Yahoo Answers: Answers and Comments for Math question! About polynomials...? [Mathematics]
Copyright © Yahoo! Inc. All rights reserved.
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
From Anonymous
enUS
Sun, 23 Mar 2008 13:59:15 +0000
3
Yahoo Answers: Answers and Comments for Math question! About polynomials...? [Mathematics]
292
38
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
https://s.yimg.com/zz/combo?images/emaillogous.png

From Moise Gunen: if 42i is zero then 4+2i is a zero
if x=1 ...
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
Sun, 23 Mar 2008 21:08:41 +0000
if 42i is zero then 4+2i is a zero
if x=1 is a double zero
then
your polynomials is
(x1)^2*((x+4)^2+4)=
(x^22x+1)(x^2+8x+20)=
x^4+6x^3+5x^232x+20

From NBL: Well, since the coefficients of the polynomial...
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
Sun, 23 Mar 2008 21:07:54 +0000
Well, since the coefficients of the polynomial are real numbers, then the root 4  2i, has its conjugate pair as a root also, i.e.
4 + 2i
((x  1)^2)(x + 4+2i)(x + 4  2i)
Multiply this out, and you will get a fourth degree polynomial with real integer coefficients.
x^4 + 6x^3 + 5x^2  32x + 20

From Anonymous: The complex roots come in conjugate pairs. So ...
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
Sun, 23 Mar 2008 21:05:04 +0000
The complex roots come in conjugate pairs. So the roots are
42i and 4+2i.
The other roots are 1 and 1.
So the polynomial would looks like
(x(42i)) * (x(4+2i)) * (x1)^2
=x^4+6*x^3+5*x^232*x+20

From Rachel C: 4 isn't a polynomial
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
Sun, 23 Mar 2008 21:08:22 +0000
4 isn't a polynomial

From Anonymous: First, you find all of the zeroes.
(complex n...
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
Sun, 23 Mar 2008 21:08:12 +0000
First, you find all of the zeroes.
(complex numbers have conjugates)
(42i), (4+2i), (+1), (+1)
Then, you make it into the factored form of the polynomial.
(x+42i)(x+4+2i)(x1)(x1)=0
Then you multiply it out.
(x^2+8x+20)(x^22x+1)=0
x^4+6x^3+5x^232x+20=0

From debbie t: #'s are everlasting they never end and a p...
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
Sun, 23 Mar 2008 21:04:32 +0000
#'s are everlasting they never end and a polynomial is in the 4th power an on & on & on 00000000000000000

From Emily K: 4
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
Sun, 23 Mar 2008 21:02:35 +0000
4

From Mike: Do your own homework.
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
Sun, 23 Mar 2008 21:03:03 +0000
Do your own homework.

From Old Man Winter: the answer is: STOP CHEATING do your homewor...
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
Sun, 23 Mar 2008 21:02:41 +0000
the answer is: STOP CHEATING do your homework on your own.

From <3 LOVE <3: 4
http://answers.yahoo.com/question/index?q...
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
https://answers.yahoo.com/question/index?qid=20080323135915AApNvGF
Sun, 23 Mar 2008 21:01:22 +0000
4
http://answers.yahoo.com/question/index?qid=20080323140050AAQkEyg&r=w