find all zeros of: x^4 - 4x^3 - 5x^2 +36x - 36 = 0 such that 2 is a zero of multiplicity 2
I found 2 and 3 are zeros thus; (x - 2)^2 (x - 3) (x^2 + x - 6)
I then applied the quadratic formula to (x^2 + x - 6) resulting in (-1 +- 5) / 2
I then put x^4 - 4x^3 - 5x^2 +36x - 36 into y= on the calculator and only found 2 and 3 to be the only zeros
If you don't mind,can you check this problem out? I want to be sure I'm doing it right, got myself an exam tomorrow and this part is where I don't feel quite confident...