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

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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

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

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

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

How do you write 64w^2+169v^264w^2+169v^2 in the form A(Bx+C)(Dx+E)A(Bx+C)(Dx+E)?