Let p(x) be a polynomial with real coefficients and such that the product of all the roots is negative. Show that if the degree of p(x) is 6, then p(x) has at least one positive root.
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1Do you have any ideas/progress you'd like to share? – jlammy Jan 24 '21 at 14:42
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So basically I reached a standstill because for a polynomial with 6 distinct zeros, it always needs to have an odd number of negative zeros to keep the product negative. But what happens in a degree 6 polynomial like (x+1)(x+1)(x+2)(x+3)(x+4)(x+5) where we have a duplicate zero? Because here product of zeros will be -1-2-3-4-5 = negative and we don't have a positive root. – Louis Jones Jan 24 '21 at 14:45
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1It sounds like you're nearly there. In the case of the duplicate root, we should count that root twice in the product, so $(-1)^2(-2)(-3)(-4)(-5)$ is positive. So repeated roots don't really make a difference. But there's one extra subtlety that you're missing -- who said the polynomial need have six real roots? What about complex ones? – jlammy Jan 24 '21 at 14:48
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Can a complex number be positive? As in is (1 - i) positive or negative? – Louis Jones Jan 24 '21 at 14:50
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Complex numbers are neither positive or negative. But the condition that your polynomial has real coefficients gives you some extra information about the (complex) roots – jlammy Jan 24 '21 at 15:18
1 Answers
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If a polynomial has real coefficients then for any complex root $z$, the conjugate $\bar{z}$, is also a root .
And $z\cdot \bar{z}=|z|^2>0$.
Therefore if the degree is $6$ it can have $0,2,4,6$ real roots, including multiple roots.
If the product of all roots is negative, then real roots can't be all negative: there must be a positive one, at least.
