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Mathematics 6 Online
OpenStudy (anonymous):

please help question in comments.. medal and fan

OpenStudy (anonymous):

OpenStudy (steve816):

"showing all your working" Wow, that's some atrocious grammar right there!

OpenStudy (anonymous):

@steve816 if you can't help,, be sure to get lost :) this is an examination question of Cambridge paper..

OpenStudy (owlcoffee):

Well, I would suggest you first find the roots of the function: \[p(z)=z^3+6z^2+24z +32\] Then it'll make the work of \(p(z^2)\) much easier.

OpenStudy (anonymous):

yes already done that.. it was a part of the question.. ive got \[z=2 , z=-2+2\sqrt{3}i, z=-2-2\sqrt{3}i\]

OpenStudy (owlcoffee):

Good, that's corect, in a factorized form it would look like: \[p(z)=(z-2)(z^2 + 4z +16)\] And now, if we want to find the roots of \(p(z^2)\) we will just use that form: \[p(z^2)=(z^2-2)(z^4+4z^2+16)\] You can use ruffinis division to factor out that 4th degree polynomial and make it into a 3rd degree.

OpenStudy (anonymous):

okay thank you.. i can do the rest

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