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.

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

Hello, Odonsky!

There's something wrong with the problem . . . is not a zero of the equation!

I divided by and got:

. .

Like you, I saw that is a zero and divided by

. .

That last factor cannot be factored (in real numbers).

. .

The four zeros are: .

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...

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?

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

From what I understand the problem has been solved correctly.