# factorise, hard

• Jan 12th 2011, 07:51 AM
andyboy179
factorise, hard
hi i need to know how to factorise this:

Attachment 20407

i know if it was 2x^2 - 121 it would = (x+11) (x-11)
so for this one would it be (2x +11) (2x-11) ???

if not could you please tell me where im going wrong!

Thanks!
• Jan 12th 2011, 07:57 AM
Amer
Quote:

Originally Posted by andyboy179
hi i need to know how to factorise this:

Attachment 20407

i know if it was 2x^2 - 121 it would = (x+11) (x-11)
so for this one would it be (2x +11) (2x-11) ???

if not could you please tell me where im going wrong!

Thanks!

if it is x^2 - 121 = (x -11)(x + 11 )
but 2x^2 - 121 is not equal to (x- 11)(x+11)

ur solution for the question is rite 4x^2 - 121 = (2x - 11)(2x +11)
• Jan 12th 2011, 10:35 AM
HallsofIvy
In general, [tex]a^2- b^2= (a- b)(a+ b)/[tex]

Actually, $2x^2- 121= (\sqrt{2}x)^2- 11^2= (\sqrt{2}x- 11)(\sqrt{2}x+ 11)$, not $(x- 11)(x+ 11)$. I presume that was a mis-type.

$4x^2= (2x)^2$ so, yes, $4x^2- 121= (2x- 11)(2x+ 11)$.