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anonymous
 one year ago
Find a thirddegree polynomial with real coefficients and with zeros –3 and –4 + i.
anonymous
 one year ago
Find a thirddegree polynomial with real coefficients and with zeros –3 and –4 + i.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0one factor is \(x+3\) for sure the hard part is finding a quadrtic with zeros \(4+i\) and\(4i\) there are three ways do to it one is hard one is easy one is real real easy you pick

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0lol pick the hard way \(\color\magenta\heartsuit\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lets 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
 one year ago
Best ResponseYou've already chosen the best response.0adding 1 gives you the quadratic as \[x^2+8x+17\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is the easy wat

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0i guess that was easy...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the real real easy way requires memorizing something namely, that if \(a+bi\) is a zero, the quadratic is \[x^22ax+(a^2+b^2)\] in your case \(a=4,b=1\) so \[x^22(4)x+(6^2+1^2)\] \[x^2+8x+17\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0typo there, but you get the idea

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0yeah \[x^22(4)x+((4)^2+1^2)\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0lot easier than trying \[(x(4+i))(x(4i))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0final job is to multiply \[(x+3)(x^2+8x+17)\]

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0\[(x(4+i))(x(4i))\] 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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