# Roots of polynomial equations

• October 26th 2009, 10:46 AM
Roots of polynomial equations
i need help

find the real root of the equation z^3 + z + 10=0 given that one root is 1-2i

since 1-2i is a root its complex conjugate 1+2i is also a root

so far i know the sum of the roots is -1
• October 26th 2009, 12:31 PM
red_dog
No, the sum of all three roots is 0.
• October 26th 2009, 12:44 PM
Logic
I came up with an idea for factoring which renders the initial knowledge of any roots unnecessary unless mine idea is faulty.
• October 26th 2009, 12:58 PM
HallsofIvy
How do you arrive at your very first statement:
$z^3+ z+ 10= 0$ if and only if $(z^3+ z^3)+ (z+ 2)= 0$?

It seems clear to me that z= -2 satisfies the first equation but does NOT satisfy the second.
• October 26th 2009, 06:39 PM
Logic
Quote:

Originally Posted by HallsofIvy
How do you arrive at your very first statement:
$z^3+ z+ 10= 0$ if and only if $(z^3+ z^3)+ (z+ 2)= 0$?

It seems clear to me that z= -2 satisfies the first equation but does NOT satisfy the second.

Well, it is:
$(z^3+ 2^3)+ (z+ 2)= 0$ actually.
Apart from that, I take advantage of the fact that 10 = 2*2*2 + 2.
If this is what you asked about?

It does not satisfy the second, because it is not a solution, I guess?
The quadratic one has complex roots. I think. :?
• October 27th 2009, 12:17 AM
pacman
deadmotor, you said the 2 roots are 1-2i and 1+2i because they are pair conjugate, but notice the equation z^3 + z + 10 = 0, sum of roots is zero.

if a, b, and c are the roots, then a + b + c = 0,

a + ( 1 + 2i) + (1 - 2i) = 0, surely you know the third root, right? which is a . . . why not evaluate it?

-------------------------------------------------

It can be factored by grouping,

z^3 + z + 10 = 0,

z^3 + 2z^2 - 2z^2 - 4z + 4z + z + 10 = 0,

z^2(z + 2) - 2z(z + 2) + 5(z + 2) = 0,

(z + 2)(z^2 - 2z + 5) = 0.

surely you know how to use the quadratic formula to find the rest of the roots . . .
• October 27th 2009, 02:02 AM
harbottle
Another possibility: if 1 + 2i and 1 - 2i are roots then

z - (1 + 2i) and z - (1 - 2i) are factors

So (z-1-2i)(z-1+2i) is a factor

Simply this and you will get a quadratic. then you need one more factor, which should now be obvious (there are not many choices now that you have the 2nd power of z).

From there you should get the third root.