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840 = 2x2x2x3x5x7
= 23x3x5x7 Using exponent form 23 for 2 x 2 x 2
The number 840 is now expressed as the product of prime numbers.
References: Factor, Prime Factor, Product.
Factorials are used in combinations and permutations. The factorial of a counting number n is the product of the first n counting numbers. The symbol for factorial n is n1 By definition 0! = 1, and also 1! = 1.
Example. Find the value of factorial 5.
51 = 1x2x3x4x5. It is the product of the first five counting numbers
Factorial 5 is 120.
Factorials can be worked out using a scientific calculator.
References: Combinations, Permutations.
When we factorize an expression we write the expression as a product of its factors using sets of brackets. Factorizing is the reverse of expanding brackets, and expanding should be well understood before factorizing is studied. There are three kinds of factorizing explained under this entry:
♦ Common factors, type 1 and type 2
♦ Quadratic factors
♦ Difference of two squares
We will also look at combinations of these kinds of factorizing.
Common Factors. Type 1 The factorized answer will contain one set of brackets.
Example 1. Factorize 2x2 + 6xy. Solution.
2x2 + 6xy = 2 • x • x 2 • 3 • x • y = 2- x- xJr2-3-x-y = 2x(x + 3y)
Writing each term in the form of products.
2x is common to both terms
The common factor 2x is written at the front of the brackets, and what is left is written inside the brackets
Expanding the brackets can check the answer. Always make sure that the common factor is as large as possible. For example, using a common factor of only 2 or a common factor of only x is not satisfactory in this example.
Example 2. Factorize 2x4 — 8x2 + 2x.
2x4 — 8a:2 + 2x = 2 • x • x • x • x — 2-2-2-x-x + 2- x
= 2x(x - x- x— 2- 2- x + l) If the common factor
is all of one term, we must leave a 1 in the brackets
= 2x(x?> — 4x + 1)
Common Factors. Type 2 The common factor may be a term in brackets instead of a single term.
Example. Factorize 2(x + y) + y(x + y).
Solution. The principle for factorizing this expression is the same as for factorizing 2b + yb = b(2 + y), where b is the term (jt +y). Write
2(*+y) + y(jr+y) = (x + y)(2 + y) (x + y) is the common factor
In harder problems it may be necessary to group the terms into pairs and then apply common factors to each group. This method is sometimes called “factors by grouping.” For example,
2x + 2y + xy + y2 = 2(x+y) + y(*+y) Using common factors on each
pair of terms 2x + 2y andxy + yy
= (x + y)( 2 + y)
(x + y) is a common factor
Note. When factorizing four terms, as in the example above, it may be necessary to regroup them into different pairs at the start if there are no obvious common factors.
Quadratic Factors The rales for factorizing quadratic expressions are easily noticed when we expand pairs of brackets in the following examples. Look for the way the two numbers in the brackets are related to the numbers in the expansion. Write
(x + 3)(jc + 7) = jc2 + 10* + 21 3 + 7 = 10 and 3 x 7 = 21
(x — 4)(x + 2) = x2 — 2x — 8 —■4 + 2 = — 2 and —4 x 2 = —8
(x - 7)0 - 1) = x2 - 8x + 7 -7 + -1 =-8 and -7 x -1 =7
Having established a pattern in expanding brackets, we can reverse the process of expanding brackets and factorize the expansions.
Example 1. Factorize x2 + Ix + 6.
x2 + Ix + 6 = (x )(x ) There are two sets of brackets, each containing
x, and spaces are left in which to insert the two numbers
The two numbers that go into the brackets must multiply to make +6 and add to make +7. There is only one possible combination that meets both criteria, and that is +6 and +1:
x2 + 7x + 6 = (x + 6)(x + 1) Expanding the two sets of brackets can
check the answer
The factors may be written the other way around as (x + l)(x + 6).
Example 2. Factorize x2 — 5x — 14.
x2 — 5x — 14 = (x )(x )
The two numbers that go into the brackets must multiply to make —14 and add to make —5. One of the numbers must be positive and one negative to obtain a product that is negative. There is only one possible combination that meets both criteria, and that is —7 and +2:
x2-5x-14=(x- 1)(x + 2)
With practice the two numbers that go into the brackets can be quickly identified.
Difference of Two Squares The expressions we factorize are of the form a2 — b2, which factorizes into the two sets of brackets (a — b)(a + b). This can be expressed as a formula:
a2 — b2 = (a — b)(a + b)
The reason for the name “difference of two squares” is because difference means minus, and the two terms a2 and b2 are squared terms.
Example 1. Factorize 9 — x2.
9 — x2 = 32 — x2 In the formula a is 3 and b is x
= (3 — x)(3 + x) Substituting a = 3 and b = x into (a — b)(a + b)
Example 2. Factorize 16.x2 — 25.