Complex Zeros in polynomials

**Ife** · Jul 10th 2009, 09:08 PM

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.

**red_dog** · Jul 10th 2009, 10:13 PM

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

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

**Ife** · Jul 10th 2009, 10:28 PM

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)...

**yeongil** · Jul 10th 2009, 10:54 PM

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

**red_dog** · Jul 10th 2009, 10:54 PM

For long division see this: Polynomial Long Division

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

**Ife** · Jul 10th 2009, 11:18 PM

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

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