1. 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
2. Zero exponents
Consider am x ao = am + o
ao = 1
Laws of logarithm
If a and b are two positive numbers there exist a thirdnumber c such that
ac = b
→c is the logarithm of b to base a
i. e = c
Definition
Logarithm of ‘x’ to base ‘a’ is the power to which ‘a’ must be raisedto give ‘x’.
If p = and q = , then = x and = y
Thus
1.
2.
3.
4. Note that:
There are two important bases of logarithms
10 and e
Series
A series is the sum of a sequentiallyorderedfinite or infiniteset 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 definedthe first but not the last term e.g. 1 + 3+ 5+ 7+ 9+ 11+ …..
In both cases the first term is 1
The sigmanotation
∑ stands for ‘’ sum of ‘’
5. e.g.
Exercise
Discuss the following and find the sum if n = 8
The sum of the first nnatural numbers
The sum of squares of the first nnatural numbers
6. The sum of the cubes of the first n natural numbers
Example
I. If an = n2 + 3n + 1 determine an expressionfor n
II. If an = n3 + 2n2 + 4nevaluate
a) a1 b) a4 c)
The sum of the first nnatural numbers
7. The sum of the squares of the first n natural numbers
Proof,
8. Exercise
1. Evaluate
Proof bythe mathematical induction
Example
Prove that n (n2 + 5) is exactlydivisible 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) Whenn = k + 1
(k + 1) ( (k +1)2 + 5) = (k + 1) (k2 + 2k + 1 + 5)
9. = (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∈ +
Principle of proof bymathematical induction
It states if s1, s2, s3…..Sn…. is a sequence of statementsand 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 inductionthat 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
10. 2 + 4 + 6 + …. 2k = k (k + 1)
Requiredto 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 inductionthat
Solution
Proof:
When n = 1,
Also n = 1 give
L.H.S = R. H. S
Let the statement be true for n =k
Let
Requiredto prove when n = k + 1
11. 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
12. 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 ofa polynomial function
If and are roots of quadratic equation
Then (x - ) (x – β) = 0
x2 – βx - x + β = 0
x2 – (β + ) x + β = 0
Given a quadratic equationas
ax2 + bx + c = 0, where a, b, c, are constant
Summary
13. 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
Solution
Sum of roots 2 β + β2 = β ( + β)
Products of root are ( 2 β) (β2 )
= 3 β3
= ( β) 3
( 2 β) (β2 ) = ( β)3
The given equation can be written as
The required equation is
=0
14. 2. The equation 3x2 – 5 + 1 = 0 has roots and β
a) Find values of
b) Find the values of
Solution
+ β = and β =
15. Roots of cubic equations
If , β, are roots of acubic equation then
(x - )(x– β)(x – γ) = 0
(x2 - x – βx + β) (x – γ) = 0
x3 – γx2 - x2 + γx – βx2 + βγx+ βx - βγ = 0
x3 – ( + β + γ) x2 + ( γ+ βγ + β) x - βγ = 0
x3 – ( + β + γ) x2 + ( γ + βγ + β) x - βγ = 0
Given cubic equation can be writtenas
ax3 + bx2 + cx+ d = 0
Equating coefficientsof x2, x and the constant terms
i) + β + γ = -; sum of
roots
ii) γ + βγ + β = ; sum of products of
roots
iii) γβ = ; products of roots
Examples
1. The equation3x3 + 6x2 – 4x + 7 = 0 has roots , β, γ. Find the equations with roots
16. 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 equation3x3 + 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
a) + 1, β + 1, γ + 1 b) 2, β2, γ2
Solution
4x3 + 7x2 – 5x – 1 = 0
18. Definition:
A polynomial is an expressionof the form
anxn + an – 1 x n – 1 + an – 2 xn -2 + …… + a1x + a0
Where an, an – 1, an – 2 …a1, ao are real numbers known as coefficients of the
polynomial
→an 0
→anxn is the leading term
→n is calledthe degree of the polynomial
Normallythe polynomial is writtenas 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 calleda dividend
19. Q (x) is calleda quotient
D (x) is calleddivisor
r (x) is calledremainder
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 ( )
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 (azero) of p(x)
Examples
1. Find the remainder when x5 + 4x4 – 6x2 + 3x + 2 is divided by x + 2
Solution
20. 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
=
= 1 – 3 – 10
2
P ( ) = -6
Factors 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)
21. 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 completelythe followingpolynomial functionx4 – 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)
22. Q (x) = x2 – 2x – 2
= (x2 – 2x + 1) – 3
= ((x – 1) + ) ((x – 1) - )
→P (x) = (x – 1) (x – 2) (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 divisionto divide 2x3 + x2 – 3x + 4 by x + 3
Solution
x – a = x + 3
So; a = -3
Then
23. Note that in the synthetic division the third row will contain the coefficients of the quotient
and the remainder
2. Use synthetic divisionto 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 coefficientsand
Let be a rational number in its lowest term
Then if is a zero of p (x) when p is a factor of a0
q Is a factor of an
24. Example
To find zero of 2x3 – x- 3
If is a zero of the expression
Then 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
Partial fraction(decompositionof fraction)
The process of decompositionof fractiondepends onone 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
correspondfractionof the form
4) If the degree of the numerator is greater than or equal to the degree of denominator,
division is encouragedand the remainder is treatedas in (1), (2) 0r (3)
Examples
1. Express in partial fraction
Solution
Let
Where A and B are constant
25. 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 = +
B
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)
26. 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
20 = -1 – 12B + 12C + 12
9 = -12B + 12C
4 = 8C
C = ½
From (ii)
3 = -4B + 4 x ½
3 = -4B + 2
27. QUESTION
4. Express x4 + x3 – x2 + 1 into partial fraction
(x – 1) (x2 + 1)
Quadratic inequalities
A quadratic inequalities is an inequality of one of the followingfour 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
Solving quadratic inequality
Solving quadratic inequality involves changing inequality signs to equal sign to obtain the
associatedquadratic equation.
E.g. x2 + x – 2 ≤ 0 – quadratic inequality
x2 + x – 2 = 0 – associatedquadratic equation
Example
1. Solving the followinginequality
x2 + x – 2 ≤ 0
Solution
x2 + x – 2 = 0
28. 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 followingquadratic 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
29. (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 followinginequalities
a) (x – 2) (x – 1) > 0
b) (3 – 2x) (x + 5) ≤ 0
c) (1 – x) (4 – x) > x + 11
d) x2 – 2x + 3 > 0
e) 3x+ 4 < x2 – 6 < 9 – 2x
Rational inequalities
Examples
1. Solve the inequality
30. 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
X = 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
31. 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
32. 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 satisfyin eachof the inequalities?
A.
33. B.
C.
D.
Absolute value inequality
Examples
Solve the followinginequalities
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 followinginequalities
34. Solution
X < 0, 0.8 < x < 1 and x > 1
Determinant of a 3 x 3 matrix
Let
35. EXAMPLE
1. Find the determinant of
Solution
QUESTION
Evaluate the followingdeterminant
A)
B)
C)
36. Solutions to system ofequation
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
37. Use Cramer’s rule to solve the system
Solution
Exercise
Solve the given system of equations by using Cramer’s rule
38. 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 identitymatrix
Transpose of matrix
Let A be a 3 x 3 matrix
The transpose of matrixA is denotedby AT
Examples
Find AT if
39. Co-factor matrix‘C’
Where
c11, c12, c13, c21, c22, c23, c31, c32, and c33 are calledminor factor
From
Example:
Find A-1 of the followingmatrix
42. Exercise
1. Use matrix inversionto 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
43. 3. Solve the followingsystem of equationusing Cramer’s rule
4. a) If the matrix verify that Det A = Det AT
b) 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 4 6 4 1
1 5 10 10 5 1
Arranging the coefficients
The arrangement given is calledthe Pascal’s triangle
44. 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
Solution
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 ascendingpowers of x hence find (0.98) 4 to four decimal places.
45. 2. The binomial theorem
If n is a positive integer
nC1an – b + nC2an – 2b2 + nCran – rbr + … + bn
Where nCr =
n somethingcombine 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
46. 2. Write down the first 4 terms of the expansionof 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
47. If is any rational number, then
Note:
1) If = 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)
Examples
Expand
The expansion is valid for |x| < 1 or -1 < x < 1
Note: the expansion is valid for and not for
To expand will be
Expand
Solution
49. 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 has roots
equal in magnitude but opposite insign.
