Class 8 Maths Factorisation What is Factorization

What is Factorization

Factors of a number are numbers that divide evenly into another number. Factorization writes a number

as the product of smaller numbers.

For example, let’s factor the number 12:

12 = 6 * 2 = 3 * 4 = 2 * 2 * 3

Again 3xy + 2x, x2 + 10x, x2 + 7x + 12, etc are not in the factor form.

Method of Common Factors:

Let us take an example.

Ex: Factorize 3x + 9

Now, we write each term as a product of irreducible factors.

3x = 3 * x

9 = 3 * 3

So, 3x + 9 = (3 * x) + (3 * 3)

= 3 * (x + 3)                     [By Distributive Law]

= 3(x + 3)

Again let we factorize 10xy + 2y

Now, the irreducible factor forms of 10xy and 4y are respectively,

10xy = 2 * 5 * x * y

2y = 2 * 2 * y

10xy + 2y = (2 * 5 * x * y) + (2 * 2 * y)

= (2y * 5x) + (2y * 2)

= 2y * (5x + 2)                           [By Distributive Law]

= 2y(5x + 2)

Factorization by regrouping terms:

Let we want to factorize x2 + xy + 8x + 8y

Since there is no common factor among all terms. So, we form groups.

We form two groups (x2 + xy) and (8x + 8y) and then factorize

x2 + xy = x * x + x * y

= x * (x + y)

8x + 8y = 8 * x + 8 * y

= 8 * (x + y)

Now, x2 + xy + 8x + 8y = x * (x + y) + 8 * (x + y)

= (x + y)(x + 8)

We can regroup the algebraic expressions in many ways.

Problem: Find the common factors of the given terms.

(i) 12x, 36                         (ii) 2y, 22xy                           (iii) 14pq, 28 p2 q2                    (iv) 2x, 3x2, 4

(i) 12x = 2 * 2 * 3 * x

36 = 2 * 2 * 3 * 3

Hence, the common factors are 2, 2 and 3 = 2 * 2 * 3 = 12

(ii) 2y = 2 * y

22xy = 2 * 11 * x * y

Hence, the common factors are 2 and y = 2 * y = 2y

(iii) 14pq * 2 * 7 * p * q

28 p2 q2 = 2 * 2 * 7 * p * p * q * q

Hence, the common factors are 2 * 7 * p * q = 14pq

(iv) 2x =2 * x * 1

3x2 = 3 * x * x * 1

4 = 2 * 2 * 1

Hence, the common factors is 1

Problem: Factorize:

(i) 15xy – 6x + 5y – 2           (ii) ax + bx – ay – by         (iii) 15pq + 15 + 9q + 25p            (iv) z – 7 + 7xy – xyz

(i) 15xy – 6x + 5y – 2 = 3x(5y - 2) + 1(5y - 2)

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

(ii) ax + bx – ay – by = x(a + b) – y(a + b)

= (a + b)(x - y)

(iii) 15pq + 15 + 9q + 25p = 15pq + 25p + 15 + 9q

= 5p(3q + 5) + 3(3q + 5)

= (3q + 5)(5p + 3)

(iv) z – 7 + 7xy – xyz = 7xy – 7 – xyz + z

= 7(xy - 1) – z(xy - 1)

= (xy - 1)(7 - z)

Factorization using Identity:

We know the identity

(a + b)2 = a2 + 2ab + b2

(a - b)2 = a2 - 2ab + b2

(a + b)(a - b) = a2 – b2

(x + a)(x + b) = x2 + (a + b)x + ab

We use these identities to do the factorization of algebraic expressions.

Problem: Factorize the following expressions:

(i) a2 + 8a + 16      (ii) p2 – 10p + 25  (iii) 25m2 + 30m + 9

(iv) 49y2 + 84yz + 36z2   (v) 4x2 – 8x + 4   (vi) 121b2 – 88bc + 16c2

Solution:

(i) a2 + 8a + 16 = a2 + (4 + 4)a + 4 * 4

= (a + 4)(a + 4)                      [x2 + (a + b) + ab = (x + a)(x + b)]

= (a + 4)2

(ii) p2 – 10p + 25 = p2 – (-5 - 5)p + (-5)(-5)

= (p - 5)(p - 5)

= (p - 5)2                             [x2 + (a + b) + ab = (x + a)(x + b)]

(iii) 25m2 + 30m + 9 = (5m)2 + 2 * 5m * 3 + 32

= (5m + 3)2                              [a2 + 2ab + b2 = (a + b)2]

(iv) 49y2 + 84yz + 36z2 = (7y)2 + 2 * 7y * 6z + (6z)2

= (7y + 6z)2                           [a2 + 2ab + b2 = (a + b)2]

(v) 4x2 – 8x + 4 = (2x)2 – 2 * 2x * 2 + 22

= (2x - 2)2                                            [a2 - 2ab + b2 = (a - b)2]

= 22(x - 1)2

= 4(x - 1)2

(vi) 121b2 – 88bc + 16c2 = (11b)2 – 2 * 11b * 4c + (4c)2

= (11b – 4c)2                                     [a2 - 2ab + b2 = (a - b)2]

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