Algebraic Expressions

ALGEBRAIC EXPRESSION

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.

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.      a^m × aⁿ =a^{m + n}

2.      a^m    =a^{m – 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

6^2-4

6^-2

2.      32a⁴b² ÷ 4a²b

It advisable to solve using this procedure

32a⁴b²

 4a²b

8a^4-2b^2-1

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 

           9^x2 – 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

Method 3: Quadratic Trinomials

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