Author Topic: how to factorise over C (Read 2822 times) Tweet Share

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

max payne · « **Reply #1 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

kensan · « **Reply #2 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

Bhootnike · « **Reply #3 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 $-1+i$

pi · « **Reply #4 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 $-1-i$

Bhootnike · « **Reply #5 on:** April 06, 2012, 09:59:53 pm »

ah yes... lol
my bad! good pick up haha

kensan · « **Reply #6 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)

$a=2$

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

pi · « **Reply #7 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

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Author Topic: how to factorise over C (Read 2822 times) Tweet Share

1i1ii1i

how to factorise over C

max payne

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pi

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