anonymous
  • anonymous
Write a polynomial with rational coefficients having roots 3, 3+i, and 3-i. Part 1: Write the factors (in the form x-a) that are associated with the roots (a) given in the problem. Part 2: Multiply the 2 factors with complex terms to produce a quadratic expression. Part 3: Multiply the quadratic expression you just found by the 1 remaining factor to find the resulting cubic polynomial.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
nah don' t do that
anonymous
  • anonymous
\[(x-(3+i))(x-(3-i))\] is a pain to multiply there is a much easier wy
anonymous
  • anonymous
you want a quadratic with zeros \(3+i\) and \(3-i\) there is an easy way to find it also a real real easy way which would you like to use?

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anonymous
  • anonymous
Well I am pretty sure I can find the easier way, but the packet I'm doing requires me to do it the hard way, unfortunately. Can you help me with the harder way?
anonymous
  • anonymous
what difference does it make? you have to end up with a quadratic, might as well just write it
anonymous
  • anonymous
Well each of the steps is graded separately as a different question. So if I skip the first two steps (or do them differently) then I get points taken off.
anonymous
  • anonymous
if the zeros are \(a\pm bi\) the quardratic is \[x^2-2ax+(a^2+b^2)\]
anonymous
  • anonymous
ok then lets multiply
anonymous
  • anonymous
So how do I find my factors for this problem?
anonymous
  • anonymous
if a root is \(r\) then a factor is \((x-r)\)
anonymous
  • anonymous
your roots are 3, 3 + i, 3 - i so your factors are \[(x-3)(x-(3+i))(x-(3-i))\]
anonymous
  • anonymous
Okay yes that's what I got
anonymous
  • anonymous
that is the answer to A
anonymous
  • anonymous
for B they want you to multiply \[(x-(3+i))(x-(3-i))\]
anonymous
  • anonymous
So the complex roots only? I see.
anonymous
  • anonymous
that is what is says right?
anonymous
  • anonymous
Yep, thanks.
anonymous
  • anonymous
Do I multiply this by FOIL?
anonymous
  • anonymous
i wouldn't
anonymous
  • anonymous
well ok i guess, but keep \(3+i\) and \(3-i\) together lets do the first outer inner last business, but to it the easy way
anonymous
  • anonymous
so don't distribute the x to the roots?
anonymous
  • anonymous
first is a no brainer, it is \(x^2\)
anonymous
  • anonymous
lets skip to the last when you multiply a complex number \(a+bi\) by its complex conjugate \(a-bi\) you get the real number \(a^2+b^2\) to the "last" is \[3^2+1^2=10\]
anonymous
  • anonymous
Oh okay I've never seen that before.
anonymous
  • anonymous
then "outer and inner"\[-(3+i)x-(3-i)x\]combine like terms, the \(i\) part goes bye bye and you are left with \(-6x\)
anonymous
  • anonymous
So did you distribute the x first, then subtract?
anonymous
  • anonymous
if you haven't seen that then you should learn it now \[(a+bi)(a-bi)=a^2+abi-abi-b^2i^2=a^2+b^2\]
anonymous
  • anonymous
yes i would call it "distribute then combine like terms" but it is the same thing
anonymous
  • anonymous
Okay thanks. Okay and so the last part is to multiply the quadratic we found by x-3?
anonymous
  • anonymous
final answer \[x^2-6x+10\] as promised
anonymous
  • anonymous
yes i will let you do that yourself, it is easy enough right?
anonymous
  • anonymous
Yeah it'll take me a while because I'm rusty but I can do it. Thank you so much!
anonymous
  • anonymous
yw

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