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
No.
As apcalculus has suggested take x = 2. Your term becomes:
$\displaystyle f(2)=6\cdot 2^4-27\cdot 2^3-54\cdot 2^2+213\cdot 2-90$
$\displaystyle f(2)=6\cdot 16-27\cdot 8-54\cdot 4+213\cdot 2-90$
$\displaystyle f(2)=96-216-216+426-90=0$
Now perform synthetic division by (x - 2).
You'll get: $\displaystyle \dfrac{f(x)}{x-2}=6x^3-15x^2-84x+45$
Now try another factor of 90, for instance x = 5. If you get zero again you've found another factor which you can cancel out.
and so on...
Hello, VNVeteran!
We have: .$\displaystyle f(x) \:=\:3(2x^4 - 9x^3 - 18x^2 + 71x - 30)$Find all the zeros of this function: .$\displaystyle f(x) \:=\:6x^4-27x^3-54x^2+213x-90$
. . by synthetic substitution, show work.
Do you mean "synthetic division"?
. . $\displaystyle \begin{array}{ccccccc}
2 & | & 2 & -9 & -18 & +71 & -30 \\
& | & & +4 & -10 & -56 & +30 \\
& & -- & -- & -- & -- & -- \\
&& 2 & -5 & -28 & 15 & 0 \end{array}$
We have: .$\displaystyle f(x) \;=\;3(x-2)(2x^3 - 5x^2 - 29x + 15)$
. . $\displaystyle \begin{array}{cccccc}
-3 & | & 2 & -5 & -28 & +15 \\
& | & & -6 & +33 & -15 \\
& & -- & -- & -- & -- \\
& & 2 & -11 & +5 & 0 \end{array}$
We have: .$\displaystyle f(x) \;=\;3(x-2)(x+3)(2x^2 - 11x + 5)$
. . Hence: .$\displaystyle f(x) \;=\;3(x-2)(x+3)(x-5)(2x-1)$
Therefore, the zeros are: .$\displaystyle x \;=\;2,\:\text{-}3,\:5,\:\tfrac{1}{2}$
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