If equation by parabola is given by, p(x) = x^2 -5x + 6 then its factors are: A. (x-3)
B. (x-2)
C. Both (a) and (b)
D. None of the above

x^2-2x-3+6=0

(x-2x)-(3x+6)=0

x(x-2)-3(x-2)=0

(x-3)(x-2)=0

x-3=0 or x-2=0

x=3or x=2

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.