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2

Define inductively $f^n(x)=f(f^{n-1}(x))$ for all $n\geq1$, so that you are solving $f^5(x)=0$. We have $f^5(x)=f(f^4(x))$ thus $f^4(x)$ is either solution of $x^2+12x+30=0$, i.e. $$f^4(x)=-3\pm\sqrt{6}.$$ Then for the same reason $f^3(x)$ is any value $z$ such that $$z^2+12z+30=-3\pm\sqrt{6}.$$ Iterating this process a few more times you get to the ...

29

We have $f(x) = (x+6)^2 - 6$, therefore $f^2 (x) = (x+6)^4 - 6$, $f^3 (x) = (x+6)^8 - 6$, and so on. So $f^5 (x) = (x+6)^{32} - 6 = 0$, giving the (real) solutions $x=-6 \pm \sqrt[32]{6}$.

0

x^3+bx^2+cx+d = 0 has 3 solutions 3 and 2+i are given. third solution z is not. this polynomial can decomposed as (x-3)(x-2-i)(x-z)=0 from here follows 1. 3*(2+i)*z = -d 2. 3z + 3(2+i) + z(2+i) = c 3. 3 + 2+i + z = -b these admits a solution: b = -5 - i - z; c = 6 + 3 i + 5 z + i z, d = -3 z(2 + i) in order b to be real z = y-i in order d to be real y ...

3

A cubic polynomial will have three roots. We were given all three roots, since $f(2+i)=0 \space \implies \space f(2-i)=0$. Now we can factor the cubic and write $$x^3+bx^2+cx+d = (x-3)(x-(2+i))(x-(2-i))$$ Once you expand the RHS you can easily find $b,c,d$.

1

$p(x)=a_2x^2+a_1x+a_0$. We want $a_2\neq [0]_3$. So we have $2$ choices for $a_2$, $3$ choices for $a_1$ and also $3$ choices for $a_0$. How many do we have in general?

0

Because, if there is a common root, then there is a common factor of the polynomial and thus the resultant vanishes. Or put another way, the resultant is a multiply of the product of factors $(x_i-y_j)$ in all combinations.

1

Depending on the base field $K$, this might be a difficult problem in general. However, for quadratic forms of rank $2$, this is easy. You just have to check that the two quadratic forms have same discriminant, and represents a commun value, for example, which is the case here $5$ is represented by both quadratic forms, and the two determinants are equal to ...

0

One way to see this isomorphism is to consider how left multiplication by an element works in $\mathbb{Q}[X]/(X^2 - D)$; the fancy name for this is the left regular representation. I think this approach shows how one might arrive naturally at this isomorphism without knowing of its existence beforehand. First note that $\mathbb{Q}[X]/(X^2 - D) \cong ... 1 Let$t=x^2$. Then we have that $$x^4+6x^2+5=0 \Longleftrightarrow t^2+6t+5=0 \Longleftrightarrow (t+5)(t+1)=0.$$ Thus,$t=-5$and$t=-1$will be solutions. But we have to be careful because we have$t=x^2$. Thus, we really need to solve$x^2=-5$and$x^2=-1$. Thus, the answers will be$x=\pm\sqrt{-5},\pm\sqrt{-1}$. 1 OK, there's a lot here, so let's try and tackle one thing at a time (in general it's best to break up big questions like this into a couple of smaller ones). A quadratic form over a vector space means that$q$is a function which takes vectors as arguments, it just means it has more than one variable, that's all. What exactly don't you get? The definition ... 1 In$\mathbb Q[X]/(X^2-D)$, every element is of the form$c+dX$since$X^2$reduces to$c+dX$where$c=D$and$d=0$, and then$X^3$reduces to$X^2 X = (c+dX)X = cX+dX^2 = cX+dD$, etc. So ask yourself: which element$c+dX$should correspond to the matrix$\begin{pmatrix} a & bD \\ b & a \end{pmatrix}$? You have a pair$(a,b)$as input and a pair ... 1 You can define a ring-homomorphism from$\mathbf Q[x]$into the (commutative) ring generated by the matrix $$U=\begin{pmatrix}0&D\\1&0 \end{pmatrix}$$ sending$1$to$I_2$and$x$to$U$. This homomorphism is surjective and its kernel is generated by$x^2-D$since$U^2=D I_2$. Hence the ring is isomorphic with$\mathbf Q[x] /(x^2-D)$. 2$\begin{pmatrix} a & bD \\ b & a \end{pmatrix} \mapsto a\cdot 1 + b \cdot \overline{X}$is an isomorphism of rings from$\lbrace\begin{pmatrix} a & bD \\ b & a \end{pmatrix} \mid a,b\in\mathbb{Q}\rbrace$to$\mathbb{Q}[X]/(X^2-D)$, where$\overline{X}$is the image of$X$in$\mathbb{Q}[X]/(X^2-D)$by the canonical "quotient" map ... 0 The function $$f: \mathbb R \longrightarrow \mathbb R, f(x):= ax^b\cdot\exp(x)$$ extends to a holomorphic function on the complex plane $$F: \mathbb C \longrightarrow \mathbb C, F(z):= az^b\cdot\exp(z).$$ Global holomorphic functions are named entire functions. An entire function which is not a polynomial is named a transcendental function. As a ... 1 To do polynomial division in such cases, you may want to consider the following approach: I shall illustrate with your title problem. -1 | (1-m^2) -2m^2 -(m^2+1) | 0 -1+m^2 1+m^2 ------------------------------- 1-m^2 -1-m^2 | 0 This gives you the factorisation$(x+1)\left((1-m^2)x-(1+m^2)\right)$from ... 2 For the quadratic equation$(1-m^2)x^2-2m^2 x-(m^2+1)=0$, the product of the roots is$\frac{-(m^2+1)}{1-m^2}$. Since$-1$is one of the roots, the other must be$\frac{1+m^2}{1-m^2}$. The same method can be used for any quadratic equation with one known root. Remark: Please check the calculation of the coefficients of the quadratics. For example, when ... 1 These are algebraic numbers, not transcendental, so there are certainly algebraic relations involving them. For example,$T_n(\cos(\pi/n)) = T_m(\cos(\pi/m)) = -1$where$T_m$and$T_n$are Chebyshev polynomials of the first kind. 0 It guarantees that$(x-t)f''(t)$is integrable, and that the fundamental theorem of calculus is applicable to it. Usually the statement of the fundamental theorem of calculus is limited to cases of a continuous integrand to avoid difficulties. EDIT: More precisely, what's important is that in the proof of the integration by parts formula, the FTC should be ... 0 You want$f$to have a continuous second derivative so that$(x - t)f''(t)$will be integrable over$[a,x]$for every$x$. Then$E_1$will make sense. If$f''$is discontinuous, say, at finitely many points, the result is still valid. More generally, if$f''$is Riemann integrable, the integration by parts formula you used in the proof will remain valid. 1 I still don't really know what you are trying to say with definition 1, but I suspect you are trying to ensure some sort of uniqueness of coefficients, and apparently define the degree of a polynomial. I think you missed the mark here, so here is a simple and correct way to define a polynomial, and it's degree. Define a polynomial of degree$n\geq0$to be ... 0 The answer to both questions is yes, provided that$a_n=1$. 1 The map$f\colon x\mapsto 2x+1$maps roots of$x^2+x+1$to roots of$x^2+3$. Indeed, $$f(x)^2+3=(2x+1)^2+3=4(x^2+x+1)=0$$ whenever$x^2+x+1=0$. On the other hand, if$2$is invertible, then the map$g\colon x\mapsto \frac12(x-1)$maps roots of$x^2+3$to roots of$x^2+x+1$. Indeed, $$g(x)^2+g(x)+1=\frac14(x-1)^2+\frac12(x-1)+1=\frac14(x^2+3)=0$$ whenever ... 1 Complete the square. We have$x^2+x+1\equiv 0\pmod{p}$iff$4x^2+4x+4\equiv 0\pmod{p}$iff$(2x+1)^2\equiv -3\pmod{p}$. Remark: Let$F$be any field that has characteristic$\ne 2$. Then we can make precisely the same argument: We have$x^2+x+1=0$if and only if$4x^2+4x+4=0$if and only if$(2x+1)^2=-3$. So if$x^2+x+1=0$has a solution, then$w^2=-3$has ... 3 The motivation is that$\sqrt{-3}$is a root of the first and the second has roots${-1\pm\sqrt{-3}\over 2}$, where these are understood in the more general sense of finite field elements, since the quadratic formula works for any field of characteristic not$2$. But then it is clear how to get the roots of the second from the first, we can assume$p>3$... 2 Hint$\ a\!-\!b\mid P(a)\!-\!P(b) = b\!-\!c.\,$By symmetry$\,b\!-\!c\mid c\!-\!a,\ \ c\!-\!a\mid a\!-\!b.\ $Chained, these yield a divisibility cycle$\ \color{#c00}j\mid k\mid n\mid \color{#c00} j,\ $so$\ k,n = \pm j.\ $But$\,j+k+n = 0\,\Rightarrow\, j=0\,\Rightarrow\,a=b\,\Rightarrow\!\Leftarrow$0 Your polynomial is irreducible over$\mathbb Q[x]$. Its only real root is irrational, namely $$\frac{2+\sqrt[3]{152-24\sqrt{33}}+2\sqrt[3]{19+3\sqrt{33}}}3\not\in\mathbb Q$$ The same goes for the other two, whose expressions can also be deduced using the cubic formula. Perhaps you are confused by thinking that cubic conjugate roots come in pairs, just like ... 2 We can write the equation as $$a=(b-a^2)(b+a^2+1).$$ But if$b$is positive the right factor$b+a^2+1$is strictly greater than$a$, so equality can never be attained. 0 Solution:$a > 1, b = \frac{1}2\left(\sqrt{4a^4+4a^2+4a+1}-1\right)$. 5 The proof follows from the following Lemma: Lemma If$0<a \leq 1 \leq b$then $$(2+a)(2+b) \geq 3(2+ab)$$ Proof: $$(2+a)(2+b) \geq 3(2+ab) \Leftrightarrow \\ 4+2a+2b+ab \geq 6+3ab \Leftrightarrow \\ 0 \geq 2-2a-2b+2ab \Leftrightarrow \\ 0 \geq 2(a-1)(b-1)$$ QED Lemma Now, lets solve the problem. Let$b_i=-\alpha_1$, and we can assume without loss of ... 0 This formula uses the Cauchy form of the remainder.$x + \theta h $is a number between$x$and$x+h$. http://mathworld.wolfram.com/CauchyRemainder.html It comes from the mean value theorem. 1 If you divide$f$by$g$, you get a remainder of$r(x)= (a-14)x+b+3$. That means you can write $$2x^4-3x^2+ax+b = (x^2-2x+3)(2x^2+4x-1) + r(x)$$ For$f$to divide$g$, we need$r(x)=0$for appropriately chosen$a$and$b$, ie. take$a=14$and$b=-3$. 0 As @Fundamental pointed out in a comment, your definition of monomial is wrong. See Wikipedia for a better definition. Or here is a simpler definition. A monomial is a constant, a variable, or the product of constants and/or variables. Details: The monomial is usually simplified to place the constant, if any, first, and to combine multiple copies of the ... 0 The only condition of the fundamental theorem of algebra (FTA) is that the polynomial must be non-constant, i.e. some$a_i \ne 0$for an$i > 0$. Then it states that there is a unique factorisation $$\sum_{i=0}^k a_i x^k = b_0 \prod_{i=1}^k (x-b_i)$$ 0 Every polynomial over$\mathbb{R}$can be factored into linear or quadratic terms. If you have a real quadratic, it can be factored into linear terms over$\mathbb{C}$. This is what is meant by the fact that$\mathbb{C}$is an algebraic completion of$\mathbb{R}$:$\mathbb{C}$has the roots of all polynomials over$\mathbb{R}$. 1 Yes. Normally you would see$-$'s in the factors though, because then the$b_n$'s are the roots of the polynomial. 0 Hint: For a polynomial function$f$and a fixed$t$, consider the polynomial division $$f(x) = (x-t)^2 q(x) + a(x-t) + b.$$ Now evaluate at$x = t$to find$b$, then differentiate and evaluate at$x = t$to find$a$. 0 There is no solution because $$a^4+a^2+a>b^2+b$$ Indeed, $$a^2+a>b^2+b$$ 3 Hint:$a^4+a^2+a > a^2+a > b^2+b$if$a > b> 1$0 If$abc = 0$, WLOG$a = 0$, and so the common root must b$ x = - \frac{c}{b}$. However, it is easy to check that$ 0 = b \frac{-c^3}{b^2} + c \frac{-c}{b} = \frac{ -c^3-c^2b } { b^2} = - \frac{ c^2 (c+b) } {b^2}$, hence we must have$ c = -b$, which gives us$x=1$. Then,$ 0 = bx^3 + cx + a = bx^3 - bx = bx(x-1)(x+1)$would have 3 real roots and we are done. ... 1 Abel's impossibility theorem states that there is no algebraic solution to polynomial equations of degree five or higher But Jordan has shown that any algebraic equation can be solved using modular functions. There are explicit formulas without the need to use Tschirnhausen or other transformations. However, application of this theorem in practice is very ... 2 For every non-zero complex number$a$and every polynomial$P$the roots of$P$and$aP$are the same. Applying this with$a=i$answers your question. 3 The conclusion of the theorem for arbitrary$g$is false for example for$R=\mathbb{Z}$, indeed for any domain that is not a field. Consider$f=X$and$g=2X$, or generally$f=X$and$g = aX$where$a$is not invertible. 1 If you fix the value$x$one variable you have the Lagrange interpolation formula $$f(x, y) = \sum_{k=1}^N f(x, y_k)\prod_{i \neq k} \frac{y_i - y}{y_i - y_j}$$ For each fixed value$y = y_k$you can construct a Lagrange polynomial for the function$f(x, y_k)$. $$f(x, y_k) = \sum_{k=1}^N f(x_k, y_k)\prod_{i \neq k} \frac{x_i - x}{x_i - x_j}$$ The ... 0 Let$f(x) = x^3 + 4x^2 - 1$. Then$f(0) = -1$, and$f(1) = 4$, so we know there is a real root$x = r$in the interval$r \in (0,1)$. Now divide: we obtain the factorization $$f(x) = (x-r)(x^2 + (r+4)x + r(r+4)).$$ Now use the quadratic formula on the quadratic term to isolate the remaining roots: $$x \in \left\{ \frac{-r-4 \pm \sqrt{16-8r-3r^2}}{2} ... 1 For a cubic equation f(x) if f'(x) is 0 at a and b. Then, if f(a)f(b)<0 then the equation has three real roots and if f(a)f(b)>0 then it has a single real root. Here a=0 and b=\frac{-8}{3} Can u take it from here? 3 The equation has three solutions. To prove this, look at the values of the right hand side at these values of x: -4, -1, 0, 1. The values alternate between positive and negative, so there is a root between each pair of x values I gave you. That gives three solutions, and a cubic can have no more than three solutions. 5 The derivative of f(x)=x^3+4x^2-1 is f\,'(x)=3x^2+8x, which is 0 at x=0 and at x=-\frac83. Since the function is cubic with a positive leading coefficient, it has a local maximum at x=-\frac83 and a local minimum at x=0. If you calculate f\left(-\frac83\right) and f(0), you should be able to tell very quickly how many real solutions the ... 1 Let us define, for two polynomials p,q of degree at most n, the bilinear form$$(p,q):=\sum_{k=1}^{n+3}p(k)q(k)$$Observe that this bilinear form is a scalar product since ||p||^2:=(p,p) is positive definite.$$||p||^2=0\implies p=0$$We need to find the closest point of A:=\{p:\ p\text{ is monic}\} to the origin. Equivalently we can find the ... -1 Hint For f\in K[X] chose g\in K[X] such that g(s) = f(s) and g(t) = -f(t). Now look at$$f = \frac12 (f + g) + \frac12(f - g)$$Here \frac12 := (1+1)^{-1} in the field K (We must thus assume \mathrm{char}(K) \ne 2). If K = \mathbb F_2 (up to isomorphism the only field with characteristic 2), it should be very easy for you since K[X] is ... 5 Let deg(P)=n>0 and assume without loss of generality that deg(Q) \leq deg(P). Consider the polynomial R=(P-Q)P' (P' denote the derivative of P). We have :$$deg(R)\leq 2n-1$$Now if$r$is a root of multiplicity$k$of$P$then$r$is a root of$P'$of multiplicity$k-1$and because$Q(r)=0$,$r$is a root of$P-Q$. hence$r$is a root of$R\$ ...

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