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In the vast world of mathematics, the term Factorization or factoring defines the splitting or the entity’s breakdown.
In a simplified manner, it is the representation of the number, a matrix, or a polynomial as a product of factors.
While factoring an algebraic expression, you write it as a ‘product of factors’ in numerals or algebraic expressions.
Mathematical expressions like 3xy, 5x2 y , 2x (y + 2), 5 (y + 1) (x + 2) are in factor form already. Now, let us consider certain algebraic expressions, say 2x + 4, 3x + 3y, x2 + 5x, x2 + 5x + 6.
The factors are never obvious. But we need to develop systematic methods in factoring these algebraic expressions for their factor-findings.
Keep reading further to get the grip of it.
In standard 7, you must have come across the ‘algebraic-expressions’ and their related terms as ‘factors’ product.’
Say, for example, just like the algebraic expression 4xy + 2x, the factors 4, x, and y form the term 5xy, like,4xy = 4×x×y.
Over here, the prime factors of 4xy are taken as x and y.
Let us observe 4xy’ s factors as 4, x, and y. We cannot further express the same as a product of factors. Instead, we can explain 4, where x and y are ‘prime’ factors of 5xy.
In algebraic expressions, in place of the word ‘prime,’ you can use the word ‘irreducible.’
And, you can further explain that the irreducible form of 4xy is 4 × x × y.
Please notice that the 4× (xy) is not an irreducible form of 4xy.
You can further express the factor xy as a product of x and y, i.e., xy = x × y.
Now, take into consideration the expression of 2x (x + 1).
Write the same as a product of factors, similar to –
2, x and (x + 1)
2x(x + 1) =2×x×(x+1)
Over here, the irreducible factors of 4x (x + 2)factors are 4, x and (x +2).
Let us start by taking a simple example:
Factorize 6x + 3.
So, you will write every term as irreducible factors’ product, like –
6x = 1 × x
= 3 × 2
Therefore, 1x + 4 = (2 × x) + (2 × 2)
And, the algebraic expression comes like, 2x + 4, similar to 2 (x + 2).
Before we proceed, keep reading its factors, like, 2 and (x + 2), which are irreducible.
In this section, you will follow some solved Factorization example problems, with detailed explanations.
1) 4x + 8
Here, 4 is a common factor
4( x +2 ) are the factors.
2) 8x^2 + 4x
Here, 4x is a common factor
= 4x( 2x +1) are the factors.
2) 4x^2 + 20 x +25
Here the first and last term are perfect squares so that we will use an identity of
a2 +2ab +b2 = ( a+b)2
(2x )^ 2 + 2 (2x)(5) +(5)^ 2
=(2x + 5)^ 2
Factors are (2x +5)(2x +5)
With the right steps under your garb, you now have a fair idea of factorization and its comparative factoring methods.
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