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17

So at the most basic level there is a factor $(x+a)(x-a)=x^2-a^2$ and the factorisation is $$(x^2-a^2)(x^2+bx+c)=x^4-2x^3+4x^2+6x-21$$ Equating powers of $x^3$ gives immediately a value for $b$. Then equating powers of $x$ gives $a^2b=-6$ so $a^2$ is known and the constant term gives $a^2c=21$ so that $c$ can be determined.


17

If the coefficients are rational and $\dfrac{1+\sqrt 5} 2$ is a root, then $\dfrac{1-\sqrt 5} 2$ is a root. To see this, suppose you substitute $\dfrac{1+\sqrt 5} 2$ for $x$ and get $0$. What would then happen if you substitute $\dfrac{1-\sqrt 5} 2$ for $x$? When you expand $x^2$ and $x^3$, then wherever $\sqrt 5$ appears, $-\sqrt 5$ would appear, and ...


10

No such cubic exists. The golden ratio is a root of the quadratic $q(x)=x^2-x-1$. If the golden ratio is a root of a cubic $p(x)$, then it must be the root of the remainder of $p(x)$ when divided by $q(x)$. That remainder has integer coefficients and is either zero or has degree $1$. But the golden ratio is not a root of polynomial of degree $1$ with ...


10

HINT: try the ansatz $$(-2+Bx+Ax^2)(1+Cx^2+Dx^4)$$


9

You seem to want the details of a numerical method to find an good approximation for $x$. Here is one way, using the Newton-Raphson method, which is well-known and uses fewer steps than other famous methods. We want to find the zeros of $$f(x)=0.25^x+0.5^x+0.75^x-1$$ (I rearranged the terms to be in ascending order, which satisfies my sense of style. I ...


9

Let $t=2-\sqrt 3$. Note that $2+\sqrt 3=\frac 1t$. Then, we have $$\begin{align}x=t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}&\Rightarrow x^3=t+\frac 1t+3\left(t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}\right)\\&\Rightarrow x^3=4+3x\end{align}$$


7

If $\pm a$ are roots of $x^4-2x^3+4x^2+6x-21=0$ then $a^4-2a^3+4a^2+6a-21=0$ $a^4+2a^3+4a^2-6a-21=0$ Adding these two equations gives you a biquadratic equation for $a$, which you can solve. (Make sure you select the roots of the original equation.) Once you have found $a$, the other roots are $b$ and $2-b$ and must satisfy $-a^2b(2-b)=-21$.


7

The result is not quite true. We first prove a correct version, and then show that there are polynomials for which the result does not hold. Let $M$ be the maximum of the $|a_i|$. We have $$\alpha^n=-\left(a_{n-1}\alpha^{n-1}+\alpha^{n-2}+\cdots +a_0\right).$$ By the Triangle Inequality, it follows that $$|\alpha|^n\le ...


5

HINT : We have $$\alpha^4+\alpha^2=1$$and $$\alpha^6+\alpha^4=\alpha^2$$


5

Let $f(x)=\left ( \frac{1}{2} \right )^{x}+\left ( \frac{1}{4} \right )^{x}+\left ( \frac{3}{4} \right )^{x}-1$. Now note that $f(1)=\frac12$ and $f(2)=-\frac18$. Hence given that $$f'(x)=-4^{-x}(3^x \log\frac43+2^x\log 2+\log4)<0$$ there is one unique root in $(1,2)$. The rest is numerics, this could be approximated to $1.73051$.


5

When people say "zeta function has zeros on negative even integers", they are talking about the analytic continuation of the Riemann zeta function, and the naive formula only works for Re(s)>1


5

This is an old question, but should be interesting to answer. If you want an aesthetic version, then the solution to, $$x^5 + x^4 - 12x^3 - 21x^2 + x + 5 = 0$$ is given by, $$x = \frac{1}{5}\left(-1+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\right)$$ and the $z_i$ are the roots of the quartic, $$z^4 + 12679 z^3 + 78678031 z^2 + 362989005529 z + 31^{10} = ...


4

Identifying with Cardano's formula $$x=\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}},$$ we find $q=-4$, then $p=-3$.


4

$$x^3=(r+s)^3=r^3+3r^2s+3rs^2+s^3=r^3+s^3+3rs(r+s)=4+3x,$$ because $r^3+s^3=4$ and $rs=1$.


4

I think you problem may add this $x$ domain,following I add $x>0$ $$f(x)=x^n+nx-1\Longrightarrow f'(x)=n(x^{n-1}+1)>0$$ and $$f(0)=-1<0,f(\dfrac{1}{n})=\dfrac{1}{n^n}>0$$ so this equation has unique real postive.. and $$f(\dfrac{1}{n+1})=\left(\dfrac{1}{n+1}\right)^n-\dfrac{1}{n+1}<0$$ so $$\dfrac{1}{n+1}<x_{n}<\dfrac{1}{n}$$ so ...


4

HINT : $$x^4-(a+b)x^3+(ab+2)x^2-(a+b)x+1=(x^2-ax+1)(x^2-bx+1)$$


4

If you do long division $x^6+2x^4 : x^4+x^2-1$ you get a quoatient of $x^2+1$ and a reminder of $1$. Therefore $$x^6+2x^4=(x^2+1)(x^4+x^2-1)+1$$ Replace now $x=\alpha$.


4

Given enough conditions, you can find the cubic. The cubic satisfying your conditions (not the second one yet) is $$(x-\phi)\left(x-\frac{1}{\sqrt{\phi}}i\right)\left(x+\frac{1}{\sqrt{\phi}}i\right)=0$$ Expanding it gives, $$x^3-\phi x^2 + \frac{1}{\phi}x-1=0$$ Obviously the coefficients are not integral. Therefore, such a cubic does not exist.


4

Setting $$x^4+x^3-3x^2+4x-2=(x^2+ax+b)(x^2+cx+d)$$ where $a,b,c,d\in\mathbb Z$ such that $|b|\lt |d|$ gives you $$1=c+a$$ $$-3=d+ac+b$$ $$4=ad+bc$$ $$(b,d)=(1,-2),(-1,2)$$ Solving these gives you $(a,b,c,d)=(-1,1,2,-2)$, i.e. $$x^4+x^3-3x^2+4x-2=(x^2-x+1)(x^2+2x-2).$$ Here, note that $x^2-x+1=0\iff x=\frac{1\pm\sqrt{3}i}{2}$.


3

Call your equation $f(z)=0$. Their roots are the same as the roots of $z^2f(z)=0$. Now this equation, rewritten as $z^4-2z^3-1=0$ is a polynomial equation of degree $4$. This has two real roots, $a,b$. Then, in theory you can divide this polynomial by $(x-a)(x-b)$ and get a quadratic equation with real coefficients. This has negative discriminant leading to ...


3

Hint: Golden ratio satisfies $p^2=p+1, p^3=p^2+p=2p+1$. So the cubic must have $(2p+1)+a(p+1)+bp-1=0$ which implies $p$ is rational.


3

The function : $$y=0.25^x+0.5^x+0.75^x-1$$ is decreasing. For example $y(1)=2$ and $y(2)=-\frac{1}{8}$. So, the root for $y=0$ is between $x=1$ and $x=2$. In this case, among many numerical methods, the dichotomic method is very simple. The successives values $x_k$ are : $$x_{k+1}=x_{k}+\frac{\delta_k}{2^k}$$ where $\delta_k=\pm 1$ the signe is $+$ if ...


3

Hint: You have a palyndromic polynomial, hence $P(x)=0$ is equivalent to: $$ \left(x^2+\frac{1}{x^2}\right)-(a+b)\left(x+\frac{1}{x}\right)+(ab+2) = 0\tag{1} $$ or, by setting $z=x+\frac{1}{x}$, $$ z^2-(a+b)z+ab = (z-a)(z-b) = 0.\tag{2} $$ Consider now that the range of $f(x)=x+\frac{1}{x}$ is $\mathbb{R}\setminus(-2,2)$.


3

If you write the equation as $x + 1 + t (c x^2 + d x^6) = 0$, then expand a root of this in powers of $t$, you get a series $$\eqalign{ x &= \sum_{n=0}^\infty C_n(c,d) t^n \cr &= -1 + (-c-d) t + (-2 c^2-8 c d-6 d^2) t^2 + \ldots }\tag{1}$$ where each $C_n(c,d)$ is a homogeneous polynomial of degree $n$ in $c,d$. For any $c,d$ this should converge ...


3

See OEIS A$257451$. There is no known closed form for this number.


3

Consider $g(x)=f(x)-m$. $g'(x)=5x^{4}-5=5(x^{4}-1)=5(x^{2}+1)(x^{2}-1)$. So g has extrema at $\pm 1$. Notice by the sign of $g'$, $g$ is increasing on $(-\infty,-1]$ and $[1,\infty)$ and decreasing on $[-1,1]$. $g(-1)=8-m$ and $g(1)=-m$. Now if $m \in (0,8)$, $g(-1)>0$ and $g(1)<0$ and you can use the fact that $g(x) \to \infty as x \to \infty$. Added ...


2

You need here two lemmas. In the polynomial ring $\mathbb{R}[x]$ there is a division with residues. if $s$ is a root of p(x) then $p(x)=g(x)(x-s)$. Can you take it from here?


2

By the Shafer-Fink inequality we have: $$ \frac{3}{1+2\sqrt{1+x^2}}\leq\frac{\arctan x}{x}\leq \frac{\pi}{1+2\sqrt{1+x^2}} \tag{1}$$ so the only positive solution of our equation is between $\frac{1}{2}\sqrt{21}$ and $\frac{1}{2}\sqrt{4\pi^2-4\pi-3}$. Since the second derivative of $\frac{\arctan x}{x}$ is positive over such interval, the sequence given by ...


2

If you can factor $$p(x) = x^6 - 65x^3 + 64 = (x^3 - 64)(x^3 - 1),$$ then you can use the formula for a difference of cubes, $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ to factor $p(x)$ into a product of two linear factors and two irreducible quadratic factors. For example, the first quadratic factor is $x^2 + 4x + 16$. You can use the quadratic formula on each ...



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