Algebra 1: Algebraic Expressions and Indices
Mathematics
Topic Seven

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Algebraic Expressions

Algebraic expressions are expressions which contain terms, variables and coefficients, and constants.  The following is an example of an algebraic expression:

x2 + 6x + 9

Terms

Terms are the elements in an algebraic expression separated by the arithmetic signs such as:

The terms in the expression:

x2 + 6x + 9
x2
6x
9

A term may consist of variables and coefficients, or a constant.

Variables

The letters in a term are called variables. In the following expression:

3x + 4y - 5z, the variables are: x, y and z.

Coefficients

Coefficients are the numbers before variables.

Example

In the following expression,

4x2 + 6x + 9, the coefficients are:

4 and 6

Constants

Constants are the terms that contain numbers only.

Example  

In the following expression,

4x2 + 6x + 9, the constant is 9.

Monomial

3x

Binomial

3x + 4y

Trinomial

2x2 + 3x + 4

Polynomial

x3 + 2x2 - 3x + 4

Conducting Operations on Algebraic Expressions

Addition and Subtraction

The rules for the addition and subtraction of algebraic expressions are as follows:

Step 1: Group like terms because only like terms can be added and subtracted.

Step 2: Add or subtract the coefficients of the grouped like terms.

Example

    1. 4x + 3x + 2x – x

      9x – x = 8x.

    2. 7x2 + 6x – 8x – 3x2

      4x2 – 2x

    3. 7x3 + 5x2 + 9x – 8x – 3x2

      7x3 + 2x2 + x

Multiplication and Division

The rules for the multiplication and division of positive and negative numbers are as follows:

Rules for multiplication:

Step 1: State the coefficients and variables of each term separately.

Step 2: Multiply the coefficients, constants and variables.

Note when multiplying variables:

x5 = x  x. x x x

Rules for division:

Step 1: State the coefficients and variables of each term separately.

Step 2: Divide the coefficients, variables and constants by the coefficients, variables and  constants in the denominator.

Step 3: Cancel the like variables in the numerator by those in the denominator.

Examples

x8 / x6  = x 8-6 =2 = x2

8x8 / 2x6  = 4x 8-6 =2 = 4x2

Substituting Numbers for Algebraic Symbols

The values of algebraic expressions are obtained by substituting numbers in place of variables, and simplifying.

Example

Given x=2 and y=3,

Calculate:

4x2 + 6x + 9,

Solution

4(2)2 – 6(3) + 9 = 16 – 18 + 9 = 7.

Removing and Inserting Brackets

Removing Brackets

The distributive law is used when removing brackets. It is summarised by the identity below:

                c(a + b) = c x a + c x b

The distributive law basically states that when removing a bracket, use the term outside the bracket to multiply each term in the bracket.

Examples

4(4x + 2) = 16x + 8

4(4x + 3x) = 16x2 + 12x

5x(2x2 – x + 2) = 10x3 – 5x2 + 10x

-2x(x + 1) = -2x2 - 2                                             +

Inserting Brackets or Factorising

The distributive law is also used to insert brackets. However, when inserting brackets, the law is used in reverse.

The distributive law in reverse is a means of factorizing. Factorization is the breakdown of numbers into factors, which when multiplied, yields the original numbers.

Rules for factorizing are as follows:

Step 1: Write the variables and coefficients/factor common to all the terms outside the brackets.

Step 2:  Divide each term inside the bracket by the term placed outside the brackets in step 1, placing the quotient of the divisions inside the brackets while leaving the common coefficient/common factor outside the bracket.

Examples

4x + 2 = 2(2x + 1)

4x2 – 8x = 4x(x – 2)

Multiplying algebraic expressions

In multiplying algebraic expressions, we can apply the distributive rule.

Example:

Multiply 2x(3xy + 7xy)

Solution:

6x2y + 14x2y

Example:

Multiply -3/5x (15xy – 4)

Solution:

(- 3/ 5x times 15xy) + (-3/ 5 times – 4)

= -9x2y + 12/ 5x

Multiplying monomials and polynomials

Example: Multiply (x + 2y) (3x − 2y + 5)

Solution

(x + 2y)(3x − 2y + 5) =

3x2 − 2xy + 5x + 6yx − 4y2 + 10y =

3x2 + 4xy + 5x − 4y2 + 10y

Note: −2xy and 6xy are added because they are Like Terms and also 6yx means the same thing as 6xy.

Dividing a Polynomial

Dividing a polynomial by something more complicated than just a simple monomial requires the use of a method called long polynomial division and it works just like the long numerical division.

Example

Divide x2 – 8x – 9 by x + 1

Step 1:

Set up the division, putting the dividend (the thing being divided into) inside and the divisor (the thing doing the dividing) outside and to the left:

Figure 1

Step 2:

Divide the leading x2 of the dividend by the leading x in front to get an x on top of the division symbol, right above the x2 inside:

Figure 2

Step 3:

We then multiply the x through the divisor, x + 1.  We first multiply the x (on top) by the x (on the side) to get x2, and carry the resulting x2 underneath, putting it directly below the x2 from the dividend.

Step 4:

Next we multiply the x on top by the 1 on the side to get 1x and we then place the 1x underneath placing it directly below the –8x in the dividend:

Figure 3

Step 5:

We now draw the horizontal "equals" bar underneath the dividend, so that the subtraction can be done.

Figure 4

Step 6:

We must then subtract the top line by the bottom line which is the same as saying to change all the signs in the second line to get:

Figure 5

The next thing is to add where the first term (the x2) will cancel out because they have the same coefficient but opposite signs while the –8x – 1x becomes –9x:

Figure 6

Step 7:

Like the normal long division, we now bring down the next figure in the dividend which is - 9 to get:

Figure 7

At this point, we start ignoring the dividend, and instead work on the bottom line of the long division.

We now look at the x from the divisor and the new leading term, the –9x, in the bottom line of the division. In dividing the –9x by the x, we will get –9 which must be placed on top, right above the –8x:

Figure 8

Step 8:

We now multiply the  – 9 on top by the leading x on the side, to get –9x which we will carry to the bottom, directly underneath the previous line's –9x to get:

Figure 9

Step 9:

We then multiply the –9 on top by the 1 on the side to get -9, and carry this –9 to the bottom, directly below the previous line's –9  to get:

Figure 10

Step 10:

We will now draw another horizontal line and change the signs at the bottom line to get:

Figure 11

Adding both lines we get:

Figure 12

Because we ended with zeros, the division is complete and the answer is x - 9

To verify the answer, we can multiply x – 9 by x + 1 and see if we will get

x2 – 8x – 9

So (x – 9) times (x + 1).

Step 1: x times x yields: x2  

Step 2: -9 times x and 1 times x yields: -9x + x = -8x.

Step 3: we are left with -9 times 1 which yields -9.

When combining these three terms, we get x2 – 8x – 9, therefore,     

x2 – 8x – 9 divided by x+1  is in fact equal to x  – 9.

Simplification of Algebraic Equations

Example:  Reduce the algebraic fraction Figure 13 to its lowest term

Solution:

Figure 14

Example: Reduce the algebraic fraction Figure 15 to its lowest terms.

Solution:

Figure 16

Example: Simplify Figure 17

Solution

Step 1: Find the lowest common factor (the Lowest Common Multiple of 6 and 3 is 6)

Figure 18

Step 2: Express denominator in the Lowest Common Multiple Form

Figure 19

Step 3: Simplify the numerator

Figure 20

Step 4: Add the numerators to get:

Figure 21

Example

Simplify Figure 22

Solution

Step 1: Find the lowest common factor (the Lowest Common Multiple of 12 and 4 is 12)

Step 2: Express denominator in the Lowest Common Multiple Form

Figure 23

Step 3: Simplify the numerator

Figure 24

Note:

  • Be careful of a minus sign before a bracket.
  • Always reduce the final answer to its lowest terms.

Indices

A power, or an index, is used to write a product of numbers very compactly.  To manipulate math expressions, the law of indices must be considered. These laws can only apply to expressions with the same base such as 54 and 52 which equals 56 (restate the common base and just add the indices or powers).  It must be noted that the law of indices cannot be used in cases where the bases (5 and 4) are different such as 54 and 47.

The following are the rules for the Law of Indices:

Rule 1: a0 = 1

Rule 2: ax  x ay  = ax+y

Rule 3: ax  ÷  ay  = ax-y

Rule 4: a-x = 1/ax

Rule 5: ax/y = y√aor (y√a)

Rule 6: (ax)y = axy  

The laws of indices:

Examples

y0 = 1

30 = 1

1000 = 1

Examples

24 × 28 = 212

54 × 5-2 = 52 (4-2) 

(yn)m = ynm

Examples

39 ÷ 34 = 35

72 ÷ 75 = 7-3

-b = 1/yb

Examples

2-3 = 1/23 = 1/8

3-1 = 1/3

ym/n = (n√y)m

Examples

161/2 = √16 = 4

82/3 = (3√8)2 = 22 = 4

Example

25 + 84

= 25 + (23)4

= 25 + 212

Next - Algebra 2: Solving for Equations