anonymous
  • anonymous
Find a third-degree polynomial with real coefficients and with zeros –3 and –4 + i.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
one factor is \(x+3\) for sure the hard part is finding a quadrtic with zeros \(-4+i\) and\(-4-i\) there are three ways do to it one is hard one is easy one is real real easy you pick
misty1212
  • misty1212
lol pick the hard way \(\color\magenta\heartsuit\)
anonymous
  • anonymous
okay, lol.

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More answers

anonymous
  • anonymous
lets do the easy way first ] work backwards starting with \[x=-4+i\] then add 4 to both sides to get \[x+4=i\] square both sides (carefully) to get \[(x+4)^2=-1\] or \[x^2+8x+16=-1\]
anonymous
  • anonymous
adding 1 gives you the quadratic as \[x^2+8x+17\]
anonymous
  • anonymous
that is the easy wat
misty1212
  • misty1212
i guess that was easy...
anonymous
  • anonymous
the real real easy way requires memorizing something namely, that if \(a+bi\) is a zero, the quadratic is \[x^2-2ax+(a^2+b^2)\] in your case \(a=-4,b=1\) so \[x^2-2(-4)x+(6^2+1^2)\] \[x^2+8x+17\]
anonymous
  • anonymous
typo there, but you get the idea
misty1212
  • misty1212
yeah \[x^2-2(-4)x+((-4)^2+1^2)\]
misty1212
  • misty1212
lot easier than trying \[(x-(-4+i))(x-(-4-i))\]
anonymous
  • anonymous
final job is to multiply \[(x+3)(x^2+8x+17)\]
beginnersmind
  • beginnersmind
\[(x-(-4+i))(x-(-4-i))\] isn't hard either. It's in the form [(x+4)+i] * [(x+4)-i] which equals: (x+4)^2 + 1

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