Me with polynomials. what is the zeros of x^2 + 2x + 4

  -1 ± i√3

Step-by-step explanation:

You can rearrange the expression to vertex form:

  x^2 +2x +1 +3

  (x +1)^2 +3

You want to find the values of x that make this zero:

  (x +1)^2 +3 = 0 . . . . . . . set the expression equal to zero

  (x +1)^2 = -3 . . . . . . . . . subtract 3 to get the square alone

  x +1 = ±√(-3) = ±i√3 . . .take the square root

  x = -1 ±i√3 . . . . . . . . . . subtract 1

The zeros are the complex numbers -1+i√3 and -1-i√3.

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3