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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: 1i1ii1i on April 06, 2012, 09:28:33 am

Title: how to factorise over C
Post by: 1i1ii1i on April 06, 2012, 09:28:33 am: how do you factorize z^4+8z^2+16z+20 over C given that (z-1+3i) is a factor
Title: Re: how to factorise over C
Post by: max payne on April 06, 2012, 10:04:46 am: You know that (z-1+3i) is a factor and that there are only real coefficients so that means that the conjugate (z-1-3i) is also a factor. I would multiply the to known factors and them divide the expression by it to get a quadratic. Then you can easily find the remaining two factors by factorising ,etc
Title: Re: how to factorise over C
Post by: kensan on April 06, 2012, 10:20:19 am: Is the answer $(z^2-2z+10)(z^2-2)$ ?
Just making sure I still know how to do this as well :P
Title: Re: how to factorise over C
Post by: Bhootnike on April 06, 2012, 09:41:23 pm: Quote from: 1i1ii1i on April 06, 2012, 09:28:33 am
how do you factorize z^4+8z^2+16z+20 over C given that (z-1+3i) is a factor

$z^4+8z^2+16z+20$ over C given that $(z-1+3i)$ is a factor

since $(z-1+3i)$ is a factor, then $(z-1-3i)$ is also a factor(root) - conjugate room thereom

From here i'd expand the two factors/roots:

$=(z-1+3i)$ $(z-1-3i)$
$=z^2 -z -3iz -z + 1 +3i +3iz -3i -9i^2$
$=z^2 -2z +10$
now divide this from original expression to find the last 2 roots..

use long division - here im using square root sign cause i cbf finding division latex code haha...

$z^2 -2z +10\sqrt{z^4+8z^2+16z+20}$

thus you should end up .. : $z^2 + 2z +2$

now simply apply whatever method you want to find the 2 roots..

ill do the q'formula :

z= $-2 +- \frac{\sqrt{4-4(1)(2)}}{2}$

ending up with: $\frac{-2+2i}{2}$
and : $\frac{-2-2i}{2}$

edit. which simplifies to $-1-i$ and $-1+i$
Title: Re: how to factorise over C
Post by: pi on April 06, 2012, 09:50:39 pm: Quote from: Bhootnike on April 06, 2012, 09:41:23 pm
Quote from: 1i1ii1i on April 06, 2012, 09:28:33 am
how do you factorize z^4+8z^2+16z+20 over C given that (z-1+3i) is a factor

$z^4+8z^2+16z+20$ over C given that $(z-1+3i)$ is a factor

since $(z-1+3i)$ is a factor, then $(z-1-3i)$ is also a factor(root) - conjugate room thereom

From here i'd expand the two factors/roots:

$=(z-1+3i)$ $(z-1-3i)$
$=z^2 -z -3iz -z + 1 +3i +3iz -3i -9i^2$
$=z^2 -2z +10$
now divide this from original expression to find the last 2 roots..

use long division - here im using square root sign cause i cbf finding division latex code haha...

$z^2 -2z +10\sqrt{z^4+8z^2+16z+20}$

thus you should end up .. : $z^2 + 2z +2$

now simply apply whatever method you want to find the 2 roots..

ill do the q'formula :

z= $-2 +- \frac{\sqrt{4-4(1)(2)}}{2}$

ending up with: $\frac{-2+2i}{2}$
and : $\frac{-2-2i}{2}$

Which further simplifies to:
$-1+i$ and $-1-i$

:P
Title: Re: how to factorise over C
Post by: Bhootnike on April 06, 2012, 09:59:53 pm: ah yes... lol
my bad! good pick up haha
Title: Re: how to factorise over C
Post by: kensan on April 06, 2012, 10:21:38 pm: Ok so what was wrong with my method haha
$(z^2-2z+10)(z^2-a)$

$-a+10=8$ (get co-efficents of terms which give z^2, then let it equal to actual z^2)

$a=2$

$(z^2-2z+10)(z^2-2)$
Title: Re: how to factorise over C
Post by: pi on April 06, 2012, 10:31:14 pm: Quote from: kenoy on April 06, 2012, 10:21:38 pm
Ok so what was wrong with my method haha
$(z^2-2z+10)(z^2-a)$

$-a+10=8$ (get co-efficents of terms which give z^2, then let it equal to actual z^2)

$a=2$

$(z^2-2z+10)(z^2-2)$

That expands to z^4-2z^3+8z^2+4z-20, not z^4+8z^2+16z+2