# Thread: Complex Zeros in polynomials

1. ## Complex Zeros in polynomials

I need a little help with this please:

Use the given zero to find the remaining zeros of the polynomial:

h(x) = x^4 - 9x^3 + 21x^2 +21x -130; zero: 3-2i

Now i kno all about the zeros when complex existing with their conjugates, i know all about factoring using the different rules and theorems. Had this been a real root, i would've had to divide using long division and testing for roots to factor further, but with this complex one, i am not sure how to do this. Can long division be done on a complex root??? I am not seeing the way of this one. I did one before that had a zero of -2i, but now that the 3 is there i am not sure what to do. Can i get some help please????

Thanks a lot.

2. If $x_1=3-2i$ then another root is $x_2=3+2i$.

Now divide the polynomial by $(x-x_1)(x-x_2)=x^2-6x+13$.

3. ## thanks, but...

ok this i know... but how do i do this division, now? and also, how do you get those well laid out math symbols on this or on the pc on the whole? can you give me a little help on that division? i think i only know how to divide by a simple divisor, one of degree 1 (eg x+1)...

4. You can do synthetic division by complex roots. It's a little tricky, but it can be done.

First, divide $x^4 - 9x^3 + 21x^2 +21x -130$ by 3 - 2i. We know that the remainder will be zero:

Code:
3 - 2i| 1  -9      21     21     -130
-------     3-2i  -22+6i   9+20i  130
--------------------------------
1  -6-2i   -1+6i  30+20i    0
Now, let's divide the quotient by 3 + 2i, the conjugate of 3 - 2i:
Code:
3 + 2i| 1  -6-2i   -1+6i  30+20i
-------     3+2i   -9-6i -30-20i
---------------------------
1  -3     -10          0
So you're left with the quadratic
$x^2 - 3x - 10 = 0$
which factors into
$(x + 2)(x - 5) = 0$

The two remaining roots are -2 and 5.

01

5. For long division see this: Polynomial Long Division

To write mathematical symbols see Latex Tutorial in Latex help section on this forum.

6. Originally Posted by yeongil
You can do synthetic division by complex roots. It's a little tricky, but it can be done.

First, divide $x^4 - 9x^3 + 21x^2 +21x -130$ by 3 - 2i. We know that the remainder will be zero:

Code:
3 - 2i| 1  -9      21     21     -130
-------     3-2i  -22+6i   9+20i  130
--------------------------------
1  -6-2i   -1+6i  30+20i    0
Now, let's divide the quotient by 3 + 2i, the conjugate of 3 - 2i:
Code:
3 + 2i| 1  -6-2i   -1+6i  30+20i
-------     3+2i   -9-6i -30-20i
---------------------------
1  -3     -10          0
So you're left with the quadratic
$x^2 - 3x - 10 = 0$
which factors into
$(x + 2)(x - 5) = 0$

The two remaining roots are -2 and 5.

01
This helps a lot, i thought it was possible, but then i have never seen it done before so i wasn't sure.. thanks a lot! i knew the solution couldn't have been that hard. THanks