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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 (πr2), r is a variable (it varies) and π 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.

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)

Trial Questions
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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