**1. Method of Common Factors**

- In this method, we have to write the irreducible factors of all the terms
- Then find the common factors amongst all the irreducible factors.
- The required factor form is the product of the common term we had chosen and the leftover terms.

**2. Factorisation by Regrouping Terms**

When there is no common term in the expressions then

- We have to make the groups of the terms.
- Then choose the common factor among these groups.
- Find the common binomial factor and it will give the required factors.

**3. Factorisation Using Identities**

Remember some identities to factorise the expression

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

(2x + 3)^{2} = (2x)^{2}+ 2(2x) (3) + (3)^{2} = 4x^{2} + 12x + 9

(2x – 3)^{2} = (2x)^{2} – 2(2x)(3) + (3)^{2} = 4x^{2 }-12x + 9

(2x + 3) (2x – 3) = (2x)^{2} – (3)^{2} = 4x^{2} – 9

**Example 1**

Factorise x^{2}– (2x – 1)^{2} using identity.

**Solution:**

This is using the identity (a + b) (a – b) = a^{2} – b^{2}

x^{2}– (2x – 1)^{2} = [(x + (2x – 1))] [x – (2x-1))]

= (x + 2x – 1) (x – 2x + 1)

= (3x – 1) (- x + 1)

**Example 2**

Factorize 9x² – 24xy + 16y² using identity.

**Solution:**

9x² – 24xy + 16y²

= (3x)^{2} – 24xy + (4y)^{2}

Now split the middle term.

= (3x)^{2} – 2(3x) (4y) + (4y)^{2}

=(3x – 4y)^{2}

Hence the factors are (3x – 4y) (3x – 4y).

**Example 3**

Factorise x^{2} + 10x + 25 using identity.

**Solution:**

x^{2} + 10x + 25

= (x)^{2} + 2(5) (x) + (5)^{2}

x^{2} + 10x + 25 = (x + 5)^{2}

**4. Factors of the form ( x + a) ( x + b)**

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

**Example:**

Factorise x^{2} + 3x + 2.

**Solution:**

If we compare it with the identity (x + a) (x + b) = x^{2} + (a + b) x +ab

We get to know that (a + b) = 3 and ab = 2.

This is possible when a = 1 and b = 2.

Substitute these values into the identity,

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