How do you write #64w^2+169v^2# in the form #A(Bx+C)(Dx+E)#?

Question

1340 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-04T07:34:30

#64w^2+169v^2=(8a-i13v)(8w+i13v)#

#64w^2+169v^2#

= #(8w)^2+(13v)^2#

It is obvious that as it is sum of two squares and nothing is common between the two monomials, we cannot have rational or real factors.

However, using definition of imaginary numbers given by #i^2=-1#, we get

#64w^2+169v^2=(8w)^2+(13v)^2=(8w)^2-i^2(13v)^2#

= #(8w)^2-(i13v)^2#

= #(8w-i13v)(8w+i13v)#