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Posted by Chester Morton / Friday 27 May 2016 / No comments
Algebraic Expressions
In
algebra, letters are used to represent numbers. Letters that are used to
represent numbers are called variables. Letters which have fixed values are
called constants. For example; in the
formulae of area of a circle (Ï€r2), r is a variable (it varies) and
Ï€ is
the constant (it is always the same).
An algebraic expression is an expression that
contains only letters, numbers and at least one of the operations (+,-, ×, ÷). For
example; 4x – 3y + 6.
NOTE: multiplication and division signs are not
usually seen in algebraic expressions, but you have to know that ab means a is being multiplied by b,
and a/b means that b is dividing a.
In
an example like 2x + 8y – 4, 2 and 8 are the coefficients of x and y
respectively. 2x, 8y and 4 are each called a term separated by an operation. 4
is called a constant term. Please note that the coefficient of any letter
standing alone is 1. For example: a
+4a, since the coefficient of a is 1, therefore the final answer is 5a.
FORMING
AN ALGEBRAIC EXPRESSION
Statements
in words are often written as algebraic expressions in mathematics. Any letter
may be used for an unknown number, but different letters must be used for each
unknown number.
Examples: Let x represent an unknown
number,
18 more than the number = x +
18
23 less than the number = x –
23
5 times the number = 5x
Quarter of the number = x/4
If 4 times a number is subtracted
from 3 and the result is multiplied by 7
=7(3 – 4x)
Trial
Questions
One less than x
A certain number plus eight is equal
to three times the number.
Half a certain number is three more
than the number.
OPERATIONS
ON ALGEBRAIC EXPRESSION
The rules of addition, subtraction, multiplication
and division can be used to simply algebraic expressions.
Addition
and Subtraction
Only like terms can be added or
subtracted to give a single term, this is often called collecting of like
terms.
Illustrative
examples
Simplify the following expressions;
1.
5x + 4 – 9y + 3x + 2y
– 7
2.
2p – 3q + 6p + 7q
Solution
1.
5x + 4 – 9y + 3x + 2y
– 7
Grouping like terms, we have
5x + 3x – 9y + 2y + 4 – 7
Simplifying like terms, we have
8x – 7y – 3
2.
2p – 3q + 6p + 7q
Grouping like terms, we have
2p + 6p
– 3q + 7q
Simplifying like terms, we have
8p +
4q
Trial Questions
1.
8x + 5y + 4 – 4x + 3y – 2
2.
6m + 7n + 5m – 9n
3.
5q2 + 6q + 4q2 + 7q
Multiplication
and Division
It is very important to first of all group all numbers
and same letters together and then apply basic rules of indices.
1.
am × an =am + n
2.
am =am – n
an
Examples under
multiplication:
1.
62 × 64
62+4
66
2.
5a2 × 4a
Grouping
the numbers and letters together, we have
5 × 4a2
× a
20a2+1
20a3
Trial Questions
1.
3x2 × 4x3
2.
2a × 5a2
3.
3xy2 × 4x3y4
Examples under
division:
1.
62 ÷ 64
62-4
6-2
2.
32a4b2 ÷ 4a2b
It
advisable to solve using this procedure
32a4b2
4a2b
8a4-2b2-1
8a2b
Trial Questions
1.
16a2b4 ÷ 8ab3
2.
27a5b2 ÷ 9a2b
3.
4a2 ÷ 4a2
EXPANDING
EXPRESSIONS
Brackets are used to group two or more terms
together. The process of simplifying such an expression is called expanding the expression. The law of distribution is applied, where
the outside term multiples each and every term in the bracket. Be very careful with the operational signs
when expanding expressions.
Illustrative
Examples:
1.
2a(6a + 7b)
2a
multiplying each term, we have
(2a ×
6a) + (2a × 7b)
12a2
+ 14ab
2.
m(2p – 4q + 2r)
m
multiplying each term, we have
(m ×
2p) – (m × 4q) + (m × 2r)
2mp – 4mq
+2mr
Trial Questions
1.
– 3 (2x – 4)
2.
3a(4a + 2b + ab)
3.
9(4m – 3n)+3(2n + m)
MULTIPLYING
TWO BINOMIALS
A binomial is an expression, consisting of two
terms. For example: (2q+p) and (m+n)
are all binomials. When multiplying two binomials, it is necessary to use the
law of distribution. That is each term in the first bracket will multiply each
term in the other bracket.
Illustrative Example:
1.
(2a – b)(a + 2b)
2a (a + 2b)
– b (a + 2b)
2a2 + 4ab – ab
– 2b2
= 2a2 + 3ab –
2b2
2.
(3x + 2y)(3x – 2y)
3x (3x – 2y)+2y(3x – 2y)
9x2 – 6xy + 6xy – 4y2
9x2
– 4y2
Trial Questions
1.
(4x + y)(2x + 3y)
2.
(a - b)(2a - 2b)
3.
(4m - 3n)(2m + n)
4.
(6p + 2q)(3p - q)
5.
(4p + 4q)(2s + 3t)
FACTORIZING AN ALGEBRAIC EXPRESSION
When each term in an expression has a common
letter or number, we divide each term by the common number or letter and
enclose the quotient in a bracket; the factor is placed outside the bracket.
The process of putting an expression in bracket is called FACTORIZATION. For example: 6x+6y have 6 to be its highest common
factor. Hence, we have the final answer to be 6(x + y).
There are
three method of solving algebraic expressions;
Method 1:
Common Factor Method
In this method, you will have to find a factor
which is common to all the terms in the expression and bring it outside the bracket,
hence leaving the uncommon factors in the bracket.
Illustrative
Examples
1.
100x – 25x2
In this expression,
the highest common factor is 25x
25x (4 – x)
We can expand the answer to be sure its
correct.
25x (4 – x)
(25x × 4) – (25x × x)
100x – 25x2
2.
5ax+25ay
The highest common
factor is 5a, hence
5a(x+5y)
1.
2×5.89 + 2×3.11
2.
4x + 2x2
3.
3a2x + b2
Method 2: Grouping Method
In this type, there are usually four terms. You
have to group them into two binomials, and then find the common factor of each
binomial and bring the common factor outside the bracket of each binomial.
NOTE: You can rearrange the terms to ensure that each
binomial has a common factor within.
Illustrative Examples:
1.
ab – by – ay + y2
(ab – by) – (ay + y2)
b(a – y) – y(a – y )
(b – y) (a – y)
2.
3x2 + 2xy – 12xz –
8yz
(3x2 +
2xy) – (12xz – 8yz)
x (3x + 2y) – 4z(3x +
2y)
(x – 4z) (3x + 2y)
Trial Questions
1.
x2 – xm + xy – my
2.
9r2 + 3rs – 15rt – 5st
3.
4p2 – 6tp + 18pq + 27tq
Trinomial is an expression with three terms. Any
expression in the form ax2+bx+c, where a, b, and c are constants are
not equal to zero is a quadratic trinomial.
For example:
2x2 +
3x +1 is a quadratic trinomial where 2 and 3 are the coefficient of x2
and x respectively, and 1 is a constant.
There are steps involved in solving a quadratic
equation or expression;
1. Multiply the coefficient of x2 by the constant.
2.
Find the possible pair of factors of the results in step 1, which when
added gives the coefficient of x.
3.
Replace the coefficient of x by the factors obtained in step 2 and use
the method of grouping to complete the factorization.
Illustrative example:
Example 1: 4x2 – 8x + 3
Step 1: The coefficient of x2 is 4 which is
multiplied by the constant, which is 3. Therefore we have, 4 × 3=12
Step 2: Possible pair of factors of 12 are (1 and 12), (2
and 6), (3 and 4), (-2 and -6), and a few other. Now, (-2 and -6) are the only
factors that adds up to -8 which is the coefficient of x.
Step 3: Replace -8x by -2x -6x in the expression and use
grouping method to factorize.
4x2 – 8x + 3 = 4x2
– 2x – 6x + 3
2x (2x – 1) –
3(2x – 1)
(2x – 3) (2x
– 1)
Example 2: 3x2
– 2x – 5
Step 1: The coefficient of x2 is 3 which is multiplied
by the constant, which is -5 in this case. Therefore we have, 3 × (-5) = -15
Step 2: The possible
pair of factors of – 15 is (- 1 and 15), (1 and – 15), (– 3 and 5) and (5 and –
3). But of these, only the pair of (3 and – 5) adds up to – 2 which is the
coefficient of x.
Step 3:
Replace – 2x by 3x and – 5 xs in the expression and use the grouping method to
factorize.
3x2
– 2x – 5 =3x2 + 3x – 5x – 5
3x(x + 1) – 5(x + 1)
(3x – 5)(x + 1)
Trial Questions
1.
X2 + 6x – 27
2.
5x2 – 13x – 6
3.
X2 – 8x + 12
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MATHEMATICS
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