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jennag
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Can someone help work through this problem with me :) f(x)= x^34x^2+16x64, zero 4i. Use the given zero to find the remaining zeros of f! I'm lost!
 one year ago
 one year ago
jennag Group Title
Can someone help work through this problem with me :) f(x)= x^34x^2+16x64, zero 4i. Use the given zero to find the remaining zeros of f! I'm lost!
 one year ago
 one year ago

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KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Rule #1 of complex roots: If you have one complex root, you have to have another corresponding one. In this example, since you have 4i as a zero, you also have to have 4i. In general, if you have a+bi as a root, you will also have abi as another root. Make sense so far?
 one year ago

jennag Group TitleBest ResponseYou've already chosen the best response.0
Yes I follow that
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Great. So that means that \(x4i\) and \(x+4i\) both divide your polynomial. Notice that \((x4i)(x+4i)=x^24i+4i16i^2=x^2+16\) since \(i^2=1\).
 one year ago

mathstudent55 Group TitleBest ResponseYou've already chosen the best response.0
@KingGeorge is correct about the the roots in a case like this one, where the polynomial has rational coefficinets. In such a case, complex roots always come in complex conjugate pairs, a + bi and a  bi.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
So \(x^2+16\) divides your polynomial. Can you try to find some factor \(x+a\) such that \((x+a)(x^2+16)=x^34x^2+16x64\)?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Also, @mathstudent55 you just need real coefficients for it to work, not necessarily rational. But it should be noted that if your coefficients are complex numbers, that fact will not hold true.
 one year ago
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