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**Points to Remember**** **

• Factorisation means write an expression as a product of its factors.

• Like prime factors, an irreducible factor, a factor which cannot be expressed further as a product of factors.

• Some expression can easily be factorised using these identities:

I. a^{2} + 2ab + b^{2} = (a + b)^{2}

II. a^{2} – 2ab + b^{2} = (a – b)^{2}

III. a^{2} – b^{2} = (a – b)(a + b)

IV. x^{2} + (a + b)x + ab = (x + a)( x+ b)

• The number 1 is a factor of every algebraic term also, but it is shown only when needed.

• When factorisation of x^{2} + (a + b)x + ab is done by splitting the middle term, the two numbers which give the product ab and (a + b) as the coefficient of x have to be chosen very carefully with correct sign.**We Know That**

(i) (a + b)^{2} = a^{2} + b^{2} + 2ab

(ii) (a – b)^{2} = a^{2} + b^{2} – 2ab

(iii) a^{2} – b^{2} = (a + b)(a – b)

(iv) 1 is a factor of every term of an algebraic expression. Unless it is specially required, we do not show 1 as a separate factor of any term.

(v) Factorisation means writing an expression as product of factors.

Note:In case of factorisation of a term of an expression, the word ‘irreducible’ is used in place of ‘prime’. For example, 6pq = 2 * 3 * pq is not the irreducible form because pq can further be factorised as p q, i.e. the irreducible form of 6pq = 2 * 3 * p * q.

**Example 1:** Write 10y as irreducible factor form.**Solution:** We have

10 = 2 * 5

xy = x *y

∴ 10xy = 2 * 5* x * y**FACTORISATION USING COMMON FACTORS****Example 1:** Factorise 18x^{2} – 14x^{3} + 10x^{4}**Solution:** We have

Obviously, the common factors of these terms are 2, and .

∴ 18x^{2} – 14x^{3} + 10x^{4}

Thus, 18x^{2} – 14x^{3} + 10x^{4} = 2x^{2}[9 – 7x + 5x^{2}]**FACTORISATION BY REGROUPING TERMS**

In certain cases the given expression cannot be factorised easily but by rearranging its terms, we can form groups leading to factorization.**Eample 1: **Factorise 9x + 18y + 6xy + 27**Solution:** Here, we have a common factor 3 in all the terms.

∴ 9x + 18y + 6xy + 27 = 3[3x + 6y + 2xy + 9]

We find that 3x + 6y = 3(x + 2y) and 2xy + 9 = 1(2xy + 9)

i.e. a common factor in both the groups does not eist,

Thus, 3x + 6y + 2xy + 9 cannot be factorised.

On regrouping the terms, we have

3x + 6y + 2xy + 9 = 3x + 9 + 2xy + 6y

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

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

Now, 3[3x + 6y + 2xy + 9] = 3[(x + 3)(3 + 2y)]

Thus, 9x + 18y + 6xy + 27 = 3(x + 3)(2y + 3)**Solved Examples:Q1: Let f(x)=2x3+16x2+44x+42 be a polynomial having one of the factors as (x2+5x+7), then the other factor of f(x) would be a multiple of:A) 1B) 2C) 3D) 4Solution: **B) Since f(x) is a cubic polynomial, and one of the factors is a polynomial of degree two, then we can say that the other factor will be a polynomial of the form ax + b; where ‘a’ nd ‘b’ are two constants and a ≠ 0. Hence, we can write:

2x

or 2x

Compairing the coefficients of x on both sides, we have 2 = a and 42 = 7b. Therefore, b = 6 and a = 2. hence the other factor is 2x + 6 or 2(x+3) which is a multiple of 2.**Q 2: Factorise: 5m ^{2} − 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: 5m

5m

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

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