Lesson 7: Expanding Products of Algebraic Expressions (Binomial Expansion)

Objective

At the end of this lesson, students should be able to:

• Expand products of algebraic expressions and simplify them.

When multiplying together expressions in brackets, it is necessary to multiply all terms in one bracket by all the terms in the other bracket.

The total area is (a +b) × (c +d)

= (a + b) (c + d)

It is made up of 4 smaller rectangles with areas ac, ad, bc and bd

So (a + b) (c + d) = ac + ad + bc + bd

OR

Example 1

Expand and simplify (x + 2) (x + 4).

Solution

All terms in one bracket must be multiplied by each term in the other bracket.

1^st (B1) × 1^st (B2) = x²

1^st (B1) × 2^nd (B2) = 4x

2^nd (B1) × 1^st (B2) = 2x

2^nd (B1) × 2^nd (B2) = 8

Example 2

Expand and simplify 2 (6p + 7q) – 3 (2p  – 5q)

Solution

2 (6p + 7q) – 3 (2p – 5q)

Collect like terms and simplify

= 12p + 14q – 6p + 15q

= 12p – 6p +14q + 15q

= 6p + 29q

 

Example 3

Expand and simplify (3x – 3y)²

 

Solution

The bracket is squared, so we have to write out the bracket twice and multiply each term.

(3x – 3y)² = (3x – 3y)(3x – 3y)

3x(3x) – 3x(3y) – 3y(3x) – 3y(-3y)

9x² – 9xy – 9xy + 9y²

9x² – 18xy + 9y²

 

Example 4

Simplify (x + 2)(x + 3)

Solution

We multiply x by (x + 3) and 2 by (x + 3)

(x + 2)(x + 3)= x(x + 3) +2 (x + 3)

= x² + 3x + 2x + 6

= x² + 5x + 6

 

Example 5

Simplify  (x + 4)(x + 3) – x²

Solution

(x + 4)(x + 3) – x² = x(x + 3) + 4(x + 3) – x²

= x . x + x . 3 + 4.  x + 4.  3 – x²

= x² + 3x + 4x + 12- x²

= (x² – x²)+ (3x + 4x) + 12

=  7x + 12

 

Example 6

Simplify  (y – 2)(2x + 5)

Solution

(y – 2)(2x + 5) = y(2x + 5)- 2(2x + 5)

= 2xy + 5y – 4x -10.