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jennag

  • one year ago

Can someone help work through this problem with me :-) f(x)= x^3-4x^2+16x-64, zero 4i. Use the given zero to find the remaining zeros of f! I'm lost!

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  1. KingGeorge
    • one year ago
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    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 a-bi as another root. Make sense so far?

  2. jennag
    • one year ago
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    Yes I follow that

  3. KingGeorge
    • one year ago
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    Great. So that means that \(x-4i\) and \(x+4i\) both divide your polynomial. Notice that \((x-4i)(x+4i)=x^2-4i+4i-16i^2=x^2+16\) since \(i^2=-1\).

  4. mathstudent55
    • one year ago
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    @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.

  5. KingGeorge
    • one year ago
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    So \(x^2+16\) divides your polynomial. Can you try to find some factor \(x+a\) such that \((x+a)(x^2+16)=x^3-4x^2+16x-64\)?

  6. KingGeorge
    • one year ago
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    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.

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