How do you factor #0.49v ^ { 2} + 1.12v + 0.64#?

1 Answer

#0.49v^2+1.12v+0.64=(.7v+.8)^2#

Explanation:

Note that we have perfect squares:

#0.49=0.7^2, 0.64=0.8^2#

Also see that #0.7xx0.8=0.56#. Doing that twice gives: #0.56+0.56=1.12#

And so we can factor this way:

#0.49v^2+1.12v+0.64=(.7v+.8)^2#