Notes on Factorization | Grade 8 > Compulsory Maths > Algebra | KULLABS.COM

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When two or more algebraic expressions are multiplied, the result is called product and each expression is called the factor of the product.

The process of finding out factors of an algebraic expression is known as factorisation.

For example:

If we factorise (bc + cd), you get c ( b + d ).

### Factorizing the difference of two squares

Let's multiply ( a + b ) and ( a - b )

( a + b ) ( a - b )

= a² - ab + ab -b²

= a² - b² ( This expression is called a difference of two squares )

Therefore, the factors of a² - b² are ( a + b ) and ( a - b)

Examples:

1. x- 49
Solution:
x- 49, this expression is the difference of two squares.
= x2- 72, which is in the form of a2- b2
= (x+7) (x-7)

2. 4y- 36y6
Solution:
In 4y- 36y6, there is a common factor of 4y2 that can be factored out first in this problem, to make the problem easier.
= 4y- 36y6
= 4y2(1 - 9y4)
= 4y2{(1)2- (3y2)2}
= 4y2(1+3y2)(1-3y2)

### Factoring perfect square trinomials

Let's multiply (a+b) and (a+b)

(a+b) (a+b)

= a2+ ab + ab + b2

= a2 + 2ab + b2

Thus, a2 + 2ab + b2 = (a + b)2 and (a + b)2 is the factorisation form of a+ 2ab + b2

Similarly, a- 2ab + b2 = (a -b)2 and (a - b)2 is the factorisation form of a- 2ab +b= (a -b)and (a - b)2 is the factorisation form of a2 - 2ab + b2

### Geometrical meaning

If we consider (a+ b) as one of the side of the square then the product of the expression will form two squares namely a2 and b2 and two congruent rectangles with each having an area of ab.

 a2 ab ab b2

Area of the entire square = (a + b)2

Area of two squares and two rectangles

= a2 + ab +ab + b2

= a2 + 2ab +b2

Thus, a+ 2ab + b2 = (a+b)2

• Factorization is the process of finding the factors.
• Factoring is the decomposition of an object, into a product of other objects, or factors, which when multiplied together give the original.
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#### Click on the questions below to reveal the answers

Solution:

Given expression =6x+3

= 2.3.x + 3

= 3(2x+1) [3 is common in both]

Solution:

Given expression =x2+4x

=x. x+4. x

=x(x+4) [x is common in both]

Solution:

Given expression =12a+3 b

=4.3. a + 3.b

=3(4a+b) [ 3 is common in both]

Solution:

Here,

Given expression =x+x3

=x+x . x.x

=x(1+x2) [x is common in both]

Solution:

Here,

Given expression =12x2+xy+xz

=2.2.3.x.x+x.y+x.z

=x(12x + y + z) [ x is common in all]

Solution:

Here,

Given expression =14xy+7y

=2.7.x.y+7.y

=7y(2x+1) [ 7y is common in both]

Solution:

Here,

x2- 4

=(x)2-(2)2

=(x-2)(x+2) [$$\therefore$$a2-b2=(a+b)(a-b)]

Solution:

Here,

Given = 9x2-y2

=(3x)2-(y)2

=(3x+y)(3x-y)

Solution:

Given = 121-25y2

= (11)2-(5y)2

= (11+5y)(11-5y)

Solution:

The given expression is x2 - 7x + 12

Find two numbers whose sum = -7 and product = 12

Clearly, such numbers are (-4) and (-3).

Now, x2 - 7x + 12

= x2 - 4x - 3x + 12

= x(x - 4) -3 (x - 4)

= (x - 4)(x - 3)

0%

3 and 5
1 and 1
3 and 1
9 and 2

20 and 4
21 and 1
7 and 3
9 and 5

11y
10x
9x
5x

10 y
10 y
18 y
2 y

2 xy
10 xy
5 xy
12 xy

20 x
24 x
25 x
2 x

30 x
50 x
20 x
80 x

60 y
50 y
25 y
35 y

2100
2800
2550
2600

980
820
900
250

120000
110000
100000
150000

56
22
63
84

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