If a number can be expressed as a product of two whole numbers, then the whole numbers are called factors of that number.
So, the factors of 6 are 1, 2, 3 and 6.
Example 1
Find all factors of 45.
Solution:
So, the factors of 45 are 1, 3, 5, 9, 15 and 45.
Common Factors
10 = 2 × 5 = 1 × 10
Thus, the factors of 10 are 1, 2, 5 and 10.
15 = 1 × 15 = 3 × 5
Thus, the factors of 15 are 1, 3, 5 and 15.
Clearly, 5 is a factor of both 10 and 15. It is said that 5 is a common factor of 10 and 15.
Example 2
Find a common factor of:
a. 6 and 8
b. 14 and 21
Solution:
Prime Numbers
If a number has only two different factors, 1 and itself, then the number is said to be a prime number.
For example, 31 = 1 × 31
31 is a prime number since it has only two different factors.
Note:
But 1 is not a prime number since it does not have two different factors.
Example 3
Express 150 as a product of prime numbers, i.e. find its prime factor decomposition.
Solution:
Note:
We try the prime numbers in order of their magnitude.
We observe that:
The prime factor decomposition of a number is unique.
This is called the Fundamental Theorem of Arithmetic. It provides us with a good reason for defining prime numbers so as to exclude 1. If 1 were a prime, then the prime factor decomposition would lose its uniqueness. This is because we could multiply by 1 as many times as we like in the decomposition.
Highest Common Factor
The highest common factor (HCF) of two numbers (or expressions) is the largest number (or expression) that is a factor of both.
Consider the highest common factor of 16 and 24.
The common factors are 2, 4 and 8. So, the highest common factor is 8.
Note:
The highest common factor is the product of the common prime factors.
In general:
To find the highest common factor of two (or more) numbers, make prime factors of the numbers and identify the common prime factors. Then the highest common factor is the product of the common prime factors.
Example 4
Find the highest common factor of 60 and 150.
Solution:
The prime factorisation of 60 is:
The prime factorisation of 150 is:
Note:
The highest common factor can also be obtained by a trial and error method.
For example, the highest common factor of 40 and 45 is 5 because 5 is the largest number which divides into both 40 and 45 exactly.
Likewise, the highest common factor of 27 and 36 is 9 because 9 is the largest number which divides into both 27 and 36 exactly.
Highest Common Factor of Algebraic Expressions
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The highest common factor (HCF) of algebraic expressions is obtained in the same way as that of numbers.
Example 5
Solution:
Factorisation using the Common Factor
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We know that:
a(b + c) = ab + ac
The reverse process, ab + ac = a(b + c), is called taking out the common factor.
Consider the factorisation of the expression 5x + 15.
Note that the common factor 5 has been taken out and placed in front of the brackets. The expression inside the brackets is obtained by dividing each term by 5.
In general:
To factorise an algebraic expression, take out the highest common factor and place it in front of the brackets. Then the expression inside the brackets is obtained by dividing each term by the highest common factor.
Example 6
Solution:
Note:
The process of taking out a common factor is of great importance in algebra. With practice you will be able to find the highest common factor (HCF) readily and hence factorise the given expression.
Example 7
Solution:
Factors by Grouping 'Two and Two'
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Now, consider the expression 7x + 14y + bx + 2by. Clearly, there is no factor common to every
term. However, it is clear that 7 is a common factor of the first two terms and b is a common factor
of the last two terms. So, the expression can be grouped into two pairs of two terms as shown.
This factorisation technique is called grouping 'Two and Two'; and it is used to factorise an
expression consisting of four terms.
Example 8
Solution:
Factorisation of a Difference of Two Squares
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Example 9
Solution:
Taking out a Common Factor
To factorise an expression, take out any common factor; and then use the formula:
Example 10
Solution:
A Calculation Short Cut
Some arithmetic expressions can be evaluated easily using the factorisation of the difference of two squares.
Example 11
Solution:
Factorisation of Quadratic Trinomials
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The Distributive Law is used in reverse to factorise a quadratic trinomial, as illustrated below.
Consider the expansion of (x + 2)(x + 3).
We notice that:
- 5, the coefficient of x, is the sum of 2 and 3.
- 6, the independent term, is the product of 2 and 3.
Note:
The product of two linear factors yields a quadratic trinomial; and the factors of a quadratic trinomial are linear factors.
Now consider the expansion of (x + a)(x + b).
Coefficient of x = a + b = Sum of a and b.
Independent term = ab = Product of a and b.
In general:
To factorise a quadratic trinomial, find two numbers whose sum is equal to the coefficient of x, and whose product is equal to the independent term.
Example 12
Solution:
Check:
Cross-Multiplication Method
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Place the linear factors one above the other as shown below.
Multiply the numbers along the arms of the cross, and then add the products. That is,
x×10 + x×1 = 10x+ x = 11x. This is not the middle term.
x×5 + x×2 = 5x+ 2x = 7x. This is the middle term.
Note:
- The solution is read across.
- 10 = 1 × 10 = 2 × 5 = –1 × –10 = –2 × –5
We did not try –1 × –10 and –2 × –5 because the middle term is positive.
- The answer can be checked using the Distributive Law. That is:
Example 13
Factorise the following:
Solution:
Multiply the numbers along the arms of the cross, and then add the products.
x×15 + x×1 = 15x+ x = 16x. This is not the middle term.
We reject this pair as the middle term is 8x. Now, try the next pair.
x×5 + x×3 = 5x+ 3x = 8x. We accept this pair as the middle term is 8x.
x× – 4 + x× –3 = – 4x – 3x = –7x. We accept this pair as the middle term is –7x.
Note:
We did not try 1 × 12, 2 × 6 and 3 × 4 because the middle term is negative. Also, we did not try
–2 × –6 and –1 × –12 because we have already obtained the middle term by using –3 × – 4.
x × 7 + x × –1 = 7x – x = 6x. We accept this pair as the middle term is 6x.
Note:
We did not try 1 × –7 because we have already obtained the middle term by using –1 × 7.
Further Quadratic Trinomials
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7x×2 + x×1 = 14x+ x = 15x. We reject this pair as the middle term is 9x.
Note:
The solution is read across.
Use of a Common Factor
If a quadratic trinomial has a common factor, take it out and place it in front of the brackets. Then use the cross-multiplication method to factorise the quadratic trinomial.
Example 14
Solution:
Check:
Recall that:
Example 15
Solution:
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