how do i factor 2y2+2y+4?

Answer 1

Two factors of 2y^2 + 2y + 4 are complex.
2y^2 + 2y + 4 =
2(y^2 + y + 2) =
2(y^2 + y + 1/4 + 7/4) =
2(y + 1/2)^2 + 7/4) =
2(y + 1/2 + i2√7)(y + 1/2 - i2√7)

Answer 2

When factoring polynomials, the first thing you always want to look for is the greatest common factor of all terms. In this case you have 2y^2 + 2y + 4, having terms 2y^2, 2y, and 4.
Since 2y^2 = 2*y^2, 2y = 2*y, and 4 = 2*2, the greatest common factor is 2.
So, 2y^2 + 2y + 4 = 2(y^2 + y + 2)

Now, we need to look at y^2 + y + 2. Since it is a quadratic trinomial with a lead coefficient of 1, it will be factorable if there exist two integers that multiply to 2 and add to 1.
The integers that multiply to 2 are 1 and 2, or -1 and -2, neither pair adding to 1. So, y^2 + y + 2 does not factor any further.

Therefore, 2y^2 +2y + 4 = 2(y^2 + y + 2) is the complete factorization. This is of course assuming you only want integer coefficients.

Answer 3

2y2+2y+4
First take out the common factor= 2
2(y2+y+2)
Then, use the Cross method, because you can see its a perfect square.
2(y+2)(y+2)
= 2(y+2)^2

Answer 4

We can factor 2y^2+2y+4= 2 (y^2+y+2)
or we can use the quadratic formula because2 y^2+2y+4 is not factorable.
x=-b+/- square root b^2-4ac/2a

a=2 b=2 c=4

(-2+/-sqrt((2)^2-4(2)(4)))/2(2)

-2(+/-sqrt(4-32))/4
2+/-sqrt(-28)/4
=2+/-sqrt(4*7)/4
=2+/-2sqrt(-7)/4
x1=1+isqrt(7)/2
and
x2=1-isqrt(7)/2