Trisaba
  • Trisaba
make a polynomial where 1. the leading coefficient is one 2. the others are real and rational 3. the zeros are 3i and 2-i
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
ok ready?
Trisaba
  • Trisaba
yes
anonymous
  • anonymous
i agree

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anonymous
  • anonymous
one zero is \(3i\) that makes another zero its conjugate \(-3i\) so the quadratic will be \[(x-3i)(x+3i)=x^2+9\]
anonymous
  • anonymous
another way we could have found \(x^2+9\) is to set \[x=3i\] square both sides and get \[x^2=-9\] then add 9 to get \[x^2+9\] that takes care of the \(3i\) part
Trisaba
  • Trisaba
i forgot about congegents lol
anonymous
  • anonymous
we still have to deal with the \(2+i\) part
anonymous
  • anonymous
there is a hard way to find the quadratic with zeros at \(2\pm i\), an easy way, and a real real easy way you pick
Trisaba
  • Trisaba
r r easy
anonymous
  • anonymous
lets do it the easy way first the real real easy way requires memorizing something (as with most math, the more you know the easier things are)
anonymous
  • anonymous
set \[x=2+i\] and work backwards subtract 2 get \[x-2=i\] then square both sides (carefully) get \[(x-2)^2=i^2\] or \[x^2-4x+4=-1\]
anonymous
  • anonymous
add 1 and get \[x^2-4x+5=0\] so your quadratic is \[x^2-4x+5\]
anonymous
  • anonymous
the real real easy way is to know that if the solution is \(a+bi\) then the quadratic is \[x^2-2ax+a^2+b^2\] in your case \(a=2,b=1\) quadratic is \[x^2-2\times 2x+2^2+1^2=x^2-4x+5\]
anonymous
  • anonymous
you still have one more job, to multiply \[(x^2+9)(x^2-4x+5)\] i will leave that to you
anonymous
  • anonymous
hope it was more or less clear, if you have a question or two let me know
Trisaba
  • Trisaba
thanks
anonymous
  • anonymous
yw

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