SOLUTION: "Write the simplest polynomial function with zeros 3-i,square root of 5 and -3." The first step I think is [x-(3-i)][x-(3+i)][x-(sq rt 5)][x-(-sq rt 5)][x-(-3)]. I am not sure wha
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Question 248377: "Write the simplest polynomial function with zeros 3-i,square root of 5 and -3." The first step I think is [x-(3-i)][x-(3+i)][x-(sq rt 5)][x-(-sq rt 5)][x-(-3)]. I am not sure what the next step or solution is.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Your first step is spot on. Now you have to do the drudgery of multiplying all of those expressions together. The only thing that will make this easier is to realize that you have two sets of conjugate pairs -- and the product of a pair of conjugates is the difference of two squares.
First FOIL \right)\left(x\ -\ \left(3\ +\ i\right)\right))
. This looks ugly, but is not such a big deal if you consider the complex numbers as a single number:
\right)\left(x\ -\ \left(3\ +\ i\right)\right)\ =\ x^2\ -\ x(3\ +\ i)\ -\ x(3\ -\ i)\ +\ (3\ -\ i)(3\ +\ i))
For the last part of that above, use the idea of the product of two conjugates being the difference of two squares -- but remember that 
, so:
\ -\ x(3\ -\ i)\ +\ (3\ -\ i)(3\ +\ i)\ =)
\right)\ =\ x^2\ -\ 6x\ +\ 10)
Then you have \left(x\ +\ \sqrt{5}\right))
is just the difference of two squares, so:
\left(x\ +\ \sqrt{5}\right)\ =\ x^2\ -\ 5)
And you now have to deal with the product:
\left(x^2\ -\ 5\right)\left(x\ +\ 3\right))
FOIL the two binomials:
\left(x\ +\ 3\right)\ =\ x^3\ +\ 3x^2\ -\ 5x\ -\ 15)
Finally, you have a trinomial times a four term polynomial:
\left(x^3\ +\ 3x^2\ -\ 5x\ -\ 15\right))
Which multiplies out to:
I'll leave verification of that last step to you. DO NOT trust me. I am a human (an old one at that) and I make mistakes sometimes. Check my work before you turn in your homework.
John
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