I cannot seem to figure out how derive the last two zeros of the equation.
Directions read 'Solve' : 3x^4 - 4x^3 + 23x^2 - 36x - 36 = 0, given that one zero is 2/3
Using the given zero, I plugged -2/3 into synthetic division which resulted in a remainder of 0, so I changed the equation to:
(x+2/3)(3x^3 - 6x^2+27x-54) = 0
Next I randomly chose a factor of 2 and carried out another synthetic division, resulting in a remainder of 0 again. Now the equation looks like:
(x+2/3)(x-2)(3x^2 + 0x + 27) = 0
**This is where I am not quite sure how to solve for remaining zeros.
Do I solve the unsolved portion by pulling out a factor, making it 3(x+3)(x+3)? That would leave me with two zeros of -3...
If anyone could advise I would greatly appreciate it.