 # Factorization of Algebraic Expressions

We know that the product of 5x² and 2x-3y = 5x²(2x-3y) = 10x³-15x²y. we say that 5x² and 2x-3y are factors of 10x³-15x²y. We write it as 10x³-15x²y = 5x²(2x-3y).

Similarly the product of 3x+7 and 3x-7 = (3x+7)(3x-7) = 9x²-49; we say that 3x+7 and 3x-7 are factors of 9x²-49. we write it as 9x²-49 = (3x+7)(3x-7). Thus, when an algebraic expression can be written as the product of two or more expressions, then each of these expressions is called a factor of the given expression. Factorization of algebraic means to obtain two or more expressions whose product is the given expression.

The process of finding two or more expressions whose product is the given expression is called the factorization of algebraic expressions. Thus, the factorisation of algebraic expressions is the reverse process of multiplication.

Here are a few examples for a better understanding:

Product                                                                              Factors

i) 7xy (5xy-3) = 35x²y²-21xy                                            35x²y²-21xy = 7xy(5xy-3)

ii) 16a²-25b² = (4a+5b)(4a-5b)                                        (4a+5b)(4a-5b) = 16a²-25b²

iii) (p+3)(p-7) = p²-4p-21                                                    P²-4p-21 = (p+3)(p-7)

iv) (2x+3)(3x-5) = 6x²-x-15                               6x²-x-15 = (2x+3)(3x-5)

Methods for Factorisation of Algebraic Expressions

Factorization using identities can be solved using three methods that can be used for the factorization of algebraic expressions, they are:

1. Taking out common factors

2. Grouping

3. Difference of two squares

Before taking up factorization, there is one thing that needs to be clear and that is THE CONCEPT OF H.C.F. Yes! So, what is H.C.F?

H.C.F. of Two or more Polynomials(with integral coefficients) is the Largest Common Factor of the Given Polynomials.

 H.C.F of two or more monomials = (H.C.F. of their numerical coefficients)x(H.C.F. of their literal coefficients) H.C.F. of literal coefficients = product of each common literal raised to the lowest power

For Example,

H.C.F. of 6x²y² and 8xy³

H.C.F. of numerical coefficients = H.C.F. of 6 and 8 = 2.

H.C.F. of literal coefficients = H.C.F. of x²y² and xy³= product of each common literal raised to the lowest power =xy²

Therefore, H.C.F. of 6x²y² and 8xy³ = 2 x xy² = 2xy²

Factorization of Algebraic Expressions By Taking out Common Factors

In case the different terms/expressions of the given polynomial have common factors, then the given polynomial can be factorized by the following procedure:

1. Find the H.C.F. of all the terms/expressions of the given polynomial

2. Then divide each term/expression of the given polynomial by H.C.F. The quotient will be enclosed within the brackets and the common factor will be kept outside the bracket.

Here are a few examples;

Example 1) Factorise the following polynomials:

i) 24x³-32x²

ii) 15ab²-21a²b

Solutions) factorizing algebraic expressions of the following:

i)  H.C.F. of 24x³ and 32x² is 8x²

24x³-32x² = 8x²(3x-4)

ii)  H.C.F. of 15ab² and 21a²b is 3ab

15ab²-21a²b = 3ab(5b-7a)

Example 2) Factorise the following:

i) 3x(y+2z)+5a(y+2z)

ii) 10(p-2q)³+6(p-2q)²-20(p-2q)

Solutions) factorising algebraic expressions of the following:

i)  H.C.F. of the expressions

3x(y+2z) and 5a(y+2z) is y+2z

3x(y+2z)+ 5a(y+2z) = (y+2z)(3x+5a)

ii) H.C.F. of the expressions 10(p-2q)³,6(p-2q)²,and 20(p-2q) is 2(p-2q)

10(p-2q)³+6(p-2q)²- 20(p-2q) = 2(p-2q)[5(p-2q)²+3(p-2q)-10]

Factorization of Algebraic Expressions By Grouping of Terms

When the grouping of terms or factorization by regrouping terms of the given polynomial gives rise to a common factor, then the given polynomial can be factorized by the factorization using common factors if followed the following procedure:

1. Arrange the terms of the given polynomial in groups so that each group has a common factor.

2. Factorize each group.

3. Pick out the factor which is common to each group.

Note: factorisation of algebraic expressions by grouping is possible only if the given polynomial contains an even number of terms.

Here are few examples:

Example 1)  Factorise the following polynomials:

i) ax-ay+bx-by

ii) 4x²-10xy-6xy+15yz

Solutions) factorising algebraic expressions of the following:

i)ax-ay+bx-by = (ax-ay)+(bx-by)

= a(x-y)+b(x-y)

= (x-y)(a+b)

ii) 4x²-10xy-6xy+15yz = (4x²-10xy)-(6xy+15yz)

= 2x(2x-5y)-3z(2x-5y)

= (2x-5y)(2x-3z)

Example 2) Factorise the following expression:

i) xy-pq+qy-px

ii) a²+bc+ab+ca

Solutions)

i) Since xy and pq have nothing in common, we do not group the terms in pairs in the order in which the given expression is written. Here we interchange -pq and -px

Therefore, xy-pq+qy-px = (xy-px)+(qy-pq)

= x(y-p)+q(y-p)

= (y-p)(x+q)

ii) a²+bc+ab+ca = a²+ab+bc+ca

= a(a+b)+c(b+a)

= a(a+b)+c(a+b)

= (a+b)(a+c)

Difference of Two Squares

When the given polynomial is expressible as the difference of two squares, then it can be factorized by using the formula: a²-b² = (a+b)(a-b)

Example 1) Factorise 25a²-64b²

Solution 1) 25a²-64b² = (5a)²-(8b)² = (5a+8b)(5a-8b)