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Your previous question concerns generalized CRT. So you know that $$\frac{{\bf R}[x]}{(x-1)(x^2+1)}\cong\frac{{\bf R}[x]}{(x-1)}\times \frac{{\bf R}[x]}{(x^2+1)}\cong {\bf R}\times{\bf C}$$ Find the idempotents in $\bf R$ and $\bf C$ to find the idempotents in ${\bf R}\times{\bf C}$, then pull them back through the isomorphisms implicit above. Indeed ...

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The following is correct, though not fully expanded: $$(x^2 + 4x + 4)(x + 2) = (x+2)^2(x+2) = (x+2)^3$$ Consider writing this as $(x + 2)(x^2 + 4x + 4)$, and then distribute (multiply) each term in the first factor, with each term of the second factor. \begin{align} (\color{blue}{\bf x + 2})(x^2 + 4x + 4) & = \color{blue}{\bf x}(x^2 + 4x + 4) + ... 7 Fun fact:x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1).$$This can be derived by setting x^4+1 equal to a product of two monic quadratics with unknown coefficients and then solving for said coefficients little by little. As a consequence,$$x^8+1=(x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1).$$We can go further. For example, set ... 7 It is possible. For example, given a prime number p and a,b\in\mathbb Z with a+b\le-2, let$$f(x)=x^3+pax^2+pbx+p\in \mathbb Z[x].$$Then by Eisenstein's criterion, f is irreducible in \mathbb Q[x], i.e. f has no rational root. However, since f(0)=p>0 and f(1)=1+(a+b+1)p<0, f has three distinct real root located in (-\infty,0), ... 6 For any polynomial p(x), we have \displaystyle\lim_{x\to\infty}\dfrac{p(x)}{e^x}=0. This can be shown, for example, by using L'Hospital's Rule, and in other ways. Your argument uses the derivative in a different way. The argument is somewhat informal, but it can be made formal. One could use induction. Let us prove by induction on degree that ... 6 Generally, if there's two roots whose sum is zero, then it means that two factors are x-a and x+a, which means that x^2-a^2 must be a factor. So clearly$$ (x^2-a^2)(x^2+bx+c)=x^4+2x^3-8x^2-18x-9=0 Find the values of a^2, b, and c that satisfy the left equality, and you'll have found factors that you can then solve for all roots. (This works ... 5 I doubt an easy proof of the irreducibility exists in general. If n is a prime, then the polynomial is Artin-Schreier and handled easily. Selmer gave a clever proof in the general case, working explicitly with the roots of the polynomial in \mathbb{C}. See E. S. Selmer, On the irreducibility of certain trinomials, Math. Scand. 4 (1956), 287-302, ... 5 Why is (x^{2} + 4x + 4)(x+2) wrong? (x^{2} + 4x + 4)(x+2) = x^{3} + 4x^{2} + 4x + 2x^{2} + 8x + 8 = x^{3} + 6x^{2} + 12x + 8 And it is the right answer. If you have problem remembering the power formulas for binomials, just use pascal's triangle to calculate the coefficients: \begin{align*} (a + b)^{0}\to &1\\ (a + ... 4 Given: (x+2)^3 we can rewrite (x+2)^3 as: (x+2)(x+2)(x+2) When expanding 3 terms we must first calculate (x+2)(x+2) before we can multiply by the third (x+2). = (x+2)(x+2)\implies x^2+2x+2x+4\implies x^2+4x+4 =  (x+2)(x^2+4x+4)\implies x(x^2+4x+4) + 2(x^2+4x+4) =  x^3+4x^2+4x+2x^2+8x+8 = x^3+6x^2+12x+8 4 Let a be a root of x^3+2x+1=0 and b=a^2 be a root of the required equation So, a^3+2a+1=0\implies a\cdot b+2a+1=0\implies a=-\frac1{b+2} As a be a root of x^3+2x+1=0, put this value of a in x^3+2x+1=0 On simplification, I get (b+2)^3-2(b+2)^2-1=0\iff b^3+4b^2+4b-1=0 So, the required equation will be y^3+4y^2+4y-1=0 4 The condition for the cubic x^3 + px + q to have 3 real roots is 4p^3+27q^2 \leq 0. This condition can be proved algebraically (search for Discriminants of a polynomial) or by calculus: A cubic has 3 real roots if and only if it has a local minimum and a local maximum and the product of those two values is \leq 0. So as Potato mentioned in the ... 4 Hint: What values does the function x^2 acquire(positive/negarive)? What is the solution of the equation x^2=0? Can you find the solution of the equation x^2+y^2=0? Now, what can you say about the equation (x-3)^2+(y-5)^2+(z-4)^2=0 ? Can you find the values of x,y,z? 4 The method you cite should work, because the polynomial is of degree 2 and hence any factors will correspond to solutions of the equation x^2+x+1=0. But you have to do it right and work over the field \mathbb F_{256}, which means you must begin by constructing that field etc. Working over the integers modulo 256 (i.e. \mathbb Z/256\mathbb Z) will not ... 4 As "Myself" observed, you are apparently misconstruing \mathbb F_{2^8} as being isomorphic to \mathbb Z/2^8, which it is not. For example, the latter has many zero-divisors, and is far from being a field. Also, in any case, for human-executable work checking 256 cases is usually a terrible method. Instead, look for some structure or meaning. Here, the ... 4 Steps: Prove that constant functions are continuous Prove that the identity function f(x)=x is continuous Prove that if f(x) and g(x) are continuous, then so are f+g and f\cdot g. This suffices to prove that all polynomials are continuous. 4 Using middle term factor,(x^2+5)^2-(9+6)(x^2+5)+6\cdot9=0\implies (x^2+5)(x^2+5-9)-6(x^2+5-9)=0\implies (x^2-4)(x^2-1)=0$$\implies x^2-1=0 or x^2-4=0 Alternatively, using quadratic equation formula for x^2+5=\frac{15\pm\sqrt{15^2-4\cdot1\cdot 54}}{2\cdot1}=\frac{15\pm 3}2=9 or 6 4 Let a = (x^2 + 5). Then$$(a-9)(a-6)=0\implies a = 9\; or \;a = 6\implies x^2+5 = 9 \; or \;6\implies x = \pm\sqrt4\; or \pm\sqrt1\implies x = \pm2\; or \pm1$$This kind of equation is called a biquadratic equation. Cheers! 3 First: looking for roots to deduce a polynomial over some field is irreducilbe works as long as the polynomials degree \,\le 3\, . Second: the polynomial \,x^2+x+1\, is defined over \,\Bbb F_2\, and it's irreducible over this field. Since \,2\mid 8\;, we have that \,\Bbb F_{2^2}\le\Bbb F_{2^8}\, , and this means that all the roots of \,x^2+x+1\, ... 3 No error on your part: If you've copied the problem correctly, then your solution is correct: k = -3. I was very careful in calculating, as I'm sure you were, in double checking, so if (x - 2) is a factor for your given polynomial, then k must be -3. Typo/misprint I suspect, in your text: a typo in the solution, or a misprint of the desired ... 3 Hints: The constant coefficient (constant term) is -9.$$\pm 1, \pm 3\pm 9 \;\text{divide}\; -9 $$If there are rational roots to the polynomial, they will be among these values. \large (\star) Two of the roots x_i, x_j be must be factors of -9: and they must be such that x_i = -x_j. Lets call this pair a, \pm a. That means (x−a) and ... 3 This is a standard example of an equation that does not have an explicit formula for its root. It can only be solved numerically. The important thing is to get a good first approximation to the root. For example, if you want to solve \frac{(1+x)^n-1}{x} = a, and you think that there might be a root close to 0, you can approximate (1+x)^n by ... 3 The first expression evaluates to e^{2u}-e^u-e^l+1, so if your question is whether there is some algebraic manipulation that brings this into the form of an exponential minus 1 for general l,u then the answer is no. In fact the range of the function x\mapsto e^x-1 in \Bbb R is (-1,\infty), and the product of two values in that range may be ... 3 The answer depends on R. Example 1: R=\mathbb{C} and you know that any polynomial, including the one in question, splits into linear factors Example 2: R=\mathbb{Q} denote$$f(x)=x^{8}+1$$then$$f(x+1)=(x+1)^{8}+1=x^{8}+8x^{7}+28x^{6}+56x^{5}+70x^{4}+56x^{3}+28x^{2}+8x+2$$is irreducible by Eisenstein with p=2 hence f(x) is irreducible. 2 Over \mathbb{C}, this polynomial splits into linear factors of 8 of the 16 solutions to x^{16} - 1 = 0. The irreducible factors of this equation are the cyclotomic polynomials. http://en.wikipedia.org/wiki/Cyclotomic_polynomial So x^8 + 1 is indeed irreducible over \mathbb{Q}, as it is a cyclotomic polynomial. Over \mathbb{R}, this polynomial ... 2 Considering \sigma(A) denotes the set of characteristic(eigen) values of A As \lambda is an eigen value, \exists 0\ne v\in V such that Av=\lambda v Note that A^kv=\lambda^k v,\forall k\in N(easy to prove using Induction) Let p(x)=\sum_{i=0}^{n}a_ix^i,a_i\in R Then we have , p(A)=\sum_{i=0}^{n}a_iA^i So we have , ... 2 If \,f\, is a polynomial in \,n\, indeterminates with coefficients in \,R\, , we want \,\phi f\, to be an element in \,S\, , and that's why there's written \,\phi f(s_1,...,s_n)\;,\;\;s_i\in S\, . For this to happen, we need the coefficients of the polynomial \,f\, in \,R[x_1,...,x_n]\, to be mapped to elements in \,S\, , and this is ... 2 Make a change in variables: y = x + 1. Then the equation becomes:$$\dfrac{y^{36} - 1}{y-1} = 20142.9/420$$The numerator y^{36}-1 is equal to (y-1)(y^{35}+y^{34}+\cdots+1) (use the formula for the sum of a geometric series). Therefore dividing by y-1 gives:$$y^{35}+y^{34}+\cdots+1 = 20142.9/420$$This is a polynomial of degree 35 in y. It ... 2 Edit: There is a mistake in this proof. I think it can be fixed with an argument from the degrees of q and q', so I'll work on that. Suppose p(x) is reducible. Then$$ p(x) = (x^i + q(x))(x^j+q'(x)) $$with i+j=n. But then$$x^iq'(x) + x^jq(x) + q(x)q'(x)=-x - 1.$$forces$$ x^iq'(x) + x^jq(x) = -x  q(x)q'(x) = -1. $$Hence, q(x), q'(x) are ... 2 Solved the equation using Tartaglia's method$$ x^3 + x^2 + 1$$Make then substitution  x = t + h$$ t^3 + 3t^2h+3th^2 + h^3 + t^2 +2th+h^2 + 1 = 0 t^3 + t^2(3h+1)+t(3h^2+2h) + (h^3 + h^2 + 1) = 0$$In order to eliminate the second degree term we add a condition  h = -\frac{1}{3}$$ t^3 -\frac{1}{3}t + \frac{29}{27} $$We make t = u+v$$ u^3 ...

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It means the roots are of the form $a, ar, ar^2$. Here it is not too difficult to see that $2, 4, 8$ are ok, just look at the constant term, which is $- (a \cdot ar \cdot ar^2) = - (a r)^3$, and check that $2, 4, 8$ fit with the other coefficients $$14 = 2 + 4 + 8, \qquad 56 = 2 \cdot 4 + 2 \cdot 8 + 4 \cdot 8.$$

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