# factorising - difference of squares

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

i have been asked to factorise the following

\$\displaystyle l^8-256\$

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

\$\displaystyle (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

\$\displaystyle l^8-265\$

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

\$\displaystyle (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 \$\displaystyle 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 \$\displaystyle l^8-256\$?

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

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

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

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

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

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