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5

You have found that $p(x)$ is a rational function. It is zero if and only if its numerator is zero. Its numerator is a polynomial of the form $Ax^{n+2}+Bx^{n+1}+C$, where $A,B,C$ are constants. The derivative of that numerator is of the form $Dx^{n+1}+Ex^n=x^n(Dx+E)$ for some constants $D,E$. It's easy to see how many real roots that derivative has. Then ...

5

Answer to the title question: Yes! There exists such a function. For example, consider the function $$f(x)=\frac{d(x,A)}{d(x,A)+d(x,B)}$$ where $A$ is the fat cantor set in $[0,1]$ and $B$ is any closed singleton set $\{b\}$ where $b \notin A$. Then $f$ is continuous whose zero set is $A$, which is nowhere dense in $[0,1]$ of positive measure! Edit: The ...

3

One way to see $a+2$ is an eigenvalue is that $$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}a+2\\a+2\\a+2\end{bmatrix}.$$ Then you can use the fact that $x-(a+2)$ divides the characteristic polynomial. More generally: if all the rows of $A$ add up to $\lambda$, then $\lambda$ is an eigenvalue.

3

use the following way $$x=\sqrt{5+\sqrt{5 + x} }$$ $$x=\sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + ....} }} }} }$$ or $$x=\sqrt{5+x }$$ $$x^2-x-5=0$$ $$x=\frac{1}{2}\pm\frac{\sqrt{21}}{2}$$ now use long division to get the other roots and then check which which one satisfies the original equation

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Note that if $x = \sqrt{x+5}$ then $x = \sqrt{\sqrt{x+5}+5}$. So, try solving $x = \sqrt{x+5}$. This is a quadratic.

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Hint: Both the real and imaginary parts of $P(ki)$ are $0$. Consider the real part and use (a).

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Hint: try to use the triangle inequality. If $|z|\leq 1$, then what can you say about $|z^3+3z|$?

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We know that $ki$ is a solution to $P(x)=0$ so in particular, $$\begin{split} (ki)^4+a(ki)^3+b(ki)^2+c(ki)+d&=0 \\ \Leftrightarrow k^4-ak^3i-bk^2+cki+d&=0 \\ \Leftrightarrow k^4-bk^2+d+ki(c-ak^2)&=0 \end{split}$$ Using part (a), this gives $k^4-bk^2+d=0$. Now add $bk^2$ on both sides and multiply through with $a^2$ to get $$a^2k^4+a^2d=a^2bk^2=... 1 Basically, you need to solve (a-\lambda)^3-3(a-\lambda)+2 =0 for \lambda. Don't expand the brackets, instead denote: t=a-\lambda. Then:$$t^3-3t+2=0 \Rightarrow (t-1)^2(t+2)=0 \Rightarrow \\ t_1=1 \Rightarrow a-\lambda =1 \Rightarrow \lambda_1 =a-1\\ t_2=-2\Rightarrow a-\lambda =-2 \Rightarrow \lambda_2=a+2.$$1 Hint: if I denotes the identity matrix, then the eigenvalues of A+cI are easily obtained from the eigenvalues of A:$$ (A+cI)v=\lambda v \iff Av=(\lambda-c)v $$What if you take c=1-a? 1 There is no guarantee that there are "10 roots with the smallest real part". That is, it is quite possible that there is an infinite sequence of roots with real parts positive but approaching 0, and no roots with real part 0. 1 You can take p(x)=x^5-4x-2. It is irreducible in \mathbb Q[x], by Eisenstein's criterion. And it is easy to deduce from the fact that p'(x)=5x^4-4 and from the intermediate value theorem that it has 3 and only 3 real roots. 1$$ p(x)=1+2x+3x^2+\ldots+(n+1)x^n;\\ xp(x)=x+2x^2+3x^3+\ldots+(n+1)x^{n+1};\\ (1-x)p(x)=1+x+x^2+\ldots+x^n-(n+1)x^{n+1}=\frac{1-x^{n+1}}{1-x}-(n+1)x^{n+1}\\ =\frac{1-x^{n+1}-(1-x)(n+1)x^{n+1}}{1-x}=0.\\ \Rightarrow1-x^{n+1}-(1-x)(n+1)x^{n+1}=0  Note that $x=1$ is clearly not a solution of the original equation. As a result, the above equation is actually ...

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