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Title: Algebra Expression
Description: This notes consists of algebra expression. In this lesson you will learn expanding brackets, use four maths operations, polynomials and factorisation. It is good for grade 11&12 learners and students doing mathematics and science related courses. It was written in University of NamibiA.
Description: This notes consists of algebra expression. In this lesson you will learn expanding brackets, use four maths operations, polynomials and factorisation. It is good for grade 11&12 learners and students doing mathematics and science related courses. It was written in University of NamibiA.
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Mathematics
Chapter 1: Algebraic expressions
Basic Mathematics
Mathematics
Chapter 1: Algebraic expressions
The word Mathematics derives from the Greek µαϑηµα meaning
”knowledge, study, learning”
...
The origins of counting can be traced back 50000 years to the
neolithic age
...
1900
BC), the Rhind Mathematical Papyrus (Egyptian c
...
1890
BC)
...
Mathematics
Chapter 1: Algebraic expressions
The study of mathematics as a ”demonstrative discipline” begins
in the 6th century BC with the Pythagoreans, who coined the term
”mathematics”
...
This is not the time and place to expand on the history of
mathematics
...
With the
arrival of computers, all kinds of mathematical techniques that
used to be of merely theoretical importance have acquired and
continue to acquire new and surprising applications
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Algebraic expressions
An algebraic expression is a meaningful mathematical term
containing numbers, variables and the ordinary operations of
arithmetic
...
In this chapter, our aim is to learn how to manipulate and
transform algebraic expressions
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
From the online etymology dictionary:
algebra (n
...
treatise on equations by Baghdad mathematician Abu Ja’far Muhammad
ibn Musa al-Khwarizmi
...
”
Al-Khwarizmi’s book (translated into Latin in 12c
...
John Dee (16c
...
The accent shifted 17c
...
The same word was used in English 15c
...
to mean ”bone-setting,” as
was Medieval Latin algebra, a usage picked up probably from Arab
medical men in Spain
...
1: Simplify the given expressions:
(i) −3x − (5 − 4x)
...
(iii) −4(x − 5) + 3(4 − 2x)
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Example 2
...
(ii)
1
5x
− 34 x
...
(iv) 2a2 − b + 3(a2 − b)
...
11
(ii) 51 x − 34 x = x( 15 − 43 ) = 1·4−3·5
20 x = − 20 x
...
(iv) 2a2 − b + 3(a2 − b) = 2a2 − b + 3a2 − 3b = 5a2 − 4b
...
Example 2
...
b d
2
...
(iii) x −
1 x +1
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Fractional algebraic expressions have variables in the
denominator; they are treated according to the rules of adding
or multiplying fractions
...
2: Simplify or combine the following
(i) a − c
...
(ii) xy − x +
y
2 − 1
...
b d
bd
2 = xy (x + y ) − 2 = x 2 y + xy 2 − 2
...
(x − 1)(x + 1)
x −1
Algebraic expressions may also be multiplied or divided by one
another
...
(ii) (a + b)2 = a2 + 2ab + b 2
...
(iv) (a + b)3 = (a2 + 2ab + b 2 )(a + b) = a3 + 3a2 b + 3ab 2 + b 3
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Example 2
...
(ii) (2 + 3)3
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Example 2
...
(ii) (2 + 3)3
...
Solutions:
(i) (x − 3)(x + 4) = x 2 + x − 12
...
(iii) (3xy − 1)3 = 27x 3 y 3 − 3(9x 2 y 2 ) + 3(3xy ) − 1 =
= 27x 3 y 3 − 27x 2 y 2 + 9xy − 1
...
(2) x 1 = x
...
(4) x m+n = x m x n
...
(6) x −n =
(7)
(8)
1
1 n
x n = ( x ) , provided x 6= 0
...
xm
xn
(9) (xy )m = x m y m
...
= 0 if n is a positive integer
...
Mathematics
Chapter 1: Algebraic expressions
Example 2
...
10
(ii) 5−13
...
(iv) (−3x −5 y −3 )4
...
2a
−1 −1
−a
b
...
10
(ii) 5−13 = 510+13 = 523
...
(iv) (−3x −5 y −3 )4 = 81x −20 y −12
...
(v) ( 3a−3m )−3 = ( 32 a(2+3)m )−3 = ( 23 )3 a−15m = 27
2a
(vi)
−a−1 b −1
−5ab −3
= 15 a−1−1 b −1+3 = 51 a−2 b 2
...
√
1
Let n be an integer
...
√
√
The symbol n is called a radical sign and n a is called a
radical ( ”radix” is Latin for ”root”)
...
Also, note
√
1
that 9 = 9 2 = 3 even though (−3)2 = 9 = 32
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Example 2
...
2
(ii) (−8) 3
...
5
(iv) 1000 4
...
(v) 4 81
16
√
3
(vi) 1252
...
2
2
(ii) (−8) 3 = ((−2)3 ) 3 = (−2)2 = 4
...
√
√
√
5
5
15
4
4
(iv) 1000 4 = (103 ) 4 = 10 4 = 1015 = 1012+3 = 1000 · 4 1000
...
16 = ( 24 ) = ( 2 )
√
1
3
(vi) 1252 = (53·2 ) 3 = 25
...
1
The following is standard procedure with rational exponents:
√
√
m
1
1
a n = (a n )m = (am ) n = n am = ( n a)m
...
Mathematics
Chapter 1: Algebraic expressions
Example 2
...
1
1
(ii) (a4 y 2 ) 4
...
2
n
1
(iv) (x n )n (x 6 ) n
...
1 1
1
(ii) (a4 y 2 ) 4 = a4· 4 y 2 · 4 = ay 8
...
2
n
1
2n
n
1
13
(iv) (x n )n (x 6 ) n = x n x 6n = x 2 x 6 = x 6
...
The general form of a polynomial is
an x n + an−1 x n−1 +
...
, an are real
numbers
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Remarks:
If two polynomials in x are added or multiplied, expanding
and combining like terms, the outcome is a polynomial in
x
...
Degree of a polynomial
Given a polynomial f (x) = an x n + an−1 x n−1 +
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Examples of polynomials
1
2
3
4
5
x 3 − 3x 2 + x − 1 is a polynomial of degree 3 with leading
coefficient 1
...
−3 is a polynomial of degree 0 with leading coefficient
−3
...
(x − 3)(−x + 2) = −x 2 + 5x − 5 is a polynomial of
degree 2 with leading coefficient −1
...
√
x 7 − 4 5 x + 8
...
x2 + x + 1
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Definition 2
...
2
A polynomial with two terms is called a binomial
...
4
A polynomial of degree 2 is called a quadratic
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Comparing polynomials
When are two polynomials
f (x) = an x n + an−1 x n−1 +
...
+ b1 x + b0 the same?
Intuitively, this is the case if and only if the coefficient of each
power x k is the same in both polynomials
...
This
complicates the definition a little
...
Then f (x) = an x n + an−1 x n−1 +
...
+ b1 x + b0 are equal if and
only if, for each k from 0 to n, ak = bk or ak = 0 and k > m
...
A nonzero polynomial is usually written as a sum of terms that
are powers of x multiplied by nonzero coeffients (i
...
monomials of the form 0x k are omitted)
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Example 2
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Example 2
...
Solution: (i) and (iii) are equal and (iv) and (v) are equal
...
e
...
Example 2
...
(ii) 2x 5 + x 4 − x 3 + 4x − 1) − (x 5 + 3x 4 − 100x 2 − 100x − 6)
...
(ii) x 5 − 2x 4 − x 3 + 100x 2 + 104x + 5
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Example 2
...
Obtain the product:
(i)
(ii)
(iii)
(iv)
(x 2 − x + 1)(x − 2)
...
(3x − 1)2
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Solutions:
(i) (x 2 − x + 1)(x − 2)
= x 3 − 2x 2 − x 2 + 2x + x − 2 = x 3 − 3x 2 + 3x − 2
...
(iii) (3x − 1)2 = (3x)2 − 2 · 3x + 12 = 9x 2 − 6x + 1
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
All roots are assumed to be defined
...
Product rule
√
p
n a
n a
2
√
=
Quotient rule
n
b
b
Mathematics
Chapter 1: Algebraic expressions
Example 1
...
Simplify
√
3
(i) 125 x 6
...
r
−32y 5
(iii) 5
...
(v) √9
...
√
√
q
3 = √ 3 = 3
...
20· 15
x 36
x
x
√
√
√
(iv) 20 = 4 · 5 = 2 5
...
3
3· 3
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Arithmetic with radicals of the same index
We say that
n = m
...
11: Combine and simplify
√
√
(i) 27 + 3
...
p p
(iii) 4 4y 3 4 12y 2
...
5
p√
2
...
√
√
√
√
3
3
− 3 81x = 23 x −√ 3 · 33 x =
(ii) 3 8x √
√
√
√
3
3
2 3 x − 3 3 3 x = 3 x(2 − 3 3)
...
√
√
√
1
2 =
(iv) √40 = ( 40
)
8
=
2
2
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Number fields
Number fields
N = {0, 1, 2, 3,
...
, −4, −3, −2, −1, 0, 1, 2, 3, 4,
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Rationalising the denominator
The process of removing radicals from a denominator is called
rationalising the denominator
...
The
√ reason for the name ”rationalising” is that many radicals, such as
2, are irrational
...
12: Simplify and/or rationalise
√
√
√
(i) ( 2 − 2 3)(2 + 4 3)
...
√
3√
...
(v) √ 1 √
...
1√
...
√
√
√ 2
(ii) (3 − 6)(3 + 6) = 32 − 6 = 9 − 6 = 3
...
= 3 3+
2
3
3 −6
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
√
√
√ √
√
(iv) ( 8 − 3)2 = 8 − 2 8 3 + 3 = 11 − 4 6
...
(v) √ 1 √ = √
3
−
2
2+ 3
( 3 + 2)( 3 − 2)
√
√
√
(vi) ( x − 9 − 3)2 = x − 9 − 2 · 3 · x − 9 + 9 = x − 6 x − 9
...
3− 2
( 3 − 2)( 3 + 2)
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Long division with polynomials
Long division with polynomials is similar to division of integers
with a remainder
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
A first example of polynomial long division
We compute 2x 4 + x 3 − x + 1 : x − 7
...
Here, the dividend is the polynomial 2x 4 + x 3 − x + 1, the divisor
is x − 7, the quotient is 2x 3 + 15x 2 + 105x + 734 and the
remainder is 5139
...
2
7
2
−1
1
0
1
14
105
735 5138
15
105
734 5139
Although we will not pursue this way of displaying the
computation, this is correct
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
The division algorithm
The result of a long division with polynomials is displayed in the
format
dividend = quotient + remainder
...
To verify your computation, check if
quotient × divisor + remainder = dividend
...
13 Obtain the quotient, performing long division
where necessary:
(i)
(ii)
(iii)
(iv)
(v)
(−9x 7 ) ÷ (−3x 2 )
...
(x 3 − 8) ÷ (x − 2)
...
(x 3 + 3x 2 − 3x + 4) ÷ (x + 4)
...
6x 3 − 4x = −3x 2 + 2
...
Expansion of algebraic expressions
Exponents
Polynomials
Mathematics
Chapter 1: Algebraic expressions
(iv): (x 3 + 3x 2 − 3x + 4) ÷ (x 2 − x + 1) =? :
We have:
x +4
x2
−x +1
x3
3x 2
+
− 3x + 4
− x3 + x2 − x
4x 2 − 4x + 4
− 4x 2 + 4x − 4
0
3
2
So x + 23x − 3x + 4 = x + 4
...
x +4
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
If the leading coefficient of the divisor polynomial is not equal to 1
or −1, we may have to expect quotients with non-integer
coefficients, as the following example shows:
1 4
2x
2x − 1
+ 41 x 3 + 18 x 2 −
x5
− x 5 + 12 x 4
−
1 4
2x
1 4
2x
23
16 x
− 3x 2
−
23
32
+1
+ 41 x 3
−
1 3
4x
1 3
4x
− 3x 2
+ 18 x 2
−
23 2
8 x
23 2
8 x
−
−
23
16 x
23
16 x
23
16 x
+1
− 23
32
9
32
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Factorisation of polynomials
Factorising is the reverse of expanding brackets, so it is, for
example, putting 2x 2 + x − 3 into the form (2x + 3)(x − 1)
...
The first step of factorising an expression is to ”take out” any
common factors which the terms have
...
There 4x and x − 2 are ”simpler”
expressions; they are called factors of 4x 2 − 8x
...
e
...
Examples of irreducible polynomials
Examples of irreducible polynomials are all polynomials of
degree 0 or 1 and quadratic polynomials such as x 2 + 1
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Evaluation of a polynomial at a given value:
Let f (x) = an x n + an−1 x n−1 +
...
Let c
be a real number
...
+ a0
...
Evaluation example:
Let f (x) = 3x 4 − 5x + 10 and c = −3
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Factor Theorem and Remainder Theorem
Consider a polynomial
f (x) = an x n + an−1 x n−1 +
...
Let c be a real number
...
In particular, x − c is a factor of f (x)
if and only if f (c) = 0
...
The Factor Theorem says that c is
a root of f (x) if and only if x − c is a factor of f (x)
...
If we apply the division algorithm to f (x) and x − c, we obtain
f (x)
r (x)
x − c = q(x) + x − c , where q(x) is the quotient and r (x) is
the remainder
...
Thus we have:
f (x)
a
x − c = q(x) + x − c , so
f (x) = q(x)(x − c) + a and
f (c) = q(c)(c − c) + a = 0 + a = a
...
Find the remainder when
x 3 − 2x 2 + 6 is divided by x − a
...
Find c
...
When x 3 + dx 2 + ex − 4 is divided by x + 1, the remainder is
−4
...
(iv) Let f (x) = x 4 + a2 x 2 − 1
...
(ii) Let g (x) = x 2 − 7x + c
...
Hence c = 7
...
We are given the information
that 0 = f (1) = 1 + d + e − 4
...
Also,
f (−1) = −4 = −1 + d − e − 4
...
It follows that
(d + e) + (d − e) = 2d = 3 + 1 = 4 and d = 2
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
(iv) The remainder when f (x) is divided by x − 2 is
f (2) = 16 + 4a2 − 14a2 + 15 So we need to solve 4a2 + 15 = 31
...
There are two possible values for
a, 2 and −2
...
The assumption
f (2) = 0 leads to 4a2 = −15, for which there is no real solution
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Factorisation
Many polynomials that you will encounter in this class can be
factorised using a few simple tricks
...
15 Factorise
(i)
(ii)
(iii)
(iv)
9x 4 − 6x 3 + 12x 2
...
aw + bc − bw − ac
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Solutions:
(i) 9x 4 − 6x 3 + 12x 2 = 3x 2 (3x 2 − 2x + 4)
...
(iii) aw + bc − bw − ac = w (a − b) − c(a − b) = (a − b)(w − c)
...
Expanding the
product, we obtain a + b = −1 and ab = −2
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Factorisation of quadratics: A method for simple cases
Given a polynomial a2 x 2 + a1 x + a0 with integer coefficients
a0 , a1 , a2 and a2 6= 0,
we try to find two integers d and e that multiply to give a2 a0 and
add to give a1
...
If we are lucky, this procedure will reveal a common factor
...
However, this is what you should try when someone gives you a
quadratic polynomial and asks you to find the roots by factorising
...
2x 2 + 7x + 3
...
15x 2 − 14x + 3
...
Expansion of algebraic expressions
Exponents
Polynomials
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Solutions:
(i) Here, we have a0 a2 = 6
...
We try −5 = −3 + (−2) and rewrite
x 2 − 5x + 6 as x 2 − 2x − 3x + 6
...
(ii): Again, a0 a2 = 6
...
Now we write
2x 2 +7x +3 = 2x 2 +6x +x +3 = 2x(x +3)+(x +3) = (2x +1)(x +3)
...
The integer divisors of −36 are
−36, −18, −12, −9, −6, −3, −4, −2, −1, 1, 2, 3 4, 6, 9, 12 18, 36
...
We rewrite 6x 2 + 5x − 6 as
6x 2 + 9x − 4x − 6 = 3x(2x + 3) − 2(2x + 3) = (3x − 2)(2x + 3)
...
The integer divisors of 45 are
−45, −15, −9, −5, −3, −1, 1, 3, 5, 9, 15, 45
...
Trying 15x 2 − 14x + 3 as 15x 2 − 15x + x + 3
will not unearth a common factor; we have better luck writing
15x 2 − 14x + 3 = 15x 2 − 5x − 9x + 3 = 5x(3x − 1) − 3(3x − 1) =
(5x − 3)(3x − 1)
...
a3 − b 3 = (a − b)(a2 + ab + b 2 )
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
The binomial formulas and some of their relatives, summary
(i) (a + b)(a − b) = a2 + ba − ba − b 2 = a2 − b 2
...
(iii) (a − b)2 = a2 − 2ab + b 2
...
(v) (a − b)3 = a3 − 3a2 b + 3ab 2 − b 3
...
(vii) a3 − b 3 = (a − b)(a2 + ab + b 2 )
...
17 Factorise:
(i) 4x 2 − 1
...
(iii) 9y 2 + 30y + 25
...
(v) x 3 − 27
...
(vii) y 3m − 1
...
(ii) x 2 − 4x + 4 = (x 2 − 2 · 2x + 22 ) = (x − 2)2
...
(iv) x 2n − 9 = ((x n )2 − 32 ) = (x n + 3)(x n − 3)
...
(vi) 8a6 + 125b 3 = ((2a2 )3 + (5b)3 ) =
= (2a + 5b)(4a2 − 20a2 b + 25b 2 )
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Miscellaneous problems
(i) Let f (x) = 12x 3 + 25x 2 + 9x − 4
...
(ii) Completely factorise the polynomial 4x 4 + 7x 2 − 2
...
(x − 2)(3x + 1)
(iv) Factorise x 4 + 8x 3 + 18x 2 + 15x + 5, given that −1 is
a root
...
You find this by trial and error
...
Next we compute:
12x 2 + 13x − 4
x +1
12x 3 + 25x 2 + 9x − 4
− 12x 3 − 12x 2
13x 2 + 9x
− 13x 2 − 13x
− 4x − 4
4x + 4
0
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Trying to factorise 12x 2 + 13x − 4, we need to find two divisors of
48 whose sum is 13
...
Thus f (x) = (x + 1)(4x − 1)(3x + 4) and the roots of f (x) are
−1, 14 , and − 43
...
Letting x 2 = u, we try to factorise the
polynomial 4u 2 + 7u − 2
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Putting x back where it belongs, we obtain
4x 4 + 7x 2 − 2 = (x 2 + 2)(4x 2 − 1)
...
Hence the
complete factorisation is
4x 4 + 7x 2 − 2 = (x 2 + 2)(2x + 1)(2x − 1)
...
(3x + 1)(x + 2)
...
This yields:
Mathematics
Chapter 1: Algebraic expressions
√
x + √2 =
x− 2
√
(x√+ 2)2 √
=
=
(x − √
2)(x + 2)
2
= x + 22 2x + 2 =
x√− 2
2
2x + 4
...
If we evaluate
this polynomial at x = −1, we obtain (−1)3 + 7 − 11 + 5 = 0
...
Mathematics
Chapter 1: Algebraic expressions
Expansion of algebraic expressions
Exponents
Polynomials
Title: Algebra Expression
Description: This notes consists of algebra expression. In this lesson you will learn expanding brackets, use four maths operations, polynomials and factorisation. It is good for grade 11&12 learners and students doing mathematics and science related courses. It was written in University of NamibiA.
Description: This notes consists of algebra expression. In this lesson you will learn expanding brackets, use four maths operations, polynomials and factorisation. It is good for grade 11&12 learners and students doing mathematics and science related courses. It was written in University of NamibiA.