Trying to find another zero of f(x) ?

[AlexHerring]

2 - 3i is a zero of f(x) = x4 - 4x³ + 14x² - 4x + 13.
What's the other one? How did you come to it?

 

[Deleted member 4993]

2 - 3i is a zero of f(x) = x4 - 4x³ + 14x² - 4x + 13.
What's the other one? How did you come to it?

Do you know - what is a complex conjugate?

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[HallsofIvy]

If a+bi is a root of a polyomial with real coefficients, then its complex conjugate, a- bi, is also a root. There are also two other roots. Once you have found the first two roots, you can reduce the fourth degree polynomial to a quadratic and use the quadratic formula to find them.

 

[AlexHerring]

Do you know - what is a complex conjugate?

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

Thank you, but since the zeros (conjugate) occur in pairs, the answer will be 2+3i

 

[Deleted member 4993]

Thank you, but since the zeros (conjugate) occur in pairs, the answer will be 2+3i

Correct....

 

[lookagain]

2 - 3i is a zero of f(x) = x4 - 4x³ + 14x² - 4x + 13.
What's the other one? How did you come to it?

"The other one what?"

There is not "the other one." There are three other zeroes. \(\displaystyle **\)
So the question's intent is incorrectly stated.


A correctly stated problem should be similar to this:


2 - 3i is a complex zero of \(\displaystyle f(x) \ = \ x^4 - 4x^3 + 14x^2 - 4x + 13.\)
What is another zero?



\(\displaystyle **\) It happens that \(\displaystyle -i \ \ and \ \ i \) are another pair of zeroes.

 

[DrPhil]

Thank you, but since the zeros (conjugate) occur in pairs, the answer will be 2+3i

Did they ask you to find the other two roots? if you divide by the product of the two roots you know, you will get a 2nd-order polynomial that is easy to solve..
The divisor you need is
[x - (2-3i)][x - (2+3i)] = . . .