anonymous
  • anonymous
Find the remaining zeros of the function: f(x) = x^3 + 3x^2 - 8x + 10 With the zero 1+i I don't think I ever actually did this in pre-calc, could someone give me a pointer on how to start it?
Mathematics
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SOLVED
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katieb
  • katieb
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blockcolder
  • blockcolder
Tip 1: If \(a+bi\) is a zero, then its conjugate \(a-bi\) is also a zero.
anonymous
  • anonymous
True, but what about the other zeros?
blockcolder
  • blockcolder
You know \(f(x)=x^3+3x^2-8x+10=(x-(1+i))(x-(1-i))(x-a)\) for some real number \(a\), since \(1\pm i\) are zeroes of \(f(x)\) You should simplify \((x-(1+i))(x-(1-i))\) first before we move on.

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anonymous
  • anonymous
I know that there are 3 zeros, and two of them are 1-i and 1+i, but I don't know how to get the third one basically. I don't need to simplify the two factors I already know, that's the end result I'm looking for. I just don't know how to get the real number a.
blockcolder
  • blockcolder
If you simplified \((x-(1+i))(x-(1-i))\), you'd see that this is equal to \(x^2-2x+2\), and this turns our problem into \(x^3+3x^2-8x+10=(x^2-2x+2)(x-a)\). A little bit of thinking should help you find the number a. The constant term of the RHS is \(-2a\), and the constant term of the LHS is 10. Thus, \(a=-5\). You should make sure that \((x^2-2x+2)(x+5)\) is really \(x^3+3x^2-8x+10\).
anonymous
  • anonymous
I guess I can't really expect a simple explanation on this site, it's almost like people answer your questions assuming you already know exactly what to do. I've never done anything like this method you're using in my math classes, so it's completely new to me, I don't understand.
blockcolder
  • blockcolder
Do you mean the latest post?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Excuse my moment of retard. I get why \[(x2−2x+2)(x+5)\] is the factored original problem, just not how you got to that. It took me a while to see it, sorry for the minor rage earlier, I've been working on math for the last 6 hours for a summer assignment.
blockcolder
  • blockcolder
The main point about that solution I did is to observe that \(x^3+3x^2-8x+10\) MUST equal \((x^2-2x+2)(x-a)\), and we must choose the correct \(a\) so that this is true. You can expand \((x^2-2x+2)(x-a)=x^3+(-a-2)x^2+(2a+2)x+(-2a)\). To make sure that this last one is equal to \(x^3+3x^2-8x+10\), you can equate coefficients. Coefficients of x^2 are (-a-2) and 3, coefficients of x are 2a+2 and -8, constant terms are -2a and 10. Thus, you can solve -a-2=3, 2a+2=-8, and -2a=10. If you solve these three equations simultaneously, they SHOULD produce the same number. If they don't, then something went wrong.

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