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Factorisation

    What is Factorisation?

    Division of Algebraic Expressions Continued ( Polynomial ÷ Polynomial)

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8th class > Factorisation > What is Factorisation?

What is Factorisation?

We know that to factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.

Expressions like 3xy, 5x2y, 2x(y + 2), 5(y + 1)(x + 2) are already in factor form. They can't be simplified any further. We can also deduce their factors by simply looking at the expression at hand.

On the other hand, consider the expressions- 2x + 4, 3x + 3y, x2+5x, x2+5x+6. Here,it becomes less obvious what the factors for these expressions are. In order to solve this, there is a need for systematic methods to factorise expressions, i.e. to find their factors. Let's look at them.

Method of common factors

  • Factorise 2x + 4 (Write each term as a product of irreducible factors)

    We can write 2x =

    and 4 = x

Therefore we get,

2x + 4 = (2 x) + (2×2)

We notice that the factor 2 is common to both the terms. Using the distributive law:

2(x + 2) = (2×x) + (2×2) (for the above case)

Therefore, we can write:

2x + 4 = 2×(x + 2) = 2(x + 2)

Thus, the expression 2x + 4 is equal to 2 (x + 2). In this form we can easily see the factors for the expression: 2 and (x + 2). These factors as we know, are .

  • Factorise 5xy + 10x and find the common factor between the two terms

5xy+10x

  • The common factor :
  • Factorised form =
  • Taking out the common factor
  • Which gives us the above answer

Therefore, 5xy + 10x = 5 x (y + 2) which is the factorised form

Let's Solve

  • Factorise the following expressions:

(A) 12a2b+15ab2

12a2b+15ab2

  • The common factor(s) product :
  • Thus, factorised form =
  • Taking out the common factor
  • Which gives us the above answer

(B) 10x2−18x3+14x4

10x2−18x3+14x4

  • The common factor(s) product :
  • Thus, factorised form =
  • Taking out the common factor
  • Which gives us the above answer

(C) 14pq+35pqr

14pq+35pqr

  • The common factor(s) product :
  • Thus, factorised form =
  • Taking out the common factor
  • Which gives us the above answer

(D) 22y−33z Factorised Form

22y−33z

  • The common factor(s) product :
  • Thus, factorised form =
  • Taking out the common factor
  • Which gives us the above answer

Factorisation by regrouping terms

Take the expression:

2xy + 2y + 3x + 3 as an example

What do we observe? Is there something different about this expression from the ones that we have considered upto this point?

We observe that the terms all together don't share a common factor. The first two terms have the common factors of 2 and y while the last two terms have a common factor of 3. So, how to solve this expression now?

Let's break the expression into smaller pieces and let's just consider the terms '2xy' and '2y'

Writing the term (2xy + 2y) in the factor form, we get:

2xy + 2y =

Similarly we can take the terms (3x + 3) which will give us:

3x + 3 =

Hence combining the results we can write:

2xy + 2y + 3x + 3 = (Take out the common factors and put remaining in brackets)

We notice that the term (x + 1) is common to both the terms on the right hand side of the equation. Upon bringing out the common term we will get:

2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1)

2xy + 2y + 3x + 3 =

The expression 2xy + 2y + 3x + 3 is now in the form of the product of its factors with the factors being (x + 1) and (2y + 3). Note that these factors are also irreducible.

What is regrouping?

As we just saw in the above expression:

2xy + 3 + 2y + 3x

it is not immediately evident what the possible factors may be. In order to further simplify this process we will rearrange the expression as

2xy + 2y + 3x + 3

thus, allowing us to form the groups (2xy + 2y) and (3x + 3) leading to factorisation. This is known as regrouping.

Regrouping may be possible in more than one ways. Suppose, we regroup and rewrite the expression as:

2xy + 3x + 2y + 3

This will also lead to us getting the factors:

2xy + 3x + 2y + 3 = x × ( ) + 1 × ( )

This gives us

=

Notice that the factors are the same (as they should be) even though they appear in different order.

Let's Solve

  1. Find the common factors of the given terms:
  1. Factorise the following expressions:

(a) 20l2m+30alm

20l2m+30alm

  • Take out the common factor(s) product :
  • Thus, factorised form =
  • Taking out the common factor out of the bracket
  • Which gives us the above answer

(b) −16z+20z3

−16z+20z3

  • Take out the common factor(s) product :
  • Thus, factorised form =
  • Taking out the common factor out of the bracket
  • Which gives us the above answer

(c) ax2y+bxy2+cxyz

ax2y+bxy2+cxyz

  • Take out the common factor(s) product :
  • Thus, factorised form =
  • Taking out the common factor out of the bracket
  • Which gives us the above answer

(d) x2yz+xy2z+xyz2

x2yz+xy2z+xyz2

  • Take out the common factor(s) product :
  • Thus, factorised form =
  • Taking out the common factor out of the bracket
  • Which gives us the above answer

(e) −4a2+4ab−4ca

−4a2+4ab−4ca

  • Take out the common factor(s) product :
  • Thus, factorised form =
  • Taking out the common factor out of the bracket
  • Which gives us the above answer

(f) 15pq+15+9q+25p

15pq+15+9q+25p

  • We can see that the factorization needs re-grouping.
  • Regrouping the expression
  • Take out the common factors from the corresponding terms
  • Thus, factorised form =
  • Which gives us the above factorised answer

(g) 15xy−6x+5y−2

15xy−6x+5y−2

  • We can see that the factorization needs re-grouping.
  • Regrouping the expression
  • Take out the common factors from the corresponding terms
  • Thus, factorised form =
  • Which gives us the above factorised answer

(h) z−7+7xy−xyz

z−7+7xy−xyz

  • We can see that the factorization needs re-grouping.
  • Regrouping the expression
  • Take out the common factors from the corresponding terms
  • Thus, factorised form =
  • Which gives us the above factorised answer

(i) ax+bx−ay−by

ax+bx−ay−by

  • The factorization doesn't need re-grouping.
  • Take out the common factors from the corresponding terms
  • Thus, factorised form =
  • Which gives us the above factorised answer

Factorisation using identities

When dealing with expressions, you might have observed that the form of the expressions we encounter are often repetitive. There are a good number of expressions which have a standard form of factorisation which have been proven over time. These identities include the following:

(a + b)2 = a2 + b2 + 2ab (I)

(a - b)2 = a2 + b2 - 2ab (II)

(a + b)(a – b) = a2 - b2 (III)

Therefore, if the expression at hand, has a form that fits the RHS of any one of these identities, then the expression corresponding to the LHS of the identity is the desired factorisation.

Let's Solve

  • Factorise the following using identities:

(a) x2+8x+16

x2+8x+16

  • The above is in the form of the identity: where a and b are two integers.
  • We see that a = x while b =
  • Thus, the factorised form =
  • Which gives us the above factorised answer

(b) 4y2−12y+9

4y2−12y+9

  • The above is in the form of the identity: where a and b are two integers.
  • We see that a = while b = 3
  • Thus, the factorised form =
  • Which gives us the above factorised answer

(c) 49p2−36

49p2−36

  • The above is in the form of the identity: where a and b are two integers.
  • We see that a = while b = 6
  • Thus, the factorised form =
  • Which gives us the above factorised answer

(d) a2−2ab+b2−c2

a2−2ab+b2−c2

  • In the above expression: the first three terms are in the form of the identity: where a and b are two integers.
  • Factorising the first three terms, we get:
  • The resulting expression is in the form: where a and b are random integers (not the ones mentioned in the current problem).
  • Thus, the resulting factorised form becomes
  • Which gives us the above factorised answer

(e) m4−256

m4−256

  • Notice that: m4=m22 and 256 = squared.
  • Thus, the above expression is in the form of the identity: where a and b are two integers.
  • Doing this, we get:
  • We see that the expression m2−16 can be further factorised using the same identity.
  • Thus, the resulting factorised form becomes
  • Which gives us the above factorised answer

Factors of the form (x + a) (x + b)

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.

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.

Let's Solve

  • Factorise the following expressions using the identities:

(1) 49y2+84yz+36z2

49y2+84yz+36z2

  • The above identity is in the form of the identity: where a and b are two integers.
  • Thus, the resulting factorised form becomes
  • So, we get the following
  • Which gives us the above factorised answer

(2) 121b2−88bc+16c2

121b2−88bc+16c2

  • The above identity is in the form of the identity: where a and b are two integers.
  • Thus, the resulting factorised form becomes
  • So, we get the following
  • Which gives us the above factorised answer

(3) l+m2−4lm

l+m2−4lm

  • Expand the first term using the identity
  • Upon simplifying, the resulting factorised form becomes
  • So, we get the following
  • Which gives us the above factorised answer

(4) a4+2a2b2+b4

a4+2a2b2+b4

  • Upon simplifying using the identity, the resulting factorised form becomes
  • So, we get the following
  • Which gives us the above factorised answer

(5) 63a2−112b2

63a2−112b2

  • Take out the common factor of which gives us the form: a2−b2
  • Upon simplifying using the identity, the resulting factorised form becomes
  • So, we get the following
  • Which gives us the above factorised answer

(6) 16x5−144x3

16x5−144x3

  • Take out the common factor of which gives us the form: a2−b2
  • Upon simplifying using the identity, the resulting factorised form becomes
  • So, we get the following
  • Which gives us the above factorised answer

(7) l+m2−l−m2

l+m2−l−m2

  • Upon expanding and simplifying using the identity, the resulting factorised form becomes
  • So, we get the following
  • Which gives us the above factorised answer

(8) x2−2xy+y2−z2

x2−2xy+y2−z2

  • Upon simplifying using the identities, the resulting factorised form becomes
  • So, we get the following
  • Which gives us the above factorised answer

(9) 25a2−4b2+28bc−49c2

25a2−4b2+28bc−49c2

  • We observe that the three terms are together form an identity.
  • Upon simplifying using the identities, the resulting factorised form becomes
  • So, we get the following
  • Further on
  • Which gives us the above factorised answer

(10) 7p2+21q2

7p2+21q2

  • Upon simplifying using the identities, the resulting factorised form becomes
  • So, we get the following as the factorised answer, upon taking the common factor.

(11) lm+l+m+1

lm+l+m+1

  • Upon simplifying using the identities, the resulting factorised form becomes
  • Further on
  • So, we get the following as the factorised answer

(12) yy+z+9y+z

yy+z+9y+z

  • Upon simplifying using the identities, the resulting factorised form becomes
  • So, we get the following as the factorised answer

(13) 5y2−20y−8z+2yz

5y2−20y−8z+2yz

  • Upon simplifying using the identities, the resulting factorised form becomes
  • Upon simplifying
  • We get the factorised answer

(14) 10ab+4a+5b+2

10ab+4a+5b+2

  • Upon simplifying using the identities, the resulting factorised form becomes
  • Upon simplifying
  • We get the factorised answer

(15) x4−y+z4

x4−y+z4

  • Upon simplifying using the identities, the resulting factorised form becomes
  • Upon simplifying
  • Further expanding using identities
  • More over we get
  • We get the factorised answer

(16) x4−x−z4

x4−x−z4

  • Upon simplifying using the identities, the resulting factorised form becomes
  • Upon simplifying
  • Further expanding using identities
  • More over we get
  • We get the factorised answer

(17) a4−2a2b2+b4

a4−2a2b2+b4

  • Upon simplifying using the identities, the resulting factorised form becomes
  • Upon simplifying
  • Further expanding using identities
  • More over we get
  • We get the factorised answer