Question

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

Answer 1

`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`