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sorry it went away
before we begin is it clear that two factors are \((x-5)\) and \((x+3)\)?
ok so the real job (before multiplying out) is the find the quadratic polynoimial with zeros at \(-1+2i\) and its conjugate \(-1-2i\)
you want the easy way, or the real real easy way?
real real easy
ok actually lets to the easy way first then the real real easy way
we can work backwards starting with \[x=-1+2i\] add 1 and get \[x+1=2i\] then square (carefully) to get \[(x+1)^2=(-2i)^2\] or \[x^2+2x+1=-4\]
add 4 to both sides and get \[x^2+2x+5=0\] and that is your polynomial
the real real easy way requires memorizing something that if \(a+bi\) is a zero of a quadratic, then it is \[x^2-2ax+(a^2+b^2)\] so in your case \(a=-1,b=2\) and the quadratic is \[x^2-2\times (-1)x+(-1)^2+2^2\] i.e. \[x^2+2x+5\]
those arent any options tho :(
your final job is to multiply \[(x-5)(x+3)(x^2+2x+5)\]
if it was me, i would cheat so as not to screw up the algebra when multiplying
you know how to do that?
wait let me try again
ok i will leave you to it, then show you how to get the answer for sure
yeah the answer i get does not match up with my choices
see if that one does
Oh thank-you! maybe i was multiplying something wrong!
that is why i said "cheat" is is easy to make a mistake when multiplying all this muck out