Author: S.LAL
Created: 20 Sep, 2010; Last Modified: 29 Apr, 2018
Algebra II - 05
Factorisation
Factorisation of an algebraic expression implies writing the expression as a product of simpler algebraic expressions, each of which is called a factor. Factorisation is also termed resolution into factors.
Principle Rules of Factorisation
Taking out the common factor in polynomials
Simple factor
When each of the terms in a polynomial contains a common factor, which is basically a monomial, it can be easily separated out. Such a common factor is called a simple factor. Simple factors are easily spotted by inspection.
Factorise (i)
3
a
2
−
6
a
b
(ii)
5
a
2
b
x
3
−
15
a
b
x
2
−
20
b
3
x
2
. [(Hall & Knight, Elementary Algebra, p. 125 Example 1, 2)]
(i) The terms of the expression have 3a as the common factor, which can be separated out. Thus,
3
a
2
−
6
a
b
=
3
a
(
a
−
2
b
)
(ii) The common factor is
5
b
x
2
. So,
5
a
2
b
x
3
−
15
a
b
x
2
−
20
b
3
x
2
=
5
b
x
2
(
a
2
x
−
3
a
−
4
b
2
)
Compound factor
Sometimes, the common factor itself is a polynomial, and can be identified and fished out only after rearranging the terms of the original polynomial into suitable groups. Such a common factor is called a compound factor.
Resolve into factors (i)
x
2
−
a
x
+
b
x
−
a
b
(ii)
6
x
2
−
9
a
x
+
4
b
x
−
6
a
b
. [(Hall & Knight, Elementary Algebra, p. 126 Example 1, 2)]
(i) Noticing that the first two terms contain a common factor x, and the last two terms a common factor b, we group the terms as follows and continue:
x
2
−
a
x
+
b
x
−
a
b
=
(
x
2
−
a
x
)
+
(
b
x
−
a
b
)
=
x
(
x
−
a
)
+
b
(
x
−
a
)
=
(
x
+
b
)
(
x
−
a
)
(ii) Using a similar thinking process, we have
6
x
2
−
9
a
x
+
4
b
x
−
6
a
b
=
(
6
x
2
−
9
a
x
)
+
(
4
b
x
−
6
a
b
)
=
3
x
(
2
x
−
3
a
)
+
2
b
(
2
x
−
3
a
)
=
(
3
x
+
2
b
)
(
2
x
−
3
a
)
Figuring out the compound factors in complex expressions becomes easier with practice.
Splitting the middle term of trinomials of degree 2
Factorisation of (x2+bx+c)
Let the factors of
x
2
+
b
x
+
c
be of the form
(
x
+
p
)
(
x
+
q
)
. Then,
x
2
+
b
x
+
c
=
(
x
+
p
)
(
x
+
q
)
=
x
2
+
p
x
+
q
x
+
p
q
=
x
2
+
(
p
+
q
)
x
+
p
q
Comparing the coefficients of like terms on both sides, we get
(
p
+
q
)
=
b
and
p
q
=
c
.
Thus, to factorise a trinomial of the form
x
2
+
b
x
+
c
,
- Find two number p and q such that
(
p
+
q
)
=
b
and
p
q
=
c
- The trinomial now stands factorised as
(
x
+
p
)
(
x
+
q
)
.
Resolve into factors (i)
x
2
+
11
x
+
24
(ii)
x
2
−
10
x
+
24
. [(Hall & Knight, Elementary Algebra, p. 127 Example 1, 2)]
(i) We need to find p and q such that
(
p
+
q
)
=
11
and
p
q
=
24
. Since the product is positive, as well as the sum, both the numbers must be positive. Clearly, p = 8 and q = 3 satisfy the conditions. Thus,
x
2
+
11
x
+
24
=
(
x
+
8
)
(
x
+
3
)
(ii) We need to find p and q such that
(
p
+
q
)
=
−
10
and
p
q
=
24
. This implies that both must be negative, since their sum is negative but the product is positive. Clearly, p = −6 and q = −4 satisfy the conditions. Thus,
x
2
−
10
x
+
24
=
(
x
−
4
)
(
x
−
6
)
Resolve into factors (i)
x
2
+
2
x
−
35
(ii)
x
2
−
3
x
−
54
. [(Hall & Knight, Elementary Algebra, p. 129 Example 1, 2)]
(i) For
(
p
+
q
)
=
2
and
p
q
=
−
54
to be true, the numbers must have opposite signs, and the greater must be positive for the sum to be positive. Clearly, p = 7 and q = −5 satisfy the conditions.
x
2
+
2
x
−
35
=
(
x
+
7
)
(
x
−
5
)
(ii) For
(
p
+
q
)
=
−
3
and
p
q
=
−
54
to be true, the numbers must have opposite signs, and the greater must be negative for the sum to be negative. p = 7 and q = −5 fit the bill.
x
2
−
3
x
−
54
=
(
x
−
9
)
(
x
+
6
)
Factorisation of (ax2+bx+c)
Let factors of
a
x
2
+
b
x
+
c
be of the form
(
r
x
+
s
)
(
t
x
+
u
)
. Then,
a
x
2
+
b
x
+
c
=
(
r
x
+
s
)
(
t
x
+
u
)
=
r
t
x
2
+
r
u
x
+
s
t
x
+
s
u
=
r
t
x
2
+
(
r
u
+
s
t
)
x
+
s
u
Comparing the coefficients of like terms on both sides, we get a = rt, b = ru + st and c = su.
If we let p = ru and q = st, then
pq = rtsu = ac and
(
p
+
q
)
=
r
u
+
s
t
=
b
Thus, we should find p and q such that
(
p
+
q
)
=
b
and pq = ac.
So, to factorise a trinomial of the form
a
x
2
+
b
x
+
c
,
- Find two number p and q such that
(
p
+
q
)
=
b
and pq = ac.
- Rewrite the expression
a
x
2
+
b
x
+
c
with the middle term split as:
(
a
x
2
+
p
x
+
q
x
+
c
)
or
(
a
x
2
+
p
x
+
q
x
+
p
q
a
)
- In this form, the compound factors can then be taken out as follows:
a
x
2
+
p
x
+
q
x
+
p
q
a
=
x
(
a
x
+
p
)
+
q
(
x
+
p
a
)
=
x
(
a
x
+
p
)
+
q
a
(
a
x
+
p
)
=
(
a
x
+
p
)
(
x
+
q
a
)
=
1
a
(
a
x
+
p
)
(
a
x
+
q
)
Resolve into factors (i)
6
x
2
+
7
x
−
3
(ii)
15
x
2
−
26
x
+
8
. [(Aggarwal & Aggarwal, Mathematics Class IX, p. 35 Ex.2 1, 2)]
(i) We need to find p and q such that
(
p
+
q
)
=
7
and pq = −18. Clearly, one of them is negative, while the positive number has a larger absolute value. p = 9 and q = −2 satisfy the conditions. Thus,
6
x
2
+
7
x
−
3
=
6
x
2
+
9
x
−
2
x
−
3
=
3
x
(
2
x
+
3
)
−
1
(
2
x
+
3
)
=
(
3
x
−
1
)
(
2
x
+
3
)
(ii) We need to find p and q such that
(
p
+
q
)
=
−
26
and pq = 120. Clearly, both of them are negative. p = −20 and q = −6 satisfy the conditions. Thus,
15
x
2
−
26
x
+
8
=
15
x
2
−
20
x
−
6
x
+
8
=
5
x
(
3
x
−
4
)
−
2
(
3
x
−
4
)
=
(
5
x
−
2
)
(
3
x
−
4
)
The difference of two squares
As we have seen in the monograph on Expansions for the simplification of
(
a
+
b
)
(
a
−
b
)
:
(
a
+
b
)
(
a
−
b
)
=
a
2
−
a
b
+
b
a
−
b
2
=
a
2
−
b
2
Thus,
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
.......(1)
In words, the difference of the squares of two numbers is equal to the product of the sum and the difference of the two numbers.
Resolve into factors (i)
25
x
2
−
16
y
2
(ii)
1
−
49
c
6
. [(Hall & Knight, Elementary Algebra, p. 133 Example 1, 2)]
| (i) |
25
x
2
−
16
y
2
=
(
5
x
)
2
−
(
4
y
)
2
=
(
5
x
+
4
y
)
(
5
x
−
4
y
)
|
| (ii) |
1
−
49
c
6
=
(
1
)
2
−
(
7
c
3
)
2
=
(
1
+
7
c
3
)
(
1
−
7
c
3
)
|
Resolve into factors (i)
x
2
+
x
y
−
2
y
−
4
(ii)
(
1
−
a
2
)
(
1
−
b
2
)
+
4
a
b
. [ICSE]
| (i) |
x
2
+
x
y
−
2
y
−
4
=
(
x
2
−
4
)
+
x
y
−
2
y
(Rearranging the terms)
=
(
x
+
2
)
(
x
−
2
)
+
y
(
x
−
2
)
=
(
x
+
y
+
2
)
(
x
−
2
)
|
| (ii) |
(
1
−
a
2
)
(
1
−
b
2
)
+
4
a
b
=
1
−
b
2
−
a
2
+
a
2
b
2
+
4
a
b
=
1
−
(
a
2
+
b
2
)
+
a
2
b
2
+
2
a
b
+
2
a
b
=
1
+
2
a
b
+
a
2
b
2
−
(
a
2
+
b
2
−
2
a
b
)
=
(
1
+
a
b
)
2
−
(
a
−
b
)
2
=
(
(
1
+
a
b
)
+
(
a
−
b
)
)
(
(
1
+
a
b
)
−
(
a
−
b
)
)
=
(
1
+
a
−
b
+
a
b
)
(
1
−
a
+
b
+
a
b
)
|
The sum and difference of two cubes
Dividing
a
3
+
b
3
by
(
a
+
b
)
gives the quotient
a
2
−
a
b
+
b
2
, while dividing
a
3
−
b
3
by
(
a
−
b
)
gives the quotient
a
2
+
a
b
+
b
2
. This leads us to the following identities:
a
3
+
b
3
=
(
a
+
b
)
(
a
2
−
a
b
+
b
2
)
.......(2)
a
3
−
b
3
=
(
a
−
b
)
(
a
2
+
a
b
+
b
2
)
.......(3)
Hence, any expression which can be written as a sum or a difference of two cubes can be resolved into factors as above.
Resolve
64
a
3
+
1
into factors. [(Hall & Knight, Elementary Algebra, p. 136 Example 2)]
64
a
3
+
1
=
(
4
a
)
3
+
(
1
)
3
=
(
4
a
+
1
)
(
(
4
a
)
2
−
4
a
⋅
1
+
(
1
)
2
)
=
(
4
a
+
1
)
(
16
a
2
−
4
a
+
1
)
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List of References
Aggarwal, RS & Aggarwal, V, Secondary School Mathematics for Class IX, Patna, India: Bharati Bhawan, 1999.
Hall, HS & Knight, SR, Elementary Algebra for Schools, metric edn, Agra, India: AK Publications, 1966.
Bibliography
Aggarwal, RS & Aggarwal, V, Secondary School Mathematics for Class IX, Patna, India: Bharati Bhawan, 1999.
Gupta, SD & Banerjee, A, ICSE Mathematics for Class 9, Patna, India: Bharati Bhawan, 2003.
Hall, HS & Knight, SR, Elementary Algebra for Schools, metric edn, Agra, India: AK Publications, 1966.
Schultze, A, Elements of Algebra, NY, USA: The Macmillan Company, 1910.
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