Factorising

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Expanding Brackets
Removing the brackets is known as distributive law. To remove the brackets, multiply each term outside the bracket.

Example:
\(3(x + 2)\)                               Multiply each term inside the bracket by 3.
\(3 (x) + 3 (2) =  3x + 6 \)

Example 2: \(2x(x + 2)\)
\( 2x (x) + 2x (2) = 2x^2 + 4x \)

Pairs of Bracket
Removing pairs of brackets using FOIL method.

FOIL Method
First – Multiply the first term in the first bracket with the first term in the second bracket.
Outside – Multiply the first term in the first bracket with the second term in the second bracket.
Inside – Multiply the second term in the first bracket with first term in the second bracket.
Last – Multiply the second term in the first bracket with the second term in the second bracket.
Example: (x + 2) (x + 3)

FOIL method will be:
\( x(x) + x(3) +2(x) + 2 (3) \)
\( x^2 + 5x + 6 \)

Example 2: \( (x + 5) (x – 2) \)
\( x (x)+ x (-2)  + 5 (x) – 2 (5) \)
\( x^2 + 3x – 10 \)

Factorising
Factorising is simplifying a quadratic expression. It is the reverse of expanding the bracket. To factorise, look for two numbers that have the sum of the second term and product of the third term.

Example: \(x^2 + 9x + 12 \)
Look for two numbers whose sum is 9 and if multiplied the answer is 12.
The factor pair of 12 are the following:
a. 12 and 1
b. 4 and 3
c. 6 and 2
In the pair of a factor, the product of 12 and 1, and 4 and 3 is 12 but the sum is not 9. The pair of the factor that suits the condition is 6 and 2. Factor of  \( x^2 + 9x + 12 \) is \( (x + 6) and (x + 2)\) .
\( x^2 + 9x + 12 =  (x + 6) (x + 2) \)

Example 2: \( x^2 + x – 30 \)
To factor the expression, think of two numbers that the product is -30 and added up is 1. Since the product is a negative, the two numbers must have a positive and a negative number.
These are all the factor of -30:
a. 15 and -5
b. -15 and 5
c. 10 and -3
d. -10 and 3
e. 6 and -5
f. -6 and 5

Only 6 and -5 have the sum of 1.

\( x^2 + x – 30 = (x + 6) (x – 5) \)

Factorising the Difference of Two Squares

Example:  \( x^2 – 36 \)
Notice that the only term is \( x^2 \) and a number. Some quadratics do not have the second term. Therefore we will look for a factor whose sum is 0.
The pair of factor of 36.
a. 36 and – 1
b. -36 and 1
c. 18 and -2
d. -18 and 2
e. 12 and  -3
f. -12 and 3
g. 9 and -4
h. -9 and 4
i. 6 and -6

\( x^2 – 36 \) can be factor by 6 and -6. These examples of factoring are called the sum and difference of square or difference of two squares.

However, not all without a second term can be factorised.

Example 2:  \(x^2 – 30 \)
Here are all the factor of -30.
a. 15 and -5
b. -15 and 5
c. 10 and -3
d. -10 and 3
e. 6 and -5
f. -6 and 5

From the pair of factor, it does not have the sum equal to zero. So, \( x^2 – 30 \) does not have a pair of factor. Therefore, it cannot be factorized.