# factorising - difference of squares

• Mar 26th 2009, 11:57 AM
sammy28
factorising - difference of squares
hi,

i have been asked to factorise the following

$l^8-256$

using the difference of squares $a^2-b^2=(a+b)(a-b)$ and by working from the lowest term $(l^2-2^2)$, i managed to get a solution but its was more down to guesswork than a systematic attempt to solve this.

$(l^4+16)(l^2+4)(l+2)(l-2)$

is there a way of dealing with these type of questions?

thanks
• Mar 26th 2009, 12:02 PM
masters
Quote:

Originally Posted by sammy28
hi,

i have been asked to factorise the following

$l^8-265$

using the difference of squares $a^2-b^2=(a+b)(a-b)$ and by working from the lowest term $(l^2-2^2)$, i managed to get a solution but its was more down to guesswork than a systematic attempt to solve this.

$(l^4+16)(l^2+4)(l+2)(l-2)$

is there a way of dealing with these type of questions?

thanks

Hi sammy28,

Isn't that supposed to be $l^8-256$?
• Mar 26th 2009, 12:03 PM
sammy28
oops, (Giggle)
• Mar 26th 2009, 12:08 PM
masters
Quote:

Originally Posted by masters
Hi sammy28,

Isn't that supposed to be $l^8-256$?

$l^8-256$ is the difference of two squares. Look at it this way:

$((l^4)^2-16^2)$

$(l^4-16)(l^4+16)$

$((l^2)^2-4^2)(l^4+16)$

$(l^2-4)(l^2+4)(l^4+16)$

$(l-2)(l+2)(l^2+4)(l^4+16)$
• Mar 26th 2009, 12:21 PM
sammy28
thanks masters (Clapping)