Arriving at Factorisation: How Is It Done?

[rbnphlp]

How does one arrive at this ?

I was wondering how does someone get from factorising
:{a^(n)-b^(n)}

to
{a-b}{a^(n-1)-a^(n-2)(b)+a^(n-3)b^(2)----+b^(n-1)}

thanks

 

[tiny-tim]

Welcome to PF!

Hi rbnphlp! Welcome to PF!

(try using the X² tag just above the Reply box )

Just multiply the long one by a on one line, and by b on another line, and subtract.

 

[rbnphlp]

Hi rbnphlp! Welcome to PF!

(try using the X² tag just above the Reply box )

Just multiply the long one by a on one line, and by b on another line, and subtract.

Sorry , but my original question is how do I arrive at the long one from (aⁿ-bⁿ),

 

[tiny-tim]

ah!

use long division, exactly as you would for decimals.

 

[jgens]

Clearly $a=b$ is a solution to $a^n-b^n$. To factor out the $a-b$ term, we use polynomial long division (or synthetic polynomial division) to find that $\frac{a^n-b^n}{a-b} = a^{n-1} + \dots + b^{n-1}$.

 

[Elucidus]

Although polynomial long division can arrive at the desired quotient, it is usually easier to demonstrate the equality by just distributing the left factor on the right-hand side across its cofactor:

$$(a-b)\cdot(a^{n-1} + a^{n-2} b + \cdots + ab^{n-2} + b^{n-1})<br /> = (a^n + a^{n-1}b + \cdots + ab^{n-1}) - (a^{n-1}b + a^{n-2}b^2 + \cdots + b^n)$$

and then simplifying.

--Elucidus

 

[rbnphlp]

ah!

use long division, exactly as you would for decimals.

oh got it..thankyou

 

[rbnphlp]

thanks evryone