Pagen13
  • Pagen13
Write a polynomial function using the given roots. 3+i, 2, and -4 (I need help understanding and helping getting the answer)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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triciaal
  • triciaal
example if the roots is -4 then (x+4) is a factor if (3 + i) is a root then -(3 + i) is also a root when you multiply factors then you get the original polynomial maybe I should start by saying when you set the polynomial equal to zero and solve then you get the roots. let me know if you have any questions.
Pagen13
  • Pagen13
So wait step by step how do I get the answer?
triciaal
  • triciaal
|dw:1449810515894:dw|

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triciaal
  • triciaal
combine like terms to simplify polynomial in my example x^3 -6 x^2 -x + 30
Pagen13
  • Pagen13
so (3+i)(x+2)(x-4)
triciaal
  • triciaal
check your signs and look again for the complex root
triciaal
  • triciaal
|dw:1449810917398:dw|
Pagen13
  • Pagen13
I am confused
triciaal
  • triciaal
let me try to explain again when you have a polynomial and set it equal to zero means when factorized there is no remainder. to multiply factors and get zero means at least one factor has a value of zero.
triciaal
  • triciaal
when you have a factor and set = 0 you solve to get the root. here work backwards from the root to the polynomial
triciaal
  • triciaal
do you understand more now?
triciaal
  • triciaal
@dan815, maybe you can explain this better
whpalmer4
  • whpalmer4
Actually, @triciaal you made a small mistake: if \(3+i\) is a root, the conjugate root is \(3-i\), not \(-(3+i) = -3 -i\)
whpalmer4
  • whpalmer4
For a polynomial with only real coefficients, any complex roots will come in conjugate pairs of the form \(a \pm bi\) where \(i = \sqrt{-1}\)
whpalmer4
  • whpalmer4
We can write any polynomial with known roots \(r_1, r_2, ... r_n\) in factored form as \[P(x) = a(x-r_1)(x-r_2)...(x-r_n)\]\(a\) is just a constant multiplier to get the polynomial with a given set of roots to pass through an arbitrary point (there are an infinite number of polynomials with the same roots — take a moment to convince yourself of that if you aren't sure). You have roots of \(r_1=3+i,\ r_3=2,\ r_4=-4\) That first root is a complex one, so we will also have its conjugate root \(r_2 = 3-i\) We can write the polynomial now: \[P(x) = a(x-r_1)(x-r_2)(x-r_3)(x-r_4)\]\[=a(x-(3+i))(x-(3-i))(x-2)(x-(-4))\]\[=a(x-3-i)(x-3+i)(x-2)(x+4)\]Now you just expand that, and choose (or find) a suitable value of \(a\) if you need the curve to go through a specific point, otherwise \(a=1\). The ugly part of that algebra might be the complex roots: \[(x-3-i)(x-3+i)=x*x-3*x+i*x -3*x -3(-3)\]\[\ \ \ -3*i-i*x-i(-3)-i+i = x^2-6x+9-i^2\]but \(i^2 = (\sqrt{-1})^2 = -1\) so that becomes \[x^2-6x+9-(-1) = x^2-6x+10\] So \[P(x) = a(x^2-6x+10)(x-2)(x+4)\]and you can do the rest of the expansion...just a bunch of somewhat tedious algebra!

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