find all the zeros of this function: f(x)=6x^4-27x^3-54x^2+213x-90.
By synthetic substitution, show work.
im in desperate help for this one.
Try plugging small numbers like -3, -2, -1, 1, 2, 3. If any of them is a zero, then one of the factors is known.
For your examples, check to see that x=-3 and x=2 are zeros, which means (x-2) and (x+3) are factors.
You can either do synthetic division of 6x^4-27x^3-54x^2+213x-90 by (x-2).
Then take the quotient and divide it synthetically by x+3.
The quotient at this stage will be a quadratic function, which should be easy to factor.
OR
...perform long division by (x-2)(x+3).
I hope this helps. Good luck!!
Spoiler: http://www.wolframalpha.com/input/?i=6x^4-27x^3-54x^2%2B213x-90%3D0+solve
The rational roots theorem only gives the possible rational zeros. In order to find out which are good you can only guess. (However there are a few theorems that can help you cut down on the possible answers. I can't find my text right now, but there is a theorem that will give you the largest possible and smallest possible zero. But as I said the process is really guess-work.)
-Dan
Yep your talking about the same thing I am, the lower and upper bound theorem.
Upper bound
You divide f(x) by x-b (where b > 0) using synthetic division. If the last row containing the quotient and remainder has no negative numbers, then b is an upper bound for the real roots of f(x) = 0
Lower bound
Divide f(x) by x-a (where a <0) using synthetic division. If the last row containing the quotient and remainder has numbers that alternate in sign (zero entries count as positive or negative), then a is a lower bound for the real roots of f(x) = 0