f(x)=x2+4x+4=(x+2)2.
If factorisation is required we look at the constant term and note its factors=1, 2. Its rational zeroes are therefore ±1 and ±2. Because there are no minus signs in the quadratic, the zeroes must be negative. So we have -1 or -2 as the possible zeroes. f(-1)=1-4+4=1, so -1 is not a zero; f(-2)=4-8+4=0, so -2 is a zero and x+2 is a factor. Dividing f(x) by x+2 using synthetic or long division, we get x+2 as the quotient, so f(x)=(x+2)(x+2)=(x+2)2.
(No question has been asked.)
This function can be graphed as a parabola with vertex at (-2,0) and y-intercept at (0,4). The axis of symmetry is x=-2 (the arms of the parabola are reflections of one another in the axis of symmetry. Because (-2,0) is the lowest point of the graph, f(x) can never be negative.