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this again ready?
i dont know how to do this at all lolol
ok then we better go slow
do you know if \(r\) is a zero, then one factor must be \((x-r)\)? so in this case you have \(4\) is a zero making one factor \((x-4)\)
clear or not? that is really what we need to start this
oh okay so (x-4) and (x+2) ?
we start with \[(x-4)(x+2)\] now we have more work to do
you also need the quadratic that has the zeros of \(-1+2i\) and it "conjugate" \(-1-2i\) next job is to find that
any ideas? "no" is a fine answer
no i dont know
ok there is a hard way, an easy way, and a really really easy way lets do the easy way first, then the real easy way
work backwards starting with \[x=-1+2i\] add \(1\) to get \[x+1=2i\]
then square both sides (carefully) to get \[(x+1)^2=(2i)^2\] or \[x^2+2x+1=-4\]
add 4 to both sides to finish with \[x^2+2x+5\]
final job is to multiply \[(x-4)(x+2)(x^2+2x+5)\]i would cheat so as not to screw up the algebra
and would that be the answer? or is there more
OOO got it. thank you
yeah that is it