# Find all the Zeros

• Nov 13th 2009, 04:04 PM
VNVeteran
Find 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, 05:08 PM
apcalculus
Quote:

Originally Posted by VNVeteran
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)

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, 08:53 AM
VNVeteran
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, 09:49 AM
earboth
Quote:

Originally Posted by VNVeteran
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.

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...
• Nov 14th 2009, 09:56 AM
VNVeteran
Quote:

Originally Posted by earboth
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...

ok so this is all the rational zeros i found.

±1,±1/2,±1/3,±1/6,±2,±2/3,±3,±3/2,±5, ±5/2,±5/3,±5/6,±6,±9,±9/2,±10,±10/3,±15,±15/2,±18,±30,±45,±45/2,±90

so i should plug all these in one by one like you showed me and then so synthetic division for all of them?????
• Nov 14th 2009, 11:01 AM
11rdc11
You could try finding what your lower and upper bounds are so you don't have to try so many solutions.
• Nov 14th 2009, 11:26 AM
Soroban
Hello, VNVeteran!

Quote:

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"?

We have: .$\displaystyle f(x) \:=\:3(2x^4 - 9x^3 - 18x^2 + 71x - 30)$

. . $\displaystyle \begin{array}{ccccccc} 2 & | & 2 & -9 & -18 & +71 & -30 \\ & | & & +4 & -10 & -56 & +30 \\ & & -- & -- & -- & -- & -- \\ && 2 & -5 & -28 & 15 & 0 \end{array}$

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: .$\displaystyle x \;=\;2,\:\text{-}3,\:5,\:\tfrac{1}{2}$

• Nov 14th 2009, 01:25 PM
VNVeteran
Quote:

Originally Posted by Soroban
Hello, VNVeteran!

We have: .$\displaystyle f(x) \:=\:3(2x^4 - 9x^3 - 18x^2 + 71x - 30)$

. . $\displaystyle \begin{array}{ccccccc} 2 & | & 2 & -9 & -18 & +71 & -30 \\ & | & & +4 & -10 & -56 & +30 \\ & & -- & -- & -- & -- & -- \\ && 2 & -5 & -28 & 15 & 0 \end{array}$

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: .$\displaystyle x \;=\;2,\:\text{-}3,\:5,\:\tfrac{1}{2}$

ok i have a question. from all the ratioanl zeros i found (±1,±1/2,±1/3,±1/6,±2,±2/3,±3,±3/2,±5, ±5/2,±5/3,±5/6,±6,±9,±9/2,±10,±10/3,±15,±15/2,±18,±30,±45,±45/2,±90)

how did you determine to choose the correct ones??? thats what i dont understand
• Nov 14th 2009, 02:23 PM
topsquark
Quote:

Originally Posted by VNVeteran
ok i have a question. from all the ratioanl zeros i found (±1,±1/2,±1/3,±1/6,±2,±2/3,±3,±3/2,±5, ±5/2,±5/3,±5/6,±6,±9,±9/2,±10,±10/3,±15,±15/2,±18,±30,±45,±45/2,±90)

how did you determine to choose the correct ones??? thats what i dont understand

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, 04:45 PM
11rdc11
Quote:

Originally Posted by topsquark
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