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

1 Answer
Dec 4, 2016

Answer:

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

Explanation:

#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)#