1. Introduction
Quantities are enclosed within brackets to indicate that they are to be treated as a single entity. If we wish to subtract, say, 3a − 2b from 4a − 5b then we do this as follows.
Example 1:
Solution:
| (4a − 5b) − (3a − 2b) | = | 4a − 5b − 3a− (−2b) |
| = | 4a − 5b − 3a + 2b | |
| = | 4a − 3a − 5b + 2b | |
| = | a − 3b |
Solution:
| (7x + 5y) − (2x − 3y) | = | 7x + 5y − 2x− (−3y) |
| = | 7x + 5y − 2x + 3y | |
| = | 7x − 2x + 5y + 3y | |
| = | 5x + 8y |
When there is more than one bracket it is usually necessary to begin with the inside bracket and work outwards:
Example 2: Simplify the following expressions by removing the brackets.
Solution: We have:
| 3a − c+(5a − 2b − [3a − c + 2b]) | = | 3a − c + (5a − 2b − 3a + c − 2b) |
| = | 3a − c + (2a − 4b + c) | |
| = | 3a − c + 2a − 4b + c | |
| = | 3a + 2a − 4b − c + c | |
| = | 5a − 4b |
Solution: Similarly, we have:
| −{3y − (2x − 3y) + (3x − 2y)} + 2x | = | −{3y − 2x + 3y + 3x − 2y} + 2x |
| = | −{3y + 3y − 2y + 3x − 2x} + 2x | |
| = | −{4y + x} + 2x | |
| = | −4y − x + 2x | |
| = | x − 4y |
Exercise 1: Remove the brackets from each of the following expressions and simplify as far as possible.
Solution:
| x − (y − z) + x + (y − z) − (z + x) | = | x − y + z + x + y − z − z − x |
| = | x + x − x − y + y + z − z − z | |
| = | x − z |
Solution:
| 2x − (5y + [3z − x]) − (5x − [y + z]) | = | 2x − (5y + 3z − x) − (5x − y − z) |
| = | 2x − 5y − 3z + x − 5x + y + z | |
| = | 2x + x − 5x − 5y + y − 3z + z | |
| = | −2x − 4y − 2z |
Solution:
| 3a + b + 7a − 2b | = | 3a + 7a + b − 2b |
|---|---|---|
| = | 3 + 7a − b | |
| = | 10a − b |
Solution:
| a − (b + [c − {a − b}])c | = | a − (b + [c − a + b]) |
| = | a − (b + c − a + b) | |
| = | a − (2b + c − a) | |
| = | a − 2b − c + a | |
| = | 2a − 2b − c |
Click on questions to reveal their solutions
2. Distributive Rule
A quantity outside a bracket multiplies each of the terms inside the bracket. This is known as the distributive rule.
Example 3:
(a)3(x − 2y) => 3x − 6y
(b)2x(x − 2y + z) => 2x2 − 4xy + 2xz
(c)7y − 4(2x − 3) => 7y − 8x + 12
This is a relatively simple rule but, as in all mathematical arguments, a great deal of care must be taken to apply it correctly.
Exercise 2: Remove the brackets and simplify the following expressions:
Solution: First note that (2x)2 = (2x) × (2x) = 4x2
| 5x − 7x2 − (2x)2 | = | 5x − 7x2 − 4x2 |
| = | 5x − 11x2 |
Solution:
| (3y)2 + x2 − (2y)2 | = | 9y2 + x2 − 4y2 |
| = | 9y2 − 4y2 + x2 | |
| = | 5y2 + x2 |
Solution:
| 3a + 2(a + 1) | = | 3a + 2a + 2 |
| = | 5a + 2 |
Solution:
| 5x − 2x(x − 1) | = | 5x − 2x2 + 2x |
| = | 7x − 2x2 |
Solution:
| 3xy − 2x(y − 2) | = | 3xy − 2xy + 4x |
| = | xy + 4x |
Solution:
| 3a(a − 4) − a (a − 2) | = | 3a2 − 12a − a2 + 2a |
| = | 3a2 − a2 + 2a − 12a | |
| = | 2a2 − 10a |
Click on questions to reveal their solution
In the case of two brackets being multiplied together, to simplify the expression first choose one bracket as a single entity and multiply this into the other bracket.
Example 4: For each of the following expressions, multiply out the brackets and simplify as far as possible.
Solution:
| (x + 5)(x + 2) | = | (x + 5) x + (x + 5) 2 |
| = | x (x + 5) + 2 (x + 5) | |
| = | x2 + 5x + 2x + 10 | |
| = | x2 + 7x + 10 |
Solution:
| (3x − 2)(2y + 3) | = | (3x − 2) 2y + (3x−2) 3 |
| = | 2y (3x − 2) + 3 (3x − 2) | |
| = | 6xy − 4y + 9x − 6 |
Try this short quiz:
Quiz 1: To which of the following does the expression (2x − 1)(x + 4) simplify?
Explanation:
| (2x − 1)(x + 4) | = | (2x − 1)x + (2x − 1)4 |
| = | (2x2 − x) + (8x − 4) | |
| = | 2x2 − x + 8x − 4 | |
| = | 2x2 + 7x − 4 |
3. FOIL
When it comes to expanding a bracketed expression like (a+c)(x+y) there is a simple way to remember all of the terms. This is the word FOIL, and stands for:
| Take products of the |
|---|
| First Outside Inside Last |
This is illustrated in the following:
| F | O | I | L | |||||
| (a+c)(x+y) | = | ax | + | ay | + | cx | + | cy |
These terms are the products of the pairs highlighted below:
| F | O | I | L |
| (a+c)(x+y), | (a+c)(x+y), | (a+c)(x+y), | (a+c)(x+y) |
There are two special cases of brackets that are worth remembering:
| (x+y)2, which is a complete square, and |
|---|
| (x+y)(x−y), which is a difference of two squares. |
These appear in the following exercises:
Exercise 3: Remove the brackets from each of the following expressions using FOIL.
Solution:
| (x + y)2 | = | (x + y)(x + y) |
| = | x2 + xy + yx + y2 using FOIL | |
| = | x2 + 2xy + y2 |
This is an IMPORTANT result and should be committed to memory. Here x is the first member of the the bracket and y is the second. The rule for the square of (x + y), i.e. (x + y)2 is:
| x2 | + | 2xy | + | y2 |
| (square the first) | + | (twice the product) | + | (square the last) |
Solution: Using FOIL again:
| (x + y)(x − y) | = | x2 + xy − xy + y2 |
| = | x2 − y2 |
The form of the solution is the reason for the name difference of two squares. This is another important result that is worth committing to memory.
Solution: Using FOIL:
| (x + 4)(x + 5) | = | x2 + 5x + 4x + 20 |
| = | x2 + 9x + 20 |
Solution: Using FOIL:
| (y + 1)(y + 3) | = | y2 + 3y + y + 3 |
| = | y2 + 4y + 3 |
Solution: Using FOIL:
| (2y + 1)(y − 3) | = | 2y2 − 6y + y − 3 |
| = | 2y2 − 5y − 3 |
Solution: This one is best done in parts. First we have:
(x − 3)2 = x2 − 6x + 9
and:
(x + 1)2 = x2 + 2x + 1
Thus:
| 2(x − 3)2 − 3(x + 1)2 | = | 2(x2 − 6x + 9) − 3(x2 + 2x + 1) |
| = | 2x2 − 12x + 18 − 3x2 − 6x − 3 | |
| = | 2x2 − 3x2 − 12x − 6x + 18 − 3 | |
| = | −x2 − 18x + 15 |
Click on questions to reveal their solutions
Quiz 2: To which of the following expressions does 9 − (x − 3)2 simplify?
Explanation: First note that (x − 3)2 = x2 − 6x + 9, so:
| 9 − (x − 3)2 | = | 9 − (x2 − 6x + 9) |
| = | 9 − x2 + 6x − 9 | |
| = | 9 − 9 − x2 + 6x | |
| = | −x2 + 6x | |
| = | 6x − x2 |
4. Quiz on Brackets
In each of the following, remove the brackets, simplify the expression, and choose the solution from the options given