Re: maths problem

Guess who wrote:
On Wed, 09 Nov 2005 13:14:47 GMT, "mm" <abc123@xxxxxxxxx> wrote:
Use the given zero to find the remaining zeros of this polynomial function.
p(x)=3x(cubed)-29x(squared)+92x+34; 5+3i
I'm taking an online course and am stuck on this principle. Your help is greatly appreciated. 
Let's cut to the chase:
5 + 3i is a given root.  The polynomial is 3rd degree, so there are
three roots, and 5 - i is another.  Multiply those together and you
get 34.  So the third root must be 1.


So
    3-29+92+34=0.

I don't think so.

You can do it by dividing the original polynomial
by three, so that the coefficient of x^3 is 1, then
using the fact that the product of *minus* the roots
is the constant term, so the third root must be -1/3.

Or you could multiply out (x-(5+3i))(x-(5-3i)) and divide
the original cubic by this quadratic, to get (3x+1),
so that the third root is -1/3.
.

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