Question

how can i understand the factorisation of polynomial??

Answer 1

In mathematics, factorization or factoring is the breaking apart of a polynomial into a product of other smaller polynomials. If you choose, you could then multiply these factors together, and you should get the original polynomial (this is a great way to check yourself on your factoring skills). One set of factors, for example, of 24 is 6 and 4 because 6 times 4 = 24. When you have a polynomial, one way of solving it is to factor it into the product of two binomials.

Answer 2

Hey there!

Factorisation! Uh..Factorisation! such a frustration right? xD But, if solved carefully and understood properly, you could solve factorisation question with ease.

$\textbf { What is factorisation? }$

Well, splitting a expression into factors is what we call as factorisation

Factorisation can be done by 2 methods

$\textbf { Method1. Splitting the middle term }$ [Factorising polynomials of degree 2]

➸ This method can be used to factorise a quadratic equation in the form ax² + bx + c, where a, b, and c are constants and a≠0

➸ This equation have 2 factors.

Here, we will split the middle term b of
ax²+bx+c into p and q, such that
p+q = b
p × q = a×c

E.g.: 6x2 - 13x + 6
6x2 - 13 x + 6 ----->(1) 

a.c = 6 × 6 = 36 

Factors of 36 = 2,18 
= 3,12 
= 4,9 
Only the factors 4 and 9 gives 13 = (4 + 9) 
For -13 , both the factors have negative sign. – 4 – 9 = - 13

Equation (1) ⇒ 6x 2 - 4x – 9x + 6 
⇒2x ( 3x – 2 ) – 3 ( 3x – 2 )

⇒(3x – 2 ) ( 2x – 3) are the factors.

$\textbf { Remainder theorom }$
If we divide a polynomial f(x) by (x-a), then remainder will be f(a)

$\textbf { Method2 : Factor theorom }$
[For polynomials of degree 2 and above]

➸This method is just the reverse of Remainder theorom.

➸ If f(a) = 0, then (x-a) is a factor of Polynomial f(x)

These are the two methods which we use to factorise a polynomial.

$\textbf { Always remember a key point, }$

  $$\boxed{No.of Degrees = No.of Factors }$$

$\textbf {By Algebraic identities}$

➸Identity I:(a + b)2 = a2 + 2ab + b2

➸Identity II: (a – b)2 = a2 – 2ab + b2

➸Identity III:a2 – b2= (a + b)(a – b)

➸Identity IV:(x + a)(x + b) = x2 + (a + b) x + ab

➸Identity V:(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

➸Identity VI:(a + b)3 = a3 + b3 + 3ab (a + b)

➸Identity VII:(a – b)3 = a3 – b3 – 3ab (a – b)

➸Identity VIII:       a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Worst part while factorising using identities is identifying which identify to use. The very practical and only way is to thorough all the equations, by heart.

Glad if helped.
Any doubts? ping me :D