What are all the real and imaginary zeros of this?

Question

y=-x^3+8x^2-21x+18

Steps as to how you get them would be nice...

Answer 1

y = (-x+2)(x-3)², so if y = 0, x = -2 or x = 3. There's nothing imaginary in this problem, unless you have some more convoluted way of solving it (or unless I've forgotten something in the last 40-some years since I used to do these regularly).

Answer 2

y = -x³ + 8x² - 21x + 18

Since the lead and final coefficient are 1 and 18, any rational factors will be found in the list ±1, ±2, ±3, ±6, ±9, ±18.

y = -x³ + 8x² - 21x + 18
y = -(x - 2)(x - 3)²

All three roots of the equation are real.

Answer 3

well, when confronted with a problem like this, I usually resort to synthetic division, the first thing I do is I guess a factor like (x-1) or (x+1) and then check with synthetic division.

if we try 2, we see that it is a factor, so we have (x-2)(-x^2+6x-9) or -1(x-2)(x^2-6x+9)
then we factor the other polynomial and see that its (x-3)(x-3) so finally we have -1(x-2)(x-3)^2 so the zeros are 2, and 3 which are both real