# Find two more zeros of P(z)=2z^3+9z^2+14z+5 if -2+i is a zero.

• Feb 4th 2011, 08:13 AM
anees7112
Find two more zeros of P(z)=2z^3+9z^2+14z+5 if -2+i is a zero.
Hello.

Find the other two zeroes of P(z) if -2+i is a zero or p(z)=2z^3+9z^2+14z+5.

My attempt:

-2+i is a zero, so -2-i is also a zero.

P(z)=2z^3+9z^2+14z+5
.....=(z-[-2-i])(z-[-2-i])(z-z3)
.....=((z+2)^2+1)(z-z3)
.....=(z^2+4z+5)(z-z3)

5*-z3= -5z3 (I do NOT mean -15z here)
......5=-5z3
.....-1=z3

Two other Zeroes are -2-i , -1

However, the correct answer given in the book is
-2-i and -0.5

Thank you :)
• Feb 4th 2011, 08:35 AM
masters
Quote:

Originally Posted by anees7112
Hello.

Find the other two zeroes of P(z) if -2+i is a zero or p(z)=2z^3+9z^2+14z+5.

My attempt:

-2+i is a zero, so -2-i is also a zero.

P(z)=2z^3+9z^2+14z+5
.....=(z-[-2-i])(z-[-2-i])(z-z3)
.....=((z+2)^2+1)(z-z3)
.....=(z^2+4z+5)(z-z3)

5*-z3= -5z3 (I do NOT mean -15z here)
......5=-5z3
.....-1=z3

Two other Zeroes are -2-i , -1

However, the correct answer given in the book is
-2-i and -0.5

Thank you :)

Hi anees7112,

I'm not at all sure what you're attempting to do with your approach.

You are correct that $2\:-\:i$ is also a factor (Complex conjugants)

to find any other zeros, I would use the Rational Roots Theorem and test

$\pm 1\,\,\pm 5\,\ \pm\frac{1}{2}\,\pm \frac{5}{2}$

You should discover that $-\frac{1}{2}$ is a zero.
• Feb 4th 2011, 03:10 PM
anees7112
Quote:

Originally Posted by masters
Hi anees7112,

I'm not at all sure what you're attempting to do with your approach.

You are correct that $2\:-\:i$ is also a factor (Complex conjugants)

to find any other zeros, I would use the Rational Roots Theorem and test

$\pm 1\,\,\pm 5\,\ \pm\frac{1}{2}\,\pm \frac{5}{2}$

You should discover that $-\frac{1}{2}$ is a zero.

Hello masters.

I will show a worked solution from the textbook which i tried to follow.

http://img153.imageshack.us/img153/4486/imgjk.jpg

And thank you , I did not know about the rational roots theorem , but googled it after you mentioned it and it helped very much. But if possible could you also show me how to do the the way the the books does it.
• Feb 4th 2011, 04:04 PM
skeeter
note the coefficient of $z^3$ in the original polynomial ...

$2(z^3 + 4.5z^2 + 7z + 2.5) = 0$

$z^3 + 4.5z^2 + 7z + 2.5 = 0$
• Feb 4th 2011, 04:14 PM
Soroban
Hello, anees7112!

Quote:

$\text{Find the other two zeroes of}\,P(z) \:=\:2z^3 + 9z^2 + 14z + 5$

$\text{if }\text{-}2+i\text{ is a zero.}$

If $\text{-}2 + i$ is a zero, then $\boxed{\text{-}2 - i}$ is also a zero.

$\text{Then: }\:z - (\text{-}2+i) \,=\,z + 2 - i\text{ is a factor of }P(z),$
. $\text{ and: }\:z - (\text{-}2-i) \,=\,z + 2 + i \text{ is a factor of }P(z).$

$\text{Hence: }\,(z+2-i)(z+2+i) \,=\,z^2 + 4z + 5\text{ is a factor of }P(z).$

$\text{We find that: }\:P(z) \;=\;(z^2 + 4z + 5)(2z + 1)$

$\text{The third zero is: }\,\boxed{\text{-}\tfrac{1}{2}}$

• Feb 4th 2011, 04:41 PM
anees7112
Thank you very much Soroban :) I understand how to do it now!
Thanks also masters and skeeter :D

Have a great day!
• Feb 4th 2011, 04:44 PM
Quote:

Originally Posted by anees7112
Hello.

Find the other two zeroes of P(z) if -2+i is a zero or p(z)=2z^3+9z^2+14z+5.

My attempt:

-2+i is a zero, so -2-i is also a zero.

P(z)=2z^3+9z^2+14z+5
.....=(z-[-2-i])(z-[-2-i])(z-z3)

Your error is in the above line
$z(z)(2z)=2z^3$
so your third factor should be

$2z-2z_3$

The remainder of your text is corrected from here

.....=((z+2)^2+1)(2z-2z3)
.....=(z^2+4z+5)(2z-2z3)

5*(-2z3)= -10z3
......5=-10z3
.....-0.5=z3

Two other Zeroes are -2-i , -0.5

Thank you :)

$[z-(-2-i)][z-(2-i)][2z-2z_3]=2z^3+9z^2+14z+5$
• Feb 4th 2011, 05:46 PM
anees7112
Quote:

$[z-(-2-i)][z-(2-i)][2z-2z_3]=2z^3+9z^2+14z+5$

Great, the method I was looking for :) Thank you very much Archie Meade! It i nice to know how the textbook solved it.
• Feb 5th 2011, 06:07 AM
HallsofIvy
Quote:

Originally Posted by anees7112
Hello.

Find the other two zeroes of P(z) if -2+i is a zero or p(z)=2z^3+9z^2+14z+5.

My attempt:

-2+i is a zero, so -2-i is also a zero.

P(z)=2z^3+9z^2+14z+5
.....=(z-[-2-i])(z-[-2-i])(z-z3)

At this point you have dropped the "2" multiplying z^3.

Quote:

.....=((z+2)^2+1)(z-z3)
You should have "= 2((z+2)^2+ 1)(z- z3)".

Quote:

.....=(z^2+4z+5)(z-z3)

5*-z3= -5z3 (I do NOT mean -15z here)
......5=-5z3
.....-1=z3
2*5*-z3= -10z3= 5 so z3= -1/2.

Quote:

Two other Zeroes are -2-i , -1

However, the correct answer given in the book is
-2-i and -0.5

Thank you :)