. H. S = 1(1 + 1) = 2
L. H.S = R. H. S
It is true for n = 1
Let the statement be true for n = k
2 + 4 + 6 + …. 2k = k (k + 1)
Required to prove when n = k + 1
2 + 4 + 6 + …… 2k + 2(k + 1) = k (k + 1) + 2(k + 1)
= k2 + k + 2k + 2
= k2 + 3k + 2
= k2 + k + 2k + 2
= k (k + 1) + 2 (k + 1)
= (k + 1) (k + 2)
Which is the same as putting n = k + 1 in the formula
Since n = 1gave a true statement, n =2, n = 3, n = 4… will be true statement as worked above
2. 2. Prove by induction that
Solution
Proof:
When n = 1,
Also n = 1 give
L.H.S = R. H. S
Let the statement be true for n =k
Let 
Required to prove when n = k + 1
Which is the same as putting n = k + 1 in the form
Since n = 1 gave a true statement
n = 2, n = 3, n = 4… will give true statement
3. Prove that 
Solution
Proof:
When n = 1
L.H. S = 3 x 1 – 2 = 1
R.H.S = 
L.H.S = R.H.S
Let the statement be true for n = k
I.e. 
required to prove when n = k + 1
required to prove when n = k + 1
Which is the same as putting n = k + 1 in the formula since n = 1 gave a true statement,
n =2, n = 3, n = 4 … will give true statement
Roots of a polynomial function
If 
and are roots of quadratic equation
and 
are roots of quadratic equationThen (x – ) (x – β) = 0
) (x – β) = 0 x2 – βx – x + β = 0
x + β = 0 x2 – (β +) x + β = 0
) x + β = 0Given a quadratic equation as
ax2 + bx + c = 0, where a, b, c, are constant
Summary
A quadratic equation is given by
x2 – (sum of factors) x + products of factors = 0
Example
1. Given and β as the roots for 4x2 + 8x + 1 = 0 form an equation whose roots are 2 β and β2
and β as the roots for 4x2 + 8x + 1 = 0 form an equation whose roots are 2 β and β2 Solution
Sum of roots 2 β + β2 = β ( + β)
2 β + β2 = β ( + β) Products of root are (2 β) (β2)
2 β) (β2) =3 β3
3 β3 = ( β) 3
β) 3 (2 β) (β2) = ( β) 3
2 β) (β2) = ( β) 3 The given equation can be written as
The required equation is
=0
=0
2. The equation 3x2 – 5 + 1 = 0 has roots and β
and β a) Find values of 
b) Find the values of 
Solution
+ β = 
and β = 
Roots of cubic equations
If, β, are roots of a cubic equation then
, β, 
are roots of a cubic equation then(x –)(x – β)(x – γ) = 0
)(x – β)(x – γ) = 0(x2 – x – βx + β) (x – γ) = 0
x – βx + β) (x – γ) = 0x3 – γx2 – x2 + γx – βx2 + βγx + βx – βγ = 0
x2 + γx – βx2 + βγx + βx – βγ = 0x3 – ( + β + γ) x2 + (γ+ βγ + β) x – βγ = 0
+ β + γ) x2 + (γ+ βγ + β) x – βγ = 0 x3 – ( + β + γ) x2 + (γ + βγ + β) x – βγ = 0
+ β + γ) x2 + (γ + βγ + β) x – βγ = 0 Given cubic equation can be written as
ax3 + bx2 + cx + d = 0
Equating coefficients of x2, x and the constant terms
i) + β + γ = 
-; sum of roots
+ β + γ = 
-; sum of roots ii) γ + βγ + β = 
sum of products of roots
γ + βγ + β = 
sum of products of roots iii) γβ = products of roots
γβ = 
products of rootsExamples
1. The equation 3x3 + 6x2 – 4x + 7 = 0 has roots, β, γ. Find the equations with roots
, β, γ. Find the equations with roots a)
Solution
From
x3 – (sum of factors) + (sum of products of factors) – products = 0
X3 – (sum of factors) x2 + (sum of products of products of factors) x – products = 0
From the equation 3x3 + 6x2 – 4x + 7 = 0
2. If the roots of the equation 4x3 + 7x2 – 5x – 1 = 0 are , β and γ find the equation whose roots are
, β and γ find the equation whose roots are a) + 1, β + 1, γ + 1 b) 2, β2, γ2
+ 1, β + 1, γ + 1 b) 2, β2, γ2 Solution
4x3 + 7x2 – 5x – 1 = 0
Remainder and factor theorem
Definition:
A polynomial is an expression of the form
anxn + an – 1 x n – 1 + an – 2 x n -2 + …… + a1x + a0
Where an, an – 1, an – 2 …a1, ao are real numbers known as coefficients of the polynomial
→an 0
0 →anxn is the leading term
→n is called the degree of the polynomial
Normally the polynomial is written as p (x) = anxn + an – 1xn– 1+ … + a1x + ao
P (x) = anxn + an – 1xn -1 + …. + a1x + ao
e.g.
∙ p (x) = 2x4 – 3x3 + 10x3 + 10x2 – x + 11
∙ p (x) = x5
∙ p (x) = 2x2 – 3x + 10
∙ p (x) = 6x3 – 22x2 – 12
∙ p(x) = 3x – 2
∙ p (x) = 17
To divide a polynomial p (x) by another polynomial D (x) means finding polynomial Q (x) and r (x)
Such that
P (x) = D (x) Q (x) + r (x)
Where p (x) is called a dividend
Q (x) is called a quotient
D (x) is called divisor
r (x) is called remainder
Note that the degree of r (x) < D(x)
The remainder theorem
When a polynomial p (x) is divided by a linear factor (x – a) the remainder is P (a)
When a linear factor is in the form kx – b then it should be put in the form k(x – ) and the remainder is then P ()
) and the remainder is then P ()Proof:
Let P (x) = (x – a) Q(x) + R
Where Q (x) is a polynomial and R is the remainder when x = a
→P (a) = (a – a) Q (a) + R
P (a) = R
R = P (a)
When R = 0
P (x) = (x – a) Q(x)
x – a is a factor or p (x)
Since p (a) = 0
‘’a’’ is a root (a zero) of p(x)
Examples
1. Find the remainder when x5 + 4x4 – 6x2 + 3x + 2 is divided by x + 2
Solution
P (x) = x5 + 4x4 – 6x2 + 3x + 2
x – a = x + 2
a = -2
p(-2) = (-2) 5 + 4(-2)4 – 6 (-2) 2 + 3x – 2 + 2
p(-2) = -32 + 64 – 24 – 6 + 2
= 66 – 62
= 4
P(-2) = 4
2. Find the remainder when 4x3 – 6x2 – 5 is divided by 2x – 1
Solution
P (x) = 4x3 – 6x2 – 5
x – a = (x – ½)
P = 4 – 6 x – 5
= 4
– 6 x 
– 5 =
= 1 – 3 – 10
2
2
P () = -6
) = -6Factors theorem
If ‘a’ is a zero of p (x) then (x – a) is a factor of p(x) i.e. p(x) = (x – a) Q (x)
Proof:
Let p (x) = (x – a) Q(x) + R
Given ‘a’ is a zero of p (x)
Then p(a) = 0
0 = (a – a) Q (a) + r
0 = r
r = 0
p (x) = (x – a) Q(x)
x – a is a factor of p (x)
Examples
Factorize completely the following polynomial function x4 – 5x3 + 6x2 + 2x – 4
Solution
Let p(x) = x4 – 5x3 + 6x2 + 2x – 4
P (1) = 1 – 5 + 6 + 2 – 4
= 0
P (2) = 24 – 5 (2)3 + 6 (2) 2 + 2 x 2 – 4
= 16 – 40 + 24 + 4 – 4
= 0
(x – 1) and (x – 2) are factors of P (x)
→P (x) = (x – 1)(x – 2) Q(x)
P (x) = (x2 – 3x + 2) Q(x)
Q (x) = x2 – 2x – 2
= (x2 – 2x + 1) – 3
= ((x – 1) +) ((x – 1) –)
) ((x – 1) –)→P (x) = (x – 1) (x – 2) (x – 1) +) (x – 1) –)
) (x – 1) –)Synthetic division
Synthetic division is the shortcut method to find the remainder when a polynomial function is divided by a factor x – a
Example
1. Use synthetic division to divide 2x3 + x2 – 3x + 4 by x + 3
Solution
x – a = x + 3
So; a = -3
Then
Then
Note that in the synthetic division the third row will contain the coefficients of the quotient and the remainder
2. Use synthetic division to divide 4x3 – 6x2 – 5 by 2x -1
Solution
x – a = 2x – 1
a = ½
Q (x) = 4x2 – 4x – 2
Remainder = -6
Rational zero theorem
Let p (x) = anxn + an – 1xn – 1 + ….. + ax + ao
Where
an, an – 1, a1, a0 are integral coefficients and
Let be a rational number in its lowest term
be a rational number in its lowest termThen if is a zero of p (x) when p is a factor of a0
a zero of p (x) when p is a factor of a0 q Is a factor of an
Example
To find zero of 2x3 – x- 3
If is a zero of the expression
is a zero of the expressionThen p is a factor of -3 ie -1, 1, -3, 3
2 i.e. 1, -1, 2, -2
We try -1, 1, -3, 3, , ½, -3/2, and 3/2
, ½, -3/2, and 3/2Partial fraction (decomposition of fraction)
The process of decomposition of fraction depends on one of the following;
1)To every linear factor ax + b in the denominator there corresponds a fraction of the form
2) To every repeated factor like (ax + b)n in the denominator there corresponds n fractions of the form 
3) To every factor of the form anxn + an – 1xn – 1+….. + a1x + a0 in the denominator there correspond fraction of the form 
4) If the degree of the numerator is greater than or equal to the degree of denominator, division is encouraged and the remainder is treated as in (1), (2) 0r (3)
Examples
1. Express in partial fraction
Solution
Let 
Where A and B are constant
3x + 7 = A (x + 4) + B (x – 2)
3x + 7 = Ax + 4A + Bx – 2B
3x + 7 = (A + B) x + 4A – 2B
3→x = (A + B) x
3 = A + B…. (i)
7 = 4A – 2B…. (ii)
2 (i) + (ii) gives
13 = 6A → A =
From (i)
3 =
18 = 13 + 6B
5 = 6B
B =
2.
Solution
x2 + 1 ≡ A (x + 1)3 + B (x + 1)2 (x – 1) + C (x + 1) (x – 1) + D (x – 1)
When x = 1
2 = 8A →A =
When x = -1
2 = -2D → D = -1
When x = 0
1 = A – B – C – D
1 = ¼- B – C – (-1)
4 = 1 – 4B – 4C + 4
1 = 4B + 4C …… (i)
When x = -2
5 = -A – 3B + 3C – 3D
5 = – – 3B + 3C + 3
– 3B + 3C + 3 20 = -1 – 12B + 12C + 12
9 = -12B + 12C
4 = 8C
C = ½
From (ii)
3 = -4B + 4 x ½
3 = -4B + 2
QUESTION
4. Express x4 + x3 – x2 + 1 into partial fraction
(x – 1) (x2 + 1)
(x – 1) (x2 + 1)
Quadratic inequalities
A quadratic inequalities is an inequality of one of the following four types
ax2 + bx + C < 0
ax2 + bx + C ≤ 0
ax2 + bx + C > 0
ax2 + bx + C ≥ 0
Where a, b and c are real numbers and a 0
0Solving quadratic inequality
Solving quadratic inequality involves changing inequality signs to equal sign to obtain the associated quadratic equation.
E.g. x2 + x – 2 ≤ 0 – quadratic inequality
x2 + x – 2 = 0 – associated quadratic equation
Example
1. Solving the following inequality
x2 + x – 2 ≤ 0
Solution
x2 + x – 2 = 0
x2 – x + 2x – 2 = 0
x (x – 1) + 2 (x – 1) = 0
X = -2 and x = 1
Testing the values
Test value -3
4≤ 0 False
Test value – 0
-2 ≤ 0 True
Test value 2
4 ≤ 0 False
→-2 ≤ x ≤ 1
2. Solve the following quadratic inequality
X2 – 3 > 2x
Solution
X2 – 3 > 2x = x2 – 2x – 3 > 0
Then,
x2 – 2x – 3 = 0
x2 + x- 3x – 3 = 0
x (x + 1) – 3(x + 1) = 0
(x + 1) (x – 3) = 0
x = -1 and x = 3
Test value -2
1 > – 4 → true
Test value 0
-3 > 0 → false
Test value 4
13 > 8 = true
x < -1 and x > 3
Exercise
Solve the following inequalities
a) (x – 2) (x – 1) > 0
b) (3 – 2x) (x + 5)
≤ 0
≤ 0
c) (1 – x) (4 – x) > x + 11
d) x2 – 2x + 3 > 0
x2 – 2x + 3 > 0e) 3x + 4 < x2 – 6 < 9 – 2x
Rational inequalities
Examples
1. Solve the inequality
Solution
1st Make one side equal to 0
Do not multiply by denominator since x – 5 is not known as positive or negative
2nd find real numbers that make either the numerator or the denominator equal to 0
I.e. -2x + 13 = 0
2x = 13
2x = 13
X = makes numerator = 0
makes numerator = 0 And x – 5 = 0
X = 5 makes the denominator = 0
Test value 4
-5 > 3 → false
Test value 6
4 > 3 → true
Test value 7
> 3 – false = 5 < x <
2. Solve the inequality
Solution
X = 6 makes the numerator = 0
X = 3 makes the denominator = 0
3. Find the possible values of x for which 
Solution
Test value P
-24 ≤ 0 →True
Test value -4
< 0 false Test value 2
0 < 0→ true
Test value 5
4 < 0 → false
→2< x < 4
Exercise
What values of x satisfy in each of the inequalities?
A.
B. 
C.
D.
Absolute value inequality
Examples
Solve the following inequalities
a) |2x – 3| < 5
b) |x – 2| > -3
c) |2x + 3| < 6
Solution
a) |2x – 3| < 5
2x – 3 < 5 or 2x – 3 > -5
2x < 8 2x > -2
x < 4 or x > -1
Examples
Solve the following inequalities
Solution
X < 0, 0.8 < x < 1 and x > 1
Determinant of a 3 x 3 matrix
Let
EXAMPLE
1. Find the determinant of
Solution
QUESTION
Evaluate the following determinant
A) 
B) 
C) 
Solutions to system of equation
Cramer’s rule
Let a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
Be a system of three equations in three variables
NOTE THAT;
System of linear equations are always in the form (Ax = b) if Ax = 0 i.e. 0 = b the system has no solution
If Ax = 0 and b = 0 i.e. 0 = 0 the system has many matrix coefficient of the system
Matrix coefficient of the system
Then,
Example
Use Cramer’s rule to solve the system
Solution
Exercise
Solve the given system of equations by using Cramer’s rule
1.
2.
The inverse of a 3 x 3 matrix
Definition
If A is a square matrix, a matrix B is called an inverse of matrix if and only if AB = I. So a matrix A has an inverse, hence A is an inverse, hence A is an invertible.
AB = I
Where I is identity matrix
Transpose of matrix
Let A be a 3 x 3 matrix
The transpose of matrix A is denoted by AT
Examples
Find AT if
Co-factor matrix ‘C’
Where
c11, c12, c13, c21, c22, c23, c31, c32, and c33 are called minor factor
From
Example:
Find A-1 of the following matrix
Solution
Examples
Solve the system, use disjoint method
Solution
Finding the cofactors from
Exercise
1. Use matrix inversion to solve the system
2. Find the inverse of the matrix A if
→Use the inverse obtained above to solve the system of the following equation
3. Solve the following system of equation using Cramer’s rule
4. a) If the matrix verify that Det A = Det AT
verify that Det A = Det ATb) Find the adjoint (A) of the matrix in a above and use it to solve the following system of equation.
BINOMIAL THEOREM
Pascal’s triangle
(a + b) o = 1
(a + b) 1 = a + b
(a + b) 2 = a2 + 2ab + b2
(a + b) 3 = a3 + 3a2b + 3ab2 + b3)
1
1 1
1 2 1
1 3 3 1
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 5 10 10 5 1
Arranging the coefficients
The arrangement given is called the Pascal’s triangle
Examples
1. Give an expanded form of 4 Taking the first three terms of the expansion find the value of the (1.025) 4 correct to 3 decimal places
4 Taking the first three terms of the expansion find the value of the (1.025) 4 correct to 3 decimal placesSolution
From Pascal’s triangle the coefficients are 1, 4, 6, 4, 1
2. Expand (2 – x) 6 in ascending powers of x. Taking x = 0.002 and using the first three terms of the expansion find the value of (1.9998) 6 as accurately as you can. Examine the fourth term of the expansion to find to how many places of decimals your answer is correctly
Solution
Coefficients = 1, 6, 15, 20, 15, 6, 1
3. Expand (1 – 2x) 5 in ascending powers of x hence find (0.98) 4 to four decimal places.
2. The binomial theorem
If n is a positive integer
nC1an – b + nC2an – 2b2 + nCran – rbr + … + bn Where nCr =
n something combine r at a time
5! = 5 x 4 x 3 x 2 x 1 = 120
nC1 =
nC2 = 
nC3 = 
Hence
Examples
1. Write down the term in x7 in the expansion
Solution
Coefficient
12C5 =
The term is
2. Write down the first 4 terms of the expansion of in ascending powers of x
in ascending powers of x Solution
9C1
9C2
9C3
9C4
3. Give the constant term in the expansion of
Solution
The required
nCr an – rbr
4. Find the ratio of the term in x5 to the term in x6 in the expansion of
Binomial series
If is any rational number, then
is any rational number, then
Note:
1) If = n where n∈+, then the series terminates at xn
= n where n∈+, then the series terminates at xn 2) If is not a positive integer then the series is infinite and converges only when |x| < 1 ( is a rational number)
is not a positive integer then the series is infinite and converges only when |x| < 1 ( is a rational number)Examples
Expand
The expansion is valid for |x| < 1 or -1 < x < 1
Note: the expansion is valid for and not for
and not for 
To expand will be 
will be 
∙ Expand
Solution
The expression is valid when 
3. Expand up to and including the term in x3
up to and including the term in x3 Solution
QUESTIONS
1. Show that if x is so small so as to neglect x3 and higher powers
2. Given that and β are roots of the equation ax2 + bx + c = 0 where a, b, and c are real and a 0. Write down the values of + β and β in terms of a, b, and c. State the conditions that the roots and β are equal in magnitude but opposite in sign. Hence find the value of k for which the equation
and β are roots of the equation ax2 + bx + c = 0 where a, b, and c are real and a 0. Write down the values of + β and β in terms of a, b, and c. State the conditions that the roots and β are equal in magnitude but opposite in sign. Hence find the value of k for which the equation 
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