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 one year ago
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
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!

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KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.0Rule #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?

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.0Great. 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 \(i^2=1\).

mathstudent55
 one year ago
Best 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.

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

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.0Also, @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.
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