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!

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- Jan 12th 2011, 07:51 AMandyboy179factorise, 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 AMAmer
- Jan 12th 2011, 10:35 AMHallsofIvy
In general, [tex]a^2- b^2= (a- b)(a+ b)/[tex]

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

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