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Math Lesson 6.4.1 - What does Factorising Mean?

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Welcome to our Math lesson on What does Factorising Mean?, this is the first lesson of our suite of math lessons covering the topic of Factorising, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

What does Factorising Mean?

Let's begin with a numerical example. For simplicity, let's take a simple sum

15 + 6 = 21

The two addends 15 and 6 have the number 3 as a common factor. Hence, we can write 15 as 3 · 5 and 6 as 3 · 2 to obtain

3 · 5 + 3 · 2 = 21

If we pull out the common factor from each number and write it only once before the rest of expression that is written in brackets, this yields

3 · (5 + 2) = 21

Obviously, the result does not change, as now we have 3 · 7 = 21. It is still the same (correct) result. We simply converted an addition into multiplication by highlighting the common factor of addends. This method is known as factorising.

The concept of factorising can be also be extended to algebraic expressions. We just need to identify the common factor(s) and then, isolate it/them from the rest of expression.

By definition, factorising is the inverse of expanding, consisting in the use of common factors to write an expression as a product of factors.

In other words, factorising means finding what to multiply together to get a new expression written in terms of these factors. This action is like "splitting" an expression into a multiplication of simpler expressions.

The simplest factorisation scheme is the one that involves three quantities, where one of them is a factor of the other two. In symbols, we write it as

ab ± ac = a ∙ (b ± c)

We can think of the term "factorisation" as "getting aside" or "isolating" the common part of an expression. The common term is factorised while the rest of each term of the expression is written in brackets multiplied by the factorised part. The signs of terms in brackets do not change unless the common factor is positive or of unknown sign; otherwise (if the common factor is negative) all terms in brackets change sign.

Example 1

Factorise the following algebraic expressions.

  1. 3x + 21x2 - 11x3
  2. 24x3 y2 - 16x2 y3

Solution 1

  1. There is only one 'x' as a common factor of the three terms in this expression. Hence, we factorise it to obtain
    3x + 21x2 - 11x3
    x ∙ 3 + x ∙ 21x - x ∙ 11x2
    x ∙ (3 + 21x + 11x2)
  2. This time, there are more common factors than in the first example. In numbers (coefficients), we have 8 a as common factor, as all numbers are multiples of 8. Moreover, there is an x2 and an y2 as common factors in variables. Therefore, we factorise all of them to obtain
    24x3 y2 - 16x2 y3
    = 8x2 y2 ∙ 3x - 8x2 y2 ∙ 2y
    = 8x2 y2 ∙ (3x - 2y)

A good indicator which shows that you have factorised an expression correctly is when you don't have any more operations to do in the expression that remains in brackets after the common factor is factorised. Thus, for both expressions in brackets obtained in the previous example there are no more operations to do in the part contained in brackets.

You may want to factorise a negative term, as said earlier. Let's illustrate what happens with the expression in this case with the expression (b) above. Thus, if we choose to factorise the term -8x2y2 instead of 8x2y2, we would obtain

24x3 y2 - 16x2 y3
= (-8x2 y2) ∙ (-3x) + (-8x2 y2) ∙ (-2y)
= -8x2 y2 ∙ (-3x + 2y)
= -8x2 y2 (2y-3x)

Obviously, the two factorised expressions 8x2 y2 ∙ (3x - 2y) and -8x2 y2 (2y - 3x) are equivalent.

We can look at the eight special identities explained in the previous tutorial in reverse to obtain some useful factorisations.

  1. Since
    (a + b)2 = a2 + 2ab + b2
  2. then
    a2 + 2ab + b2 = (a + b)2 = (a + b)(a + b)
  3. Since
    (a - b)2 = a2 - 2ab + b2
  4. then
    a2 - 2ab + b2 = (a - b)2 = (a - b)(a - b)
  5. Since
    (a - b)(a + b) = a2 - b2
  6. then
    a2 - b2 = (a - b)(a + b)
  7. Since
    (a + b)3 = a3 + 3a2 b + 3ab2 + b3
  8. Then
    a3 + 3a2 b + 3ab2 + b3 = (a + b)3
  9. Since
    (a - b)3 = a3 - 3a2 b + 3ab2 - b3
  10. then
    a3 - 3a2 b + 3ab2 - b3 = (a - b)3
  11. Since
    (a - b)(a2 + ab + b2 ) = a3 - b3
  12. then
    a3 - b3 = (a - b)(a2 + ab + b2 )
  13. Since
    (a + b)(a2 - ab + b2 ) = a3 + b3
  14. then
    a3 + b3 = (a + b)(a2 - ab + b2 )
  15. Since
    (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
  16. then
    a2 + b2 + c2 + 2ab + 2ac + 2bc = (a + b + c)2

Example 2

Factorise the following algebraic expressions.

  1. 4x3 - 12x2 + 9x
  2. x3 - x - 6

Solution 2

  1. First, it is easy to identify an 'x' as a common factor. Thus, we write
    4x3 - 12x2 + 9x
    = x ∙ (4x2 - 12x + 9)
    Now, if you look at the expression in brackets carefully, you will identify the expanded form of the special algebraic identity (2x - 3)2 because
    (2x - 3)2 = (2x)2 - 2 ∙ 2x ∙ 3 + 32
    = 4x2 - 12x + 9
    Hence, we can write the factorised form of the original expression as
    4x3 - 12x2 + 9x = x ∙ (2x - 3)2
  2. We can write this expression as
    x3 - x - 6 = x3 - 8 - x + 2
    because the first two terms represent the expanded form of the algebraic identity
    (x - 2)(x2 + 4x + 4) = x3 - 23 = x3 - 8
    On the other hand, we can write in brackets the last two terms as well, by leaving the 'minus' sign out of brackets. In this way the expression in brackets changes sign to become
    x3 - x - 6 = x3 - 8 - x + 2
    = (x - 2)(x2 + 4x + 4) - (x - 2)
    Now, we can see that there is a new factorization available; we can factorize (x - 2) as it is a common factor of the big expression. Thus since (x - 2) = 1 · (x - 2), we obtain
    (x - 2)(x2 + 4x + 4) - (x - 2)
    = (x - 2) ∙ [(x2 + 4x + 4) - 1]
    = (x - 2) ∙ (x2 + 4x + 4 - 1)
    = (x - 2) ∙ (x2 + 4x + 3)

More Factorising Lessons and Learning Resources

Expressions Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
6.4Factorising
Lesson IDMath Lesson TitleLessonVideo
Lesson
6.4.1What does Factorising Mean?
6.4.2The Meaning of Quadratics
6.4.3Factorising Quadratics of the Form x2 + bx + c (a = 1)
6.4.4Factorising Quadratics of the Form ax2 + bx + c (a ≠ 1)

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