anonymous
  • anonymous
Using the given zero, find all other zeros of f(x). -2i is a zero of f(x) = x4 - 32x2 - 144
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
if \(2i\) is a zero then so is \(-2i\) therefore one factor is \(x+2i\) and another is \(x-2i\) multiply the factors, get \[x^2+4\]
anonymous
  • anonymous
therefore this sucker factors as \[x^4-32x^2-144=(x^2+4)(\text{something})\]
anonymous
  • anonymous
2i, 12, -12 2i, 6i, -6i 2i, 6, -6 2i, 12i, -12i

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More answers

anonymous
  • anonymous
on the other hand you can solve this directly without knowing any zeros, put \(u=x^2\) and solve \[u^2-32u-144=0\] by factoring
anonymous
  • anonymous
those are my options
anonymous
  • anonymous
once you find \(u\) then replace it by \(x^2\) and solve
anonymous
  • anonymous
you know how to factor this ?
anonymous
  • anonymous
hint, one of the factors is \(x^2+4\)
anonymous
  • anonymous
is the answer a?
anonymous
  • anonymous
probably want me to check?
anonymous
  • anonymous
yes please :)
anonymous
  • anonymous
actually i change my mind probably not
anonymous
  • anonymous
\[x^4-31x-144=0\\ (x^2+4)(x^2+bx+c)=0\]can factor easily how many times does \(4\) go in to \(144\)?
anonymous
  • anonymous
36
anonymous
  • anonymous
actually \(-36\)
anonymous
  • anonymous
so this factors as \[(x^2+4)(x^2-36)=0\]
anonymous
  • anonymous
final job it to solve \[x^2-36=0\] for \(x\)
anonymous
  • anonymous
6, -6
anonymous
  • anonymous
yup
anonymous
  • anonymous
Oh i see what i did wrong now, thanks!
anonymous
  • anonymous
yw

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