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.(Doh)

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- Nov 13th 2009, 05:04 PMVNVeteranFind all the Zeros
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.(Doh) - Nov 13th 2009, 06:08 PMapcalculus

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 - Nov 14th 2009, 09:53 AMVNVeteran
i did the rational zero test got over 20 possible roots for the function. i plug them back in to test and did not get a value of zero so does this mean there is no solution??? can someone verify please.

- Nov 14th 2009, 10:49 AMearboth
- Nov 14th 2009, 10:56 AMVNVeteran
- Nov 14th 2009, 12:01 PM11rdc11
You could try finding what your lower and upper bounds are so you don't have to try so many solutions.

- Nov 14th 2009, 12:26 PMSoroban
Hello, VNVeteran!

Quote:

Find all the zeros of this function: .

. . by synthetic substitution, show work.

Do you mean "synthetic division"?

. .

We have: .

. .

We have: .

. . Hence: .

Therefore, the zeros are: .

- Nov 14th 2009, 02:25 PMVNVeteran
- Nov 14th 2009, 03:23 PMtopsquark
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 - Nov 14th 2009, 05:45 PM11rdc11

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