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# (a) Show that a polynomial of degree $3$ has at most three real roots.(b) Show that a polynomial of degree $n$ has at most $n$ real roots.

## (a) Suppose that a cubic polynomial $P(x)$ has roots $a_{1}<a_{2}<a_{3}<a_{4},$ so $P\left(a_{1}\right)=P\left(a_{2}\right)=P\left(a_{3}\right)=P\left(a_{4}\right)$By Rolle's Theorem there are numbers $c_{1}, c_{2}, c_{3}$ with $a_{1}<c_{1}<a_{2}, a_{2}<c_{2}<a_{3}$ and $a_{3}<c_{3}<a_{4}$ and$P^{\prime}\left(c_{1}\right)=P^{\prime}\left(c_{2}\right)=P^{\prime}\left(c_{3}\right)=0 .$ Thus, the second-degree polynomial $P^{\prime}(x)$ has three distinct real roots, which isimpossible.(b) We prove by induction that a polynomial of degree $n$ has at most $n$ real roots. This is certainly true for $n=1$. Supposethat the result is true for all polynomials of degree $n$ and let $P(x)$ be a polynomial of degree $n+1$. Suppose that $P(x)$ hasmore than $n+1$ real roots, say $a_{1}<a_{2}<a_{3}<\dots<a_{n+1}<a_{n+2} .$ Then $P\left(a_{1}\right)=P\left(a_{2}\right)=\dots=P\left(a_{n+2}\right)=0$By Rolle's Theorem there are real numbers $c_{1}, \ldots, c_{n+1}$ with $a_{1}<c_{1}<a_{2}, \ldots, a_{n+1}<c_{n+1}<a_{n+2}$ and$P^{\prime}\left(c_{1}\right)=\cdots=P^{\prime}\left(c_{n+1}\right)=0 .$ Thus, the $n$ th degree polynomial $P^{\prime}(x)$ has at least $n+1$ roots. This contradiction showsthat $P(x)$ has at most $n+1$ real roots.

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##### Kristen K.

University of Michigan - Ann Arbor

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it's s o We're being asked to show that upon your degree three hundred at most three little group And the way we can imagine this we have some sort of functioning X Q plus Babalola doesn't really matter and we take the derivative of this function We'LL get something that works like in this case will be to be x squared But it would always leave us with x square because this will be n minus one. That's how you take this by the polynomial the polynomial derivative rule where this will be you bring this down and subtract one And so any quadratic function generally looks something like this And this is you can think about this as an application of Rose Stone. You can assume that these two points these two routes that usually occur have have at most this is the best case scenario. You can assume that this is a this is B and then ah, since f of a is equal to effort be And since a polynomial f A B since upon normal functions always continuous, indefensible honest interval We know data satisfies roast dirham and roasted him states that there will be a sea such that the derivative zero zero and in this case to see is occurring right here and last. I heard of that Attention mine, you know, does not, um that's a better line so that there is a tangent line zero. And since we initially made the assumption that a and B already route and that the F crime tells us another, this gives us three real possible route. And since any problem of degree three of derivative gives us a quadratic function, are part of degree to mean that there's a little three possible route and this is proven by roaster. We just showed that this is the best case scenario. This is at most regroups. You could have zero routes, one roof if the if the polynomial is somewhere out here or like out here, just not even touching the X axis or or in this case, it well. But this is a better one, just as well that can give you. But this will not give you over this one. Not to be rude, but this is the most since the scenario will weaken, have three roofs. And this shows that any partner of degree and hasn't most. And roof because we're using essentially the same logic. And, um, you can test to stop further by doing contradiction so you can assume that that upon only of degree three has forward. And if you try to figure that hour, we tried to think of such a scenario. And I bet you won't, because it is not possible because we just proved that Rose Tim shows that this function of degree three hasn't most regroup. So any partner of degree and can only have at most and real route and the supplies for anything, any number. And I mean any integer number of end. What?

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##### Kristen K.

University of Michigan - Ann Arbor

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