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a) Show that if $P(x)$ is a polynomial such that $P(a)=P'(a)=0$ then there exists a polynomial $Q(x)$ such that $P(x)=(x-a)^2Q(x)$.

b) Show that if $P(x)$ is a quartic polynomial then there exists at most one line $\ell$ that is tangent to the graph of $P(x)$ at two places.

How am I suppose to begin this question? I think I should work backwards from $P(x)=(x-a)^2Q(x),$ but I am not sure how to use that $P(x)$ and $Q(x)$ are polynomials and $P(a)=P'(a)=0$ to do this. I think somehow I need to introduce at $P(a)$ or $P'(a)$ term into that equation, but I am not sure how to do this. Is this correct and the best approach? Please help me figure out what to do next/ how to start.

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  • $\begingroup$ What is a polynomial? What is the degree of P? (I know that you don't know exactly what it is, but you know what it isn't) $\endgroup$ Commented Jan 7, 2020 at 18:29

2 Answers 2

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Two hints:

1 - Consider changing the variable from $x$ to $u=x-a$. So $P(x)=T(u)$. Then you'll see if $T(0)=0$, then it has no independent term. If $T'(0)=0$, then $T(u)$ also does not have a first order term. Can you continue afterwards?

2 - Consider a quartic of the form $P(x) = a x^4+b x^3 + c x^2 + dx +e$, then consider a line of the form $L(x) = kx+m$. Try solving for $L(x)=P(x)$ and $L'(x)=P'(x)$ simultaneously.

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Here's another approach for the first question — using the factor theorem (a special case of the polynomial remainder theorem, a.k.a. Bézout's little theorem). Since $P(a)=0$, the factor theorem tells us that $P(x)=(x-a)T(x)$ for some polynomial $T(x)$. Now take the derivative of both sides of this equation, using the Product Rule for the right-hand side, and then plug in $a$. When you do that, you will see that $T(a)=0$ too, so apply the factor theorem again.

(P.S. For the second part, I have nothing to add to the other answer.)

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