# Finding all the Zeros of a Polynomial Equation (already solved for two zeros)

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• Jun 19th 2013, 06:23 PM
Odonsky
Finding all the Zeros of a Polynomial Equation (already solved for two zeros)
Hi everyone,

I cannot seem to figure out how derive the last two zeros of the equation.

Directions read 'Solve' : 3x^4 - 4x^3 + 23x^2 - 36x - 36 = 0, given that one zero is 2/3

Using the given zero, I plugged -2/3 into synthetic division which resulted in a remainder of 0, so I changed the equation to:

(x+2/3)(3x^3 - 6x^2+27x-54) = 0

Next I randomly chose a factor of 2 and carried out another synthetic division, resulting in a remainder of 0 again. Now the equation looks like:

(x+2/3)(x-2)(3x^2 + 0x + 27) = 0

**This is where I am not quite sure how to solve for remaining zeros.

Do I solve the unsolved portion by pulling out a factor, making it 3(x+3)(x+3)? That would leave me with two zeros of -3...

If anyone could advise I would greatly appreciate it.

Thank you.
• Jun 19th 2013, 08:04 PM
Soroban
Re: Finding all the Zeros of a Polynomial Equation (already solved for two zeros)
Hello, Odonsky!

There's something wrong with the problem . . . $\displaystyle \tfrac{2}{3}$ is not a zero of the equation!

Quote:

$\displaystyle \text{Solve: }\:3x^4 - 4x^3 + 23x^2 - 36x - 36 \:=\: 0,\:\text{ given that one zero is }{\color{red}-}\tfrac{2}{3}$

I divided by $\displaystyle (3x+2)$ and got:
. . $\displaystyle f(x) \:=\:(3x+2)(x^3 - 2x^2 + 9x - 18)$

Like you, I saw that $\displaystyle x=2$ is a zero and divided by $\displaystyle (x-2).$
. . $\displaystyle f(x) \:=\:(3x+2)(x-2)(x^2 + 9)$

That last factor cannot be factored (in real numbers).
. . $\displaystyle x^2+9\:=\:0 \quad\Rightarrow\quad x^2 \:=\:-9 \quad\Rightarrow\quad x \:=\:\pm\sqrt{-9} \:=\:\pm3i$

The four zeros are: .$\displaystyle -\tfrac{2}{3},\;2,\;\pm3i$
• Jun 20th 2013, 11:03 AM
Odonsky
Re: Finding all the Zeros of a Polynomial Equation (already solved for two zeros)
Thank you Soroban! Your reply is very clear.

After posting this thread, I worked the latter half of the problem out using the quadratic formula whereas a=3, b=0 & c=27 and arrived at the same solution of +/- 3i.

Interesting how math works in so many different ways...
• Jun 20th 2013, 01:38 PM
HallsofIvy
Re: Finding all the Zeros of a Polynomial Equation (already solved for two zeros)
So it doesn't bother you that the whole problem is wrong?
• Jun 20th 2013, 02:23 PM
Odonsky
Re: Finding all the Zeros of a Polynomial Equation (already solved for two zeros)
From what I understand the problem has been solved correctly.