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Chapter 12: Factorisation

    Introduction

    Factors of Algebraic Expressions

    Need of Factorisation

    Method of Common Factors

    Factorisation by grouping the terms

    Factorisation using Identities

    Factors of the form (x+a)(x+b) = x² + (a+b)x + ab

    Division of Algebraic Expressions

    Division of a Monomial by another Monomial

    Division of an Expression by a Monomial

    Division of Algebraic Expressions Continued (Polynomial ÷ Polynomial)

    Exercise 12.1

    Exercise 12.2

    Exercise 12.3

    Exercise 12.4

    What We Have Discussed?

    Easy Level Worksheet

    Moderate Level Worksheet

    Hard Level Worksheet

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Chapter 12: Factorisation > Factors of the form (x+a)(x+b) = x² + (a+b)x + ab

Factors of the form (x+a)(x+b) = x² + (a+b)x + ab

Upto now the expressions we have looked at consist of mostly perfect squares i.e. a+b2 or a−b2. However, this isn't possible all the time. Other times they might not fit into the a2−b2 form as well. For eg:

x2+5x+6

y2−7y+12

z2−4z−12 etc.

So, how can these expressions be factorized? One type that may be possible is the type of

x2+a+bx+ab

The identity is:

(x + a) (x + b) = x^2 + (a + b) x + ab

(x + a)(x + b) = x2 + (a+b)x + ab (IV)

Let's try to use this identity for factorization as we have done earlier.

Example 9: Factorise x2+5x+6

Upon comparing the above with the Identity (IV), we get:

ab = 6 and a + b = 5

From here, we can obtain a and b to give us the factorisation (x + a)(x + b).

Using Trial and Error Method:

Since ab = 6, then a and b must be factors of 6.

If we try a = 6, b = 1 we get:

a + b = 7 (not 5)

So, these values are not right.

If we try a = 2, b = 3:

a + b = 5, giving us the required value.

Thus,

The factorised form is

In general, for factorising an algebraic expression of the type x2+px+q:

we try to find two factors a and b of q (the constant term) such that

ab = q and a + b = p

Then, the expression becomes -

x2+a+bx+ab

= x2+ax+bx+ab

= x(x + a) + b(x + a)

= (x + a) (x + b)

which gives the required factors of a and b.

Example 10: Find the factors of y2−7y+12.

We note 12 = 3 × and 3 + 4 = .

Therefore, y2−7y+12 = y2 – 3y + + 12

= (y –3) – (y –3) =

Note, this time we did not compare the expression with that in Identity (IV) to identify a and b. After sufficient practice you may not need to compare the given expressions for their factorisation with the expressions in the identities; instead you can proceed directly as we did above.

Example 11: Obtain the factors of z2−4z−12.

Instructions

We have: ab = i.e. this means one of a and b is .
Further, a + b = , which means that the one with larger numerical value is .
Say, possible values are a = , b = 2, so that a + b = – 4 as required.
Hence, z2−4z−12 = z2 + 2z −12

= z + =

Example 12: Find the factors of 3m2+9m+6.

Instructions

We notice that is a common factor of all the terms.
Therefore, 3m2+9m+6 =
Now, m2 + 3m + 2 = m2 + m + + 2 (as 2 = 1 × )
= m(m + 1)+ 2( m + 1) =
Therefore, 3m2+9m+6 = 3(m + 1)(m + 2)