Algebraic Expressions

Mathematics J.S.S 2 Second Term

Theme: Algebraic Process

Sub Theme: Algebraic Operation

WEEK 1

Algebraic Expressions

Performance Objectives

Students should be able to;

1. Expand a given algebraic expressions
2. Factorize simple algebraic expressions

Expansion and Simplification of Algebraic Expressions

Content

Remember that in algebra, letters stand for numbers. The number can be whole or fractional, positive or negative.

1. Just as 5a is short for 5 x a, so -5a is short for (-5) x a.

2. Just as m is short for 1 x m, so –m is short for (-1) x m.

3. Algebraic terms and numbers can be multiplied together. For example,

4 X (-3x) = (+4) x (-3) X x

= -(4 x 3) X x = -12 X x = -12x

(-2y) x (-8y) = (-2) X y X (-8) X y

= (-2) X (-8) X x X y

= +(2 X 8) X y²

= +16y² or just 16²

4. Division with directed numbers is also possible. For example,

18a ÷ (-6) = (+18) X a/(-6)

= – (18/6) X a

= (-3) X a = -3a

-33x² /-3x = (-33) X x X x/(-3) X x

= +(33/3) X x = 11x

Expanding algebraic expression

The expression (a + 2)(b – 5) means (a + 2) X (b – 5) means (a + 2) X (b – 5). The terms in the first bracket, (a + 2), multiply each term in the second bracket, (b – 5). Just as:

X(b – 5) = bx – 5x

So, writing (a + 2) instead of x,

(a + 2)(b – 5) = b(a + 2) – 5(a + 2)

The brackets on the right-hand side can now be removed.

(a + 2)(b – 5) = b(a + 2) – 5(a + 2)

= ab + 2b – 5a – 10

ab + 2b – 5a – 10 is the product of (a + 2) X (b – 5). We often say that the expansion of (a + 2)(b – 5) is:

ab + 2b – 5a – 10

Example

Expand the following:

a. (a + b)(c + d)

b. (6 – x)(3 + y)

Solution

a. (a + b)(c + d) = c(a + b) + d(a + b)

= ac _ bc + ad + db

b. (6 – x)(3 + y) = 3(6 – x) + y(6 – x)

= 18 – 3x + 6y – xy

We sometimes call this binomial expansion, since each bracket contain two terms (bi-nomial means two-names).

 

OR

Let us evaluate the expression below:

4 × (5 + 3) or 4(5 + 3)

We have,

4 × (5 + 3) = 4 × 8 = 32.

Similarly,

4 × (5 + 3) = 4 × 5 + 4 × 3

= 20 + 12 = 32.

Using letters (alphabets) in place of numbers,

a(b + c)  or  a × (b + c) = ab + ac

(b + c)a  = ba + ca.

You observed that the term outside the bracket is used to multiply all the terms inside the bracket.

Examples:

Expand the following algebraic expression.

1. 4(2a+2b)

2. 3(2n + 3m – 4y)

Solutions:

1. 4 × 2a + 4 × 2b = 8a + 8b

2. 3 × 2n + 3 × 3m – 3 × 4y = 6n + 9m – 12y.

Now consider the expression of the form,

(a + b)(c + d)

The expansion will be: a(c + d) + b(c + d) = ac + ad + bc + bd

Examples:

Expand the following and simplify where necessary.

1. (2a – 3b) (3a – 2b)

2. (p + 2q)(3p +8q)

Solutions:

1. (2a – 3b) (3a – 2b) = 2a(3a – 2b) – 3b(3a – 2b)

= 6a² – 4ab – 9ab + 6b²

= 6a² – 13ab + 6b²

2. (p + 2q)(3p +8q) = p(3p +8q) + 2q(3p +8q)

= 3p² + 8pq + 6pq + 16q²

= 3p² + 15pq + 16q²

To expand or remove brackets, use the following rules:

(i) a(x + y) = ax + ay

(ii) a(x − y) = ax − ay

(iii) −a(x − y) = −ax + ay

(iv) −a(x + y) = −ax − ay

Note the effects of negative terms outside brackets in (iii) and (iv)

Factorization of Simple Algebraic Expressions

You can also factorize quadratic expressions. Remember that factorizing an expression simplifies it in some way. Factorizing is the reverse of expanding brackets

If every term of an expression has a common factor, then the expression can be factorized.

Examples:

Factorize these expansions:

(i) −ax − 12x

(ii) 2xyz + 5xy

(iii) 2abc² + 4ab²c

(iv) 2abc + 4ab²c

Solutions:

(i) −ax − 12x = −x(a + 12)

(ii) 2xyz + 5xy = xy(2z + 5)

(iii) 2abc² + 4ab²c = 2ab(c + 2b)

(iv) 2abc + 4ab²c = 2abc(1 + 2b)

We note that in the factorization of expression involving two terms, we need to find factors that are common to all the expression.

Example:

Factorize the following:

(a) 12x = 8y

(b) a(a − 1) + 5(a − 1)

(c) abc + abd

Solution:

(a) 12x = 8y = 4 × 3x + 4 × 2y

= 4(3x + 2y)

(b) a(a − 1) + 5(a − 1) = a × (a − 1) + 5 × (a − 1)

= (a − 1) (a + 5)

(c) abc + abd = a × b × c + a × b × d

= ab(c + d)