# Finding the remaining zeros of each polynomial function

• Jun 4th 2010, 09:49 AM
RBlax
Finding the remaining zeros of each polynomial function
Use the given zero to find the remaining zero's of each polynomial function.

P(x)= x^3+3x^2+x+3; -i

I'm pretty lost with this stuff. I've been watching Patrick's videos trying to remember my high school days but wow it's been a while hehe.

It looks like Conjugate Pair Theorem would be the way to solve. If I figure it out I'll post my solution.

If you guys could give me some direction to help me get back into the game that would be AWESOME!!

Thanks a TON
• Jun 4th 2010, 10:06 AM
earboth
Quote:

Originally Posted by RBlax
Use the given zero to find the remaining zero's of each polynomial function.

P(x)= x^3+3x^2+x+3; -i

I'm pretty lost with this stuff. I've been watching Patrick's videos trying to remember my high school days but wow it's been a while hehe.

If you guys could give me some direction to help me get back into the game that would be AWESOME!!

Thanks a TON

1. Since there are only real coefficients in the term of the function there must exist a 2nd solution x = +i. Thus $\displaystyle (x^2+1)$ must be a factor in the term:

2. $\displaystyle P(x)= x^3+3x^2+x+3 = x^2(x+3) + (x+3) = (x+3)(x^2+1)$

3. A product equals zero if one factor equals zero. Use this property to determine the last missing zero of P.
• Jun 4th 2010, 10:14 AM
Soroban
Hello, RBlax!

Quote:

Use the given zero to find the remaining zeros.

. . $\displaystyle P(x)\:=\: x^3+3x^2+x+3.\quad x = -i$

Since $\displaystyle x = -i$ is a zero of $\displaystyle P(x)$, then $\displaystyle x = +i$ is also a zero.
. .
(Complex roots always appear in conjugate pairs.)

Then: .$\displaystyle [x - (-i)]\text{ and }[x - (+i)]$ are factors of $\displaystyle P(x).$

. . That is: .$\displaystyle (x + i)(x - i) \:=\:x^2+1$ is a factor of $\displaystyle P(x).$

Dividing, we find that: .$\displaystyle P(x) \:=\:(x^2+1)(x+3)$

Therefore, the zeros of $\displaystyle P(x)$ are: .$\displaystyle i,\:-i,\:-3$

Edit: Too slow . . . again!
.
• Jun 4th 2010, 10:34 AM
RBlax
This isn't using the The Conjugate Pair Theorem though right?

I think what they are wanting me to do looks something like...

$\displaystyle P(x)= x^3+3x^2+x+3; -i$

then use synthetic division (which I begin to become lost)

Edit: I'm slow too haha

But thanks a ton for the help.