#How do you solve this one ?

23 messages · Page 1 of 1 (latest)

sly pecanBOT
winter glade
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ok idk if this helps but i think since all the coefficients of P(x) are real then if it has a nonreal root then the conjugate will also be a root

twin crown
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wheres this from

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lemme try

viral dome
twin crown
viral dome
twin crown
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i'm trying the problem let's see how far i get 😔

viral dome
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Ohh

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Alright

viral dome
twin crown
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assume that P(x) has all real roots

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that don't repeat

viral dome
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Hmm

twin crown
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so you can write P(x) as a product of its factors

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suppose

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P(x)= (x-r1)...(x-rn) cause it is a polynomial of degree n

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k<n

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for the second polynomial to be able to divide P(x), it will need to contain all factors from (x-r1)...(x-rk)

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but we know that P(x) has all real coefficients with a non-zero constant

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so the condition given, that the product of coefficients of the second polynomial must be 0, is not satisfied

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so p(x) having all real roots cannot be tru