Factorization can be done by three methods

**1. By taking out the common factor**

If we have to factorize x^{2} – x then we can do it by taking x common.

x(x – 1) so that x and x –1 are the factors of x^{2} – x.

**2. By grouping**

ab + bc + ax + cx = (ab + bc) + (ax + cx)

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

= (a + c)(b + x)

**3. By splitting the middle term**

x^{2} + bx + c = x^{2} + (p + q) + pq

= (x + p)(x + q)

This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.

**Example: 1**

Factorize 6x^{2} + 17x + 5 by splitting the middle term.

**Solution:**

If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.

Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.

6x^{2} + 17x + 5 =6 x^{2} + (2 + 15) x + 5

= 6x^{2} + 2x + 15x + 5

= 2x (3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

**Example: 2**

Factorize 8x^{3} + 27y^{3} + 36x^{2}y + 54xy^{2}

**Solution:**

The given expression can be written as

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

= (2x)^{3} + (3y)^{3} + 3(2x)^{2}(3y) + 3(2x)(3y)^{2}

= (2x + 3y)^{3}

= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.

**Example: 3**

Factorize 4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4xz.

**Solution:**

4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4xz = (2x)^{2 }+ (–y)^{2} + (z)^{2} + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)

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

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