What are the real zeroes of x3 + 6 x2 – 9x – 54?

First we have to know how many zeros can we have we check this by seeig whta is our highest degree in this case it will be 3 so we will have 3 real zeros lets check how many will be positive and how many will be negative, and how many will be complex

we chekc how many postives zeros there is by counting how many times does teh sign change with teh normal polynomial it only changes 1 time in normal form, now lets check the negatives, you do this by changing the signs of the numbers with odd exponenets this gives you -x3+6x2+9x-54 so we know it changes 2 times, so we haev 2 real negatives 0 but since the number is 2 we have to subtract to from it (when u have a number higher than 2 you always have to subtract 2 from it, till u get the smallest numer which is 0) so we know we have 2 or 0 real negative zeros, now you have to check them by getting the numbers u multiply to get teh first and last digit, well teh first on is 1 so it can only be multiplied by that same thing 1 but the last one is 54, 54 can be muliplied by 1,2,3,6,9,18,27, and 54 so now u have to try and check those out to see which ones when theyare divide give you 0
Lets start with 1
E.J. 1| 1 6 -9 -54
        ___1_7_-2
          1  7  -2  -56
This didn't work so keep going   I'll just go ahead and try 3,-3, adn -6

3| 1  6  -9  -54
   __ 3_27_54
   1   9  18  0
This one works

-3| 1  6  -9  -54
    __-3_-9__54
     1  3   -18   0
this works

-6|  1  6  -9  -54
    ___-6_0__54
      1   0  -9   0

this works, so all your zeros will be 3,-3, and -6

Hope this helps

Just a second zzz