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Old 07-26-2007   #1
jessie
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Default Algebra help!!?

find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros:5-i, sqrt24f(x)= ???Thanks, that's what I keep coming up with, but the sqrt24 doesn't check as 0. Or does it? That's the main reason I asked this. Am I missing something?
 
Old 07-26-2007   #2
candycane
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do you mean 5 minus i? as in x=5-i?
 
Old 07-26-2007   #3
Puggy
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(5 - i) , sqrt(24)In order for a polynomial function to have rational coefficients, all complex and radical roots come in conjugate pairs. The conjugate of 5 - i is 5 + i, and the conjugate of sqrt(24) is -sqrt(24). With that said, our zeros are (5 - i) (5 + i) (sqrt(24)) (-sqrt(24))For every zero r, (x - r) is a factor. Thus,f(x) = (x - (5 - i)) (x - (5 + i)) (x - sqrt(24)) (x + sqrt(24))Multiplying this out and simplifying,f(x) = (x - 5 + i) (x - 5 - i) (x - sqrt(24)) (x + sqrt(24))f(x) = ( (x - 5)^2 - i^2 ) ( x^2 - 24 )f(x) = ( (x - 5)^2 - (-1) ) (x^2 - 24)f(x) = ( (x - 5)^2 + 1 ) (x^2 - 24)f(x) = x^2 (x - 5)^2 - 24(x - 5)^2 + x^2 - 24f(x) = x^2 (x^2 - 10x + 25) - 24(x^2 - 10x + 25) + x^2 - 24f(x) = x^4 - 10x^3 + 25x^2 - 24x^2 + 240x - 600 + x^2 - 24f(x) = x^4 - 10x^3 + 2x^2 + 240x - 624
 
Old 07-26-2007   #4
pkrunner
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If some number "r" is a root, then part of the polynomial will be (x-r) because that gives us a root of r (when x=r, then the polynomial is 0).Knowing this, and also knowing that imaginary roots always come in pairs, we arrive at this:3 roots total: 5-i, 5+i, and sqrt 24.(x-(5-i))*(x-(5+i))*(x-sqrt 24)=[x^2-x(5-i)-x(5+i)+(25-i^2)]*(x-sqrt 24)[x^2-5x+xi-5x-xi+(25+1)]*(x-sqrt 24)(x^2-10x+26)*(x-sqrt 24)Because it has rational coefficients, we should toss in (x+sqrt 24) to get rid of the (x-sqrt 24)Difference of two squares gives usx^2-10x+26)*(x^2-24)Multiply that out using distributive property, and you should get your answer if all that is done properly.
 
 

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