# Complex Zeros in polynomials

• Jul 10th 2009, 09:08 PM
Ife
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. (Happy)
• Jul 10th 2009, 10:13 PM
red_dog
If \$\displaystyle x_1=3-2i\$ then another root is \$\displaystyle x_2=3+2i\$.

Now divide the polynomial by \$\displaystyle (x-x_1)(x-x_2)=x^2-6x+13\$.
• Jul 10th 2009, 10:28 PM
Ife
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)...
• Jul 10th 2009, 10:54 PM
yeongil
You can do synthetic division by complex roots. It's a little tricky, but it can be done.

First, divide \$\displaystyle 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
\$\displaystyle x^2 - 3x - 10 = 0\$
which factors into
\$\displaystyle (x + 2)(x - 5) = 0\$

The two remaining roots are -2 and 5.

01
• Jul 10th 2009, 10:54 PM
red_dog
For long division see this: Polynomial Long Division

To write mathematical symbols see Latex Tutorial in Latex help section on this forum.
• Jul 10th 2009, 11:18 PM
Ife
Quote:

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

First, divide \$\displaystyle 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
\$\displaystyle x^2 - 3x - 10 = 0\$
which factors into
\$\displaystyle (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 (Clapping)