<span>y^2 + 5y + 6
=</span><span>(y + 2)(y + 3)
----------------------------------</span>
The fraction he added is 6/30, or 1/5.
I'm not 100% sure about the equation you want me to use but I did the work this way:
5/6+x/30=31/30 (the denominator has to equal 30 because you are adding. You need a common denominator which is why you need to change 5/6. We can label the numerator as a hidden value x)
25/30+x=31/30
You can then work backwards and basically subtract 25 from 31 and get 6; so x=6
{(1,3) (2,4) (3,5) (4,3)}
Since all four terms don't have a common factor we will group them in pairs with common factors.
6uv - 3v^3 + 4u - 2v^2
6uv + 4u - 3v^3 - 2v^2
Take out 2u from the first two terms and -v^2 from the last two terms. The goal is to have a common factor once we take out the GCF's.
2u( 3v + 2) - v^2(3v + 2)
Now we have two terms:
[2u(3v+2)] + [-v^2(3v+2)]
These two terms have a GCF of (3v + 2). Take that out:
(3v + 2)(2u - v^2)
Answer:
8
Step-by-step explanation:
Eight Equeal Tri's
