## Factorisation And Division Of Algebraic Expressions

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Expressing a mathematical expression as a product of its factors is called Factorisation.

A number can be expressed as product of its factors. Similarly, an algebraic expression can be expressed as a product of its factors. These factors can be numbers, algebraic variables, algebraic terms or algebraic expressions.

A number written as a product of prime factors is said to be in the prime factor form.

When a number is expressed as a product of factors, 1 is not mentioned as a factor, unless it is specially required.

56 = 2×2×2×7

72 = 2×2×2×3×3

### Irreducible form

An algebraic term is a product of its factors. When a term is expressed as a product of  its factors which can not be further reduced into factors then the term is said to be in irreducible form.

3xy = 3⋅x⋅y

4x2= 2⋅2⋅x⋅x⋅y

21xy2z = 3⋅7⋅x⋅y⋅y⋅z

## Methods of Factorisation

### Method of common factors

➢ Write each term as a product of irreducible factors

➢ Take the common factor out of parenthesis as observed in distributive property

ab +ac = a(b+c)

➢ Combine the terms to remove their irreducible form.

➢ Factor form of an expression has only one term.

12+36x
= 2⋅2⋅3 + 2⋅2⋅3⋅3⋅x
= 2⋅2⋅3(1+3⋅x)
= 12(1+3x)  (Required factor form)

### Factorisation by regrouping terms

➢ Rearrange the terms in expression to form groups with common factors.

➢ Take out common factors in the groups which will reduce the number of terms in the expression.

➢ Then a common factors can be taken out to lead the factorization further.

➢ Regrouping is possible in more than one way.

x2 + x+ 8x + 8
x2 + 8xx+ 8y
= x(x+8) + y(x+8)
= (x+8)(x+y)

15 xy – 6x + 5y – 2
= 15 xy + 5– 6– 2
= 5y(3x+1) – 2(3x+1)
= (3x+1)(5y– 2)

### Factorisation using Identities

(a + b)2 = a2 + 2ab + b2

(a − b)2 = a2 − 2ab + b2

(a + b)(a − b) = a2 − b2

(x + a)(x + b) = x2 + (a + b)x + ab

➢ In all these standard identities, the expression on LHS are in factor form.

➢ If we can change the expression to be factorised in a form that fits the RHS of one of the identities, then the expression corresponding to the LHS of the identity can be directly applied to get the desired factorisation.
a2 + 8a + 16
=a2 + 2⋅a⋅4 + 42
= (a+4)2

p2 – 10p + 25
p– 2⋅p⋅5 + 52
= (− 5)2

49x2 – 36
= (7x)2 – 62
= (7x+6)(7x–6)

p2 + 6p + 8
=p2 + 2p + 4p + 2⋅4
p2 + (2 + 4)p + 2⋅4
= (+ 2)(+ 4)

In general, for factorising an algebraic expression of the type
x2 + p+ q, we find two factors a and b of q (i.e., the constant term) such that,
ab = q and, a + b = p

Then, the expression becomes
x+ (a + b) x + ab
x2 + ax + bx + ab
= x(+ a) + b (x + a)
= (x + a) (x + b) which are the required factors.

## Division of Algebraic Expressions

➢ Factorize the expressions in the numerator and the denominator.

➢ Cancel the factors common to both the numerator and the denominator.

### Division of a monomial by another monomial

24xy2z3 ÷ 6yz2
= 24xy2z3 / 6yz2
= 2⋅2⋅23x⋅y⋅y⋅z⋅z⋅z23y⋅z⋅z
= 2⋅2⋅x⋅y⋅z
= 4xyz

### Division of a polynomial by a monomial

(3a8 − 4a+ 5a4) ÷ a4
= (3a8 − 4a6 +5a4)/ a4
a4(3a4 − 4a2 +5)/ a4
= 3a2 − 4a+ 5

### Division of a polynomial by another polynomial

(m2 − 14m − 32) ÷ (+ 2)
= (m2 − 14m −32)/(m + 2)
= (m+ 2m – 16m  − 2⋅16)/(m + 2)
m(+ 2)− 16(m + 2)/(m + 2)
(+ 2)( − 16)/(+ 2)
= − 16