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# Factorisation Using Identities

Can an equation have factors? We can certainly multiply an equation with a number or one equation by another equation. So it must have factors. How is this Factoring done? What are the important points that need to be understood? Let us find out!

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## Factoring Algebraic Equations

Factorisation of an algebraic expression is similar to finding factors of real numbers the only difference here is we use identities instead of prime numbers. Factoring algebraic expressions seem tricky, but with apt knowledge of identities, it becomes easy and fast. Let’s see how?

### Factors of Algebraic expressions

In an algebraic expression, while factoring, terms are formed as products of factors. For example, in the algebraic expression 7xy+3, The term 7xy has been formed by the factors 7, x and y, i.e., 7xy = 7×y

Here we observe that the factors 7, x and y of 7xy cannot further be expressed as a product of factors. While factoring algebraic expressions the “prime” word is replaced by “irreducible”, therefore we may also say that 7, x and y are the prime factors or the irreducible forms of 7xy.

It should, however, be noted that 7×(xy) is not an irreducible form of 7xy since the factor xy can be further expressed as a product of x and y, i.e., xy = x×y. Let’s consider another expression 5x (x+4). It can be written as a product of factors. 5, x and (x+2). 5x(x+4) = 5×x×(x+4). The factors 5, x and (x+4) are irreducible factors of 5x (x+4).

### Factoring Algebraic Expressions

As already said above, when we factorise an algebraic expression, we write it as the product of irreducible factors. These factors may be numbers, algebraic variables or algebraic expressions. Expressions like 5xy, 7x2y, 2x(y+3), 11(y+1) (x+2) are already in an irreducible factor form.

On the other hand in expressions like 6x+8, 5x+5y, x2+7x, x2+3x+6 we need to determine factors, for which we develop systematic methods to factorise these expressions. In the space below we will reduce various forms of algebraic expressions using irreducible factors.

Example 1: Factorise 21a2b + 27ab2

Solution: We have:      21a2b = 7*3*a*a*b
27ab2 = 3*3*3*a*b*b
The two terms have 3, a and b as common factors.
Therefore,       21a2b + 27ab2 = (3*a*b*7*a) + (3*a*b*3*b*3)
= 3*a*b*[(7*a) + (9*b)]                      (combining the terms)
= 3ab*(7a + 9b)
= 3ab (7a + 9b)

### Basics of Algebra

• Expressions: Expressions are formed from variables and constants. The expressions 3y-7 is formed from the variable y and constants 3 and 7. The expression 5xy + 9 is formed from variables x and y and constant 5 and 9. You can form as many expressions as you wish using variables and constants.
• Terms, factors and Coefficients: An algebraic expression is a combination of terms, factors and coefficients. Hence, in the expression, 3x+7; 3x and 7 are terms; 3, x and 7 are factors and 3 and 7 are numeric coefficients.
• Monomials, binomials and Polynomials Expression that contain only one term is called a monomial, Expression that contains two terms is called a binomial. Therefore, an expression containing three terms is a trinomial and so on.

In general, an expression containing, one or more terms with a non-zero coefficient (with variables having non-negative exponents) is called a polynomial. A polynomial may contain any number of terms, one or more than one.

• Examples of monomials:        9x2, 5xy, -7z, 5xy2, 10y, -9, 82mnp, etc.
• Examples of binomials:           a + b, 4l + 5m, a + 4, 5 -3xy, z2 – 4y2, etc.
• Example of trinomials:           a + b + c, 2x + 3y – 5, x2y – xy2 + y2, etc.
• Examples of polynomials:       a + b + c + d, 3xy, 7xyz – 10, 2x + 3y + 7z, etc.

### What is an Identity?

An equality true for every value of the variable in an expression is called identity. While factoring an algebraic expression using identities we use these to reach an irreducible form of the expression.

Consider the Equality              (a + 1 ) (a +2) = a2 + 3a + 2

We shall evaluate both sides of this equality for some value of a, say a = 5.

For a =5,                    LHS = (a + 1) (a +2) = (5 + 1) (5 +2) =(6) * (7) =42

RHS = a2 + 3a + 2 = (5)2 + 3(5) + 2 = 25 +15 +2 =40 + 2 = 42

Thus, for a = 5, also LHS = RHS.

When we use identical numbers for both LHS and RHS we find that for any value of ‘a’ LHS = RHS. Such equality, true for every value of the variable in it is called an identity. Thus, (a + 1) (a + 2) = a2 + 3a + 2 is an identity. These identities are used as irreducible factors for factoring algebraic expressions.

### Standard Identities

We shall now study three identities which are very useful in our work. These identities are obtained by multiplying a binomial by another binomial, Let us first consider the product (a +b) (a + b) or (a + b)2

(a + b)2 = (a +b) (a + b)
= a(a + b) + b(a +b)
= a2 + ab + ba + b2         (since ab = ba)
= a2 + 2ab + b2
Thus                             (a + b)2 = a2 + 2ab + b2                              (I)

Clearly, this is an identity, since the expression on the RHS is obtained from the LHS by actual multiplication. One may verify that for any value of a and any value of b, the values of the two sides are equal. Next we consider (a – b)2 = (a – b) (a – b) = a(a – b) –b (a – b)
We have                            = a2 – ab – ba + b2 = a2 – 2ab + b2
Or                         (a – b)2 = a2 – 2ab + b2                          (II)

Finally , consider (a + b) (a – b). we have (a + b) (a – b)=a(a – b) + b(a –b)
Or a2– ab + ba – b2 = a2 – b2 (since ab= ba)
Or  (a +b) (a – b) =a2 – b2                                                            (III)
The identities (I), (II), (III) are known as standard identities. We shall now work out one more useful identity.

(x +a) (x + b) = x(x + b) + a(x + b)
Hence, we can write = x2 + bx +ax + ab
Or  (x + a) (x + b) = x2 + (a +b)x +ab                                          (IV)
We can now see that Identity (IV) is the general form of the other three identities also.

### How do we apply identities for factoring?

Identities can be applied to factorise algebraic expressions. Generally used binomials and polynomials, here we shall learn using identities for binomial expressions only. The steps to be followed are:

• The first step we follow is to recognize the suitable identity. For example take the expression: 9x2 –18 x + 36, the expressions seem to be like a– 2ab + b2
• In the second step, we rewrite the expression in the form of its identity, i.e   (3x)2 – 2(3x)(6)+ (6) [ Here a=3x and b=6]

• The last step uses identities to find the factors for the given expression. Since the LHS for the identity a2 -2ab+b2  is (a-b)2 we write the factors using the identity as: (3x-6)2  = (3x-6) (3x-6). So the factor for 9x2 –18 x + 36 is  (3x-6) (3x-6).

Learn more about Factorisation using Division.

## Solved Examples For You

Question 1: Factorise: 5m2 8m 4:

A) (5m + 2)((m + 2)    B) (5m – 2)(m – 2)    C) (5m – 2)(m + 2)    D) (5m + 2)(m – 2)

Solution: D) The given expression is: 5m2 – 8m – 4. Therefore, it can be written as:

5m2 – 10m + 2m – 4 = 5m(m – 2) + 2(m –  2)

Hence we can write this = (5m + 2)(m – 2)

Question 2: What are the various types of factoring?

Answer: The various types of factoring that exist are: greatest common factor, grouping, difference in two squares, sum or difference in two cubes, trinomials, and general trinomials.

Question 3: What is meant by factoring in maths?

Answer: To factor a number means breaking this number up into multiple numbers that one can multiply together to attain the original number. For example: 6 = 3 x 2, so factors of 6 happen to be 3 and 2 , 9 = 3 x 3, so factors are 3 and 3.

Question 4: How can one solve factoring?

Answer: One can solve factoring with the following steps:

• Move all terms to one side
• Factor the equation fully.
• Set factors equal to zero and then solve.
• List each solution to the original equation.

Question 5: Who is credited with the discovery of factorization?

Answer: Factorization was discovered by ancient Greek mathematicians in the case of integers. They presented proof of the fundamental theorem of arithmetic. This theorem asserts that every positive integer is possible to be factored into a product of prime numbers, whose further factorization is not possible into integers greater than 1.

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