# Tag Info

## Hot answers tagged polynomials

30

Transform the equation. Since all the roots are symmetric, say $$y=\frac {1+x}{1-x}\implies x(y+1)=y-1\implies x=\frac {y-1}{y+1}$$ Substitute this expression in place of $x$ in the original equation and simplify. \require{cancel}\begin{align}f(y)&=\biggl(\frac {y-1}{y+1}\biggr)^3-\frac ... 20 This is to make nice rules such as \text{deg }(PQ) = \text{deg }P + \text{deg }Q\\ \text{deg }(P+Q) \le \max(\text{deg }P , \text{deg }Q) $$So the only value that makes it possible is$$\text{deg }0= -\infty$$15 Define g(x)=\frac{a_n}{n+1}x^{n+1}+\dotsb+\frac{a_1}2x^2+a_0x then g(0)=g(1)=0,hence by Rolle's theorem there is some c in the interval (0,1) such that g'(c)=0 as we want. 12 Since the polynomial x^3-x-1 is irreducible over\def\Q{\Bbb Q}~\Q by the rational root test, one approach would be to identify the element \frac{1+a}{1-a} where a is the image of x in the field K=\Q[x]/(x^3-x-1), and to compute its minimum polynomial over~\Q; since the Galois group of the splitting field of x^3-x-1 permutes its roots ... 12 Without loss of generality, we may assume that p is monic. Since p has no real roots, n=2m for some m\ge 1, and there exist quadratic monic polynomials q_1,\dots,q_m with no real roots such that p=\prod_{k=1}^m q_k. Therefore, by Cauchy-Schwarz inequality,$$ \left(\frac{p'}{p}\right)^2=\left(\sum_{k=1}^m \frac{q_k'}{q_k}\right)^2\le m\cdot ...

9

The polynomial has no real roots, because it is equal to $(x^2-2)^2+12$. The remaining possibility is thus that it is a product of two quadratic factors. By Gauss' Lemma these need to have integer coefficients, so we are looking for a possibile factorization like $$p(x)=x^4-4x^2+16=(x^2+ax+b)(x^2+cx+d)$$ with some integers $a,b,c,d$. Modulo $3$ we have the ...

9

The associated quadratic polynomial $t^2-4t+16$ has negative discriminant, so there's no real root. Then the polynomial can be factorized over the reals as a product of degree two polynomial. You get them by a process similar to completing the square: \begin{align} x^4-4x^2+16 &=x^4+8x^2+16-12x^2\\ &=(x^2+4)^2-(\sqrt{12}\,x)^2\\ ...

8


Only top voted, non community-wiki answers of a minimum length are eligible