If you're wondering how I did the long division, what I essentially did was get the first value (at the start, it was 3x^4) and divided it by the first value of the divisor (which in x+2 was x) to get 3x^3 in our example. I then subtracted the polynomial by the whole divisor multiplied by, for example, 3x^3 and repeated the process.
For this to be a perfect factor, (x+2)*something must be equal to (22+c)x-2. Focusing on how to cancel out the 2, we have to add 2 to it. To add 2 to it, we have to multiply (x+2) by 1. However, there's a catch, which is that we subtract whatever we multiply (x+2) by, so we have to multiply it by -1 instead. We still need to cross out (22+c)x. Multiplying (x+2) by -1, we get (-x-2) but by subtracting the whole thing from something means that we have to add -(-x-2)=x+2 to something to get 0. x+2-x-2=0, xo (22+c)x-2 must equal -x-2, meaning that (22+c)=-1 and c=-23
We know that the function f(x+m) equals to the function f(x) translated 'm' times to the left. Therefore, to translate the function f(x) = x^2 twelve units to the right, we have to:
