I wanted to see what was the answer and it gave me this -x^3 (Square root of X+1) (x+1) (x^2-x+1). The only thing I wonder is why is -x^3 added?
Work the exercise, and you'll find out why there is a factor of -x^3. If not, then please show your work.
I am asked that once I factor them to set them to 0 and solve the equation in other words solve for the unknown.
I had this problem x^2+4xy+4y^2. Once you factor it, it comes out to (x+2y)^2. My question is if I set it to 0 would it be possible for me to get the unknown or do I just leave it like that.
Huh? If you are asked to solve for x, then do it. If you are not asked to solve for x, then do not do it.
In other words, just follow the specific instructions that come with each exercise.
By the way, if you were to set (x+2y)^2 equal to zero, then you would be saying that x+2y=0.
There are an infinite number of xy-pairs that make x+2y equal to zero.
Here are three such pairs: (0, 0) (1, -1/2) (-4, 2)