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17

Since $(x-1)(x^4+x^3+x^2+x+1)=x^5-1$, the roots are the fifth roots of $1$, excluding $1$. Note that the set of fifth roots of $1$ is a group of prime order, so it is cyclic and any element is a generator. Thus, if $\omega$ is any of the roots of the polynomials, the full set of roots is given by $\omega,\omega^2,\omega^3,\omega^4$. In particular all roots ...

7

If $a$ is one root, then we get $$\frac{x^3+x^2-2x+1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{1}$$ Using the quadratic equation to solve this yields $$\frac{-1-a\pm\sqrt{9-2a-3a^2}}2\tag{2}$$ for the other two roots. After the Question Change Changing the constant term only changes $(1)$ slightly: $$\frac{x^3+x^2-2x-1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{3}$$ and ...

7

HINT: Let $p(x)$ be the polynomial. Note that $$(a+b+c)(a+b+d)(a+c+d)(b+c+d)=(7-d)(7-c)(7-b)(7-a)=p(?)\;.$$

7

By the fundamental theorem of algebra, the polynomial $a_0 + a_1x + \cdots + a_nx^n$ has $n$ (not necessarily distinct) roots, say $\alpha_1,\ldots,\alpha_n$. The polynomial $\frac{a_0}{a_n} + \frac{a_1}{a_n}x + \cdots + x^n$ also has roots $\alpha_1,\ldots,\alpha_n$. Factoring the latter polynomial and expanding, \begin{align*} \frac{a_0}{a_n} + ...

6

It is the product of positive and strictly increasing functions. Alternatively: The derivative of $p$ has degree $n-1$ and at least one root in each interval $(a_i,a_{i+1})$ (Rolle). Hence it has no root besides these, especially it does not change signs beyond $x=a_n$.

5

Observe that $$P_n(z)=(z-\alpha_1)(z-\alpha_2)...(z-\alpha_n)$$ giving $$(\alpha_1+1)(\alpha_2+1)...(\alpha_n+1)=(-1)^nP_n(-1)$$ that is $$(\alpha_1+1)(\alpha_2+1)...(\alpha_n+1)=(-1)^n\left((-1)^n+a_{n-1}(-1)^{n-1}+...-a_1+a_0\right).$$ Hoping this helps.

5

For $a$ and $b$ positive numbers we have \begin{align*} \lim_{x\to 0^+}\frac{\sin \sqrt{ax}}{\sin \sqrt{bx}}=\lim_{x\to 0^+}\frac{\frac{\sqrt{a}\sin \sqrt{ax}}{\sqrt{ax}}}{\frac{\sqrt{b}\sin \sqrt{bx}}{\sqrt{bx}}} &=\frac{\sqrt{a}}{\sqrt{b}}\lim_{x\to 0^+}\frac{\frac{\sin \sqrt{ax}}{\sqrt{ax}}}{\frac{\sin ...

4

You say since the properties of powers are preserved but this is not true, and you have given a proof that exactly the opposite is true: There is no rule that, for $a$ negative, one has $a^{bc} = (a^b)^c$. Indeed, we can see that this is not a rule even without using cube roots: $$-1 = -1^{1/1} = -1^{2/2} =" (-1^2)^{1/2} = \sqrt{1} = 1.$$ This ...

4

The discriminant of the polynomial $p(x)=x^3+x^2-2x-1$ is $49$, which is a perfect square. It has no rational roots, so it is irreducible in $\Bbb{Q}[x]$. Together these facts imply that the Galois group of the polynomial is cyclic of order three. If $a$ is one of its zeros, we thus see that $\Bbb{Q}[a]$ is its splitting field. This means that the other ...

3

Hint: By Vieta, sum of all roots is $0$, and the only real root is $4$. P.S. For a simple argument that there is only one real root, apply Descartes' rule of signs on $x^5-1024$.

3

HINT: Notice, $$x^3+y^3=(x+y)(x^2+y^2-xy)$$ Then, we have $$(x+y)(x^2+y^2-xy)=0$$

3

We know the following to be true: $$\tag1x_1+x_2+x_3+x_4 = -4$$ $$\tag2x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1x_3+x_2x_4 = 6$$ $$\tag3x_1 x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2 = -a$$ $$\tag4x_1x_2x_3x_4 = b$$ Now we can square $(1)$ to get \begin{align} (x_1+x_2+x_3+x_4)^2 = &x^2_1+x^2_2+x^2_3+x^2_4\\ ... 3 The "if" part is clear, so we'll deal with the "only if". Moreover, we take the polynomials to be "true" quadratics ---that is, p and q are non-zero--- since otherwise the proposition is false (as @Winther mentions in a comment to the OP). First, note that quadratics with a common root could have both roots in common, in which case they are ... 3 If the two quadratic polynomials f(x) = px^2 + qx + r and g(x) = qx^2 + rx + p have a common root then this is also a root of the linear polynomialh(x) = qf(x) - pg(x) = (q^2-pr)x - (p^2-qr)$$which means that either q^2-pr=0 and p^2-qr=0 for which p=q=r or that x = \frac{p^2-qr}{q^2-pr} is the common root. Inserting this into either f(x) ... 3 As requested... Suppose W(x) is a real polynomial with n real zeros. We may assume n\ge 2, for otherwise there is nothing to show. For the moment, assume that the roots are distinct - which, I confess (sigh, apologies), I had assumed in the above comment. So, let x_1 < \cdots < x_n, be roots of W, i.e., W(x_i) =0, for 1\le i \le n. ... 2 For n=1,2,3,4 there are general formulas. For all other n, there is no general method for solving such an equation. 2 Notice,$$\sqrt [3]{-1}=(-1)^{1/3}=\left(\cos \pi+i\sin \pi\right)^{1/3}=\left(\cos (2k\pi+\pi)+i\sin (2k\pi+\pi)\right)^{1/3}=\cos \left(\frac{2k\pi+\pi}{3}\right)+i\sin\left(\frac{2k\pi+\pi}{3} \right)$$Now, setting k=0, 1, 2, we will get three roots. as follows setting k=0, we get$$\sqrt [3]{-1}=\cos ...

2

Since $P_n$ is a monic polynomial, $$P_n(z) = \prod_{i=1}^{n}(z-\alpha_i)\tag{1}$$ hence by evaluating at $z=-1$, $$(-1)^n P_n(-1) = \prod_{i=1}^{n}(1+\alpha_i).\tag{2}$$

2

The nonzero powers of an even number are even, the powers of an odd number are odd. Then the value of $f$ for an even argument is the sum of $a_0$ and even terms so that $$f(e)=a_0+a_1e'+a_2e''+\dots+a_ne^{(n)}$$ has the parity of $a_0=f(0)$. And the value of $f$ for an odd argument is the sum of all coefficients times an odd factor so that ...

2

Since $f(0)=a_0$ is odd and $f(1)=a_0+\ldots +a_n$ is odd, we have that $a_1+\ldots +a_n$ is even. Hence $$f(2k)=a_0+2(a_1k)+2(2a_2k^2)+2(4a_3k^3)+\ldots +2(2^{n-1}k^na_n)$$ is odd for all $k$, and $$f(2k+1)=\sum_{k=1}^na_k\sum_{i=1}^k{k\choose i}(2k)^i.$$ Since all the inner summands are even except when $i=0$ we see that ...

2

You are almost done: $a_0$ is odd and $a_0+a_1+\cdots+a_n$ is odd imply that $a_1+\cdots+a_n$ is even. Next, for each $x\in\mathbb{N}$, verify that $a_jx^j$ and $a_jx$ have the same parity whenever $j\geq 1$: $$2|a_jx\iff [(2|a_j)\text{ or }(2|x)]\iff[(2|a_j)\text{ or }(2|x^j)]\iff 2|a_jx^j.$$ Now the claim follows: for any $x\in\mathbb{N}$, ...

2

The maximum modulus principle implies that the maximum of $|f(z)|$ can only occur on the boundary. Another way to say it: for all $z\in D^\circ$, we have $|f(z)|<\max_{z\in\partial D}|f(z)|$. Since you have $|f(z)|=1$ for all $z\in D$, $f(z)$ must be constant.

2

Since $x^{5}=1024$ implies that $x=\sqrt[5]{1024}=4=x_0$ is one of the roots of $x^{5}-1024=0$, it follows that $$p(x)=x^{5}-1024=(x-4)q(x),$$ where $q(x)$ is a forth degree polynomial with real coefficients and leading term equal to $x^{4}$. Hence it is of the form $$q(x)=x^{4}+bx^{3}+cx^{2}+dx+e.$$ The coefficients $b,c,d, e$ can be found by ...

2

$\bf{My\; Solution::}$ Given $$x^5 = 1024 = 2^{10} = 4^5\Rightarrow x^5-4^5 = 0$$ So $$(x^5-4^5) = (x-5)\cdot (x^4+x^3\cdot 4+x^2\cdot 4^2+x\cdot 4^3+4^4) = 0$$ Above we use the formula $\displaystyle x^n-a^n = (x-a)\cdot (x^{n-1}+x^{n-2}\cdot a+x^{n-3}\cdot a^2+.......+a^{n-1})$ So We Get $x=4(\bf{Real \; Root})$ and $$x^4+x^3\cdot 4+x^2\cdot 4^2+x\cdot ... 2 Yes, you assume correctly that the update can also be written as$$ w_k=-\frac{f(z_k)}{g'(z_k)}. $$And correct, the derivative of g can be written as sum of products according to its construction as product of linear factors$$ g(x)=(x-z_1)…(x-z_n) $$(note that the leading coefficient of g is 1, so that this must also hold for f)$$ ...

2

From the identity: $\; x^5-1=(x-1)(x^4+x^3+x^2+x++1)$, they are the complex $5$-th roots of unity: $$\mathrm e^{\tfrac{2\mathrm i k\pi}{5}},\quad k=1,2,3,4.$$

2

Let me try. You have $$(a+b+c)(a+b+d)(a+c+d)(b+c+d) = (7-a)(7-b)(7-c)(7-d) = 7^4 - 7^3(a+b+c+d) + 7^2(ab+ac+ad+bc+bd+cd)-7(abc+abd+bcd+acd) + abcd$$

1

This is a community wiki answer to remove this question from the unanswered list: Bobby Ocean posted a counter-example at Mathoverflow: Set $p(z) = (z^4+2z^2+2) (z-1)$. Then $D:= \{ z : |p(z)|<1.45\}$ is connected, but some of the zeroes of $p'''$ and $p''''$ are outside $D$.

1

Hint: We have $$\sqrt[3]{7-4i}=(7-4i)^{1/3}$$ $$r=\sqrt{7^2+(-4)^2}=\sqrt{65}$$ $$\tan \theta=\left|\frac{-4}{7}\right|=\frac{4}{7}$$ Since, the point $(7, -4)$ lies in the fourth quadrant hence we have $$\sin \theta=\frac{\frac{4}{7}}{\sqrt{1+\left(\frac{4}{7}\right)^2}}=\frac{4}{\sqrt{65}}$$ \cos ...

1

Generally, you may use the fact that the only irreducible polynomials are those of first and second degree with negative discriminant. So if you have a polynomial of higher degree, equating it to a product of quadratic and linear functions, finding the roots of these will amount to finding the roots of the original polynomial. Then again, when you go beyond ...

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