TOPIC 6: ALGEBRA ~ ADV MATHEMATICS FORM 5 -

Table of Contents

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 = 1

Laws 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

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

Thus

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∈

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

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

and

are roots of quadratic equation

Then (x –

) (x – β) = 0

x2 – βx –

x +

β = 0

x2 – (β +

) x +

β = 0

Given 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

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

2. The equation 3x2 – 5 + 1 = 0 has roots

and β

a) Find values of

b) Find the values of

Solution

+ β =

and

β =

Roots of cubic equations

, β,

are roots of a cubic 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 written as

ax3 + bx2 + cx + d = 0

Equating coefficients of x2, x and the constant terms

+ β + γ =

-; sum of roots

ii)

γ + βγ +

β =

; sum of products of roots

iii)

γβ =

; products of roots

Examples

1. The equation 3x3 + 6x2 – 4x + 7 = 0 has roots

, β, γ. Find the equations with roots

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

+ 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

→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 (

)

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)

= 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)

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) –

)

→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 division to divide 2x3 + x2 – 3x + 4 by x + 3

Solution

x – a = x + 3

So; a = -3

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

Then if is

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

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 (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 =

+ B

18 = 13 + 6B

5 = 6B

B =

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

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)

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

Solving 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

c) (1 – x) (4 – x) > x + 11

x2 – 2x + 3 > 0

e) 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

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

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?

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

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

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

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 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

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