Factorise #(A^2- 1) (B^2- 1) + 4AB#?

2 Answers
Watchmojo J.
Mar 1, 2018

#(A-1)(A+1)(B-1)(B+1)+4AB#

Explanation:

When multiplying out 2 brackets, you use this technique.

![https://www.youtube.com/watch?v=d0gKPnKy6YQ]
(useruploads.socratic.org)

In this case, this is called the difference of 2 squares, which always produces this

#(x-y)(x+y)#

#= x^2 - yx + yx -y^2#

#= x^2 -y^2#

It will always give the square of one number minus the square of another.

It is the same for #(A^2 - 1)# and #(B^2 - 1)#

=#((A)^2 - (1)^2)#

#((B)^2 - (1)^2)#

Therefore, both can be factorised back into 2 brackets

#(A-1)(A+1)#

and

#(B-1)(B+1)#

So,

#(A^2 - 1)(B^2 - 1) + 4AB#

can be factorised to

#(A-1)(A+1)(B-1)(B+1)+4AB#

George C.
Mar 1, 2018

#(A^2-1)(B^2-1)+4AB = (AB-A+B+1)(AB+A-B+1)#

Explanation:

#(A^2-1)(B^2-1)+4AB#

#= A^2B^2-A^2-B^2+1+4AB#

#= (A^2B^2+2AB+1)-(A^2-2AB+B^2)#

#= (AB+1)^2-(A-B)^2#

#= ((AB+1)-(A-B))((AB+1)+(A-B))#

#= (AB-A+B+1)(AB+A-B+1)#

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