# Thread: Find all the Zeros

1. ## 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.

2. 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.

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

3. 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.

4. 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:

$f(2)=6\cdot 2^4-27\cdot 2^3-54\cdot 2^2+213\cdot 2-90$

$f(2)=6\cdot 16-27\cdot 8-54\cdot 4+213\cdot 2-90$

$f(2)=96-216-216+426-90=0$

Now perform synthetic division by (x - 2).

You'll get: $\dfrac{f(x)}{x-2}=6x^3-15x^2-84x+45$

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...

5. Originally Posted by earboth
No.

As apcalculus has suggested take x = 2. Your term becomes:

$f(2)=6\cdot 2^4-27\cdot 2^3-54\cdot 2^2+213\cdot 2-90$

$f(2)=6\cdot 16-27\cdot 8-54\cdot 4+213\cdot 2-90$

$f(2)=96-216-216+426-90=0$

Now perform synthetic division by (x - 2).

You'll get: $\dfrac{f(x)}{x-2}=6x^3-15x^2-84x+45$

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?????

6. You could try finding what your lower and upper bounds are so you don't have to try so many solutions.

7. Hello, VNVeteran!

Find all the zeros of this function: . $f(x) \:=\:6x^4-27x^3-54x^2+213x-90$
. . by synthetic substitution, show work.
Do you mean "synthetic division"?
We have: . $f(x) \:=\:3(2x^4 - 9x^3 - 18x^2 + 71x - 30)$

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

We have: . $f(x) \;=\;3(x-2)(2x^3 - 5x^2 - 29x + 15)$

. . $\begin{array}{cccccc}
-3 & | & 2 & -5 & -28 & +15 \\
& | & & -6 & +33 & -15 \\
& & -- & -- & -- & -- \\
& & 2 & -11 & +5 & 0 \end{array}$

We have: . $f(x) \;=\;3(x-2)(x+3)(2x^2 - 11x + 5)$

. . Hence: . $f(x) \;=\;3(x-2)(x+3)(x-5)(2x-1)$

Therefore, the zeros are: . $x \;=\;2,\:\text{-}3,\:5,\:\tfrac{1}{2}$

8. Originally Posted by Soroban
Hello, VNVeteran!

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

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

We have: . $f(x) \;=\;3(x-2)(2x^3 - 5x^2 - 29x + 15)$

. . $\begin{array}{cccccc}
-3 & | & 2 & -5 & -28 & +15 \\
& | & & -6 & +33 & -15 \\
& & -- & -- & -- & -- \\
& & 2 & -11 & +5 & 0 \end{array}$

We have: . $f(x) \;=\;3(x-2)(x+3)(2x^2 - 11x + 5)$

. . Hence: . $f(x) \;=\;3(x-2)(x+3)(x-5)(2x-1)$

Therefore, the zeros are: . $x \;=\;2,\:\text{-}3,\:5,\:\tfrac{1}{2}$

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

9. 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

10. 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