Find the value of c so that (x+2) is a factor of the polynomial p(x)=3x^4-x^2+cx-2

Using polynomial long division, we get
         
         3x^3+6x^2+11x
         
(x+2) | 3x^4-x^2+cx-2
          -(3x^4+6x^3)
         
             6x^3-x^2+cx-2
          - (6x^3+12x^2)
         
        11x^2+cx-2
      -(11x^2+22x) 
       
        (22+c)x-2. 

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

The answer for this is 9/1253decimal form 0.007