# Tag Info

21

Clearly there is no negative root as all terms are positive for $x < 0$. The question remains if there are positive roots. Here is a simple way which often works. Case 1: $0 < x <1$. $$P(x) = (15-x) + (x^2-x^7) + x^8 > 0$$ as each term is positive. Case 2: $x > 1$. Similarly $$P(x) = (x^8-x^7) + (x^2-x) + 15 > 0$$ as $x=1$ is not a ...

15

Since $f(x) - x$ is a polynomial of degree $6$ and has $6$ roots $1, 2, 3, 4, 5, 6$ by condition, we can factorize $f(x) - x$ as: $$f(x) - x = C(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6).$$ Plug in $x = 0$ in the above expression, we have $3 - 0 = C\times 6!$, hence $C = \dfrac{3}{6!}$. Therefore, $$f(7) = 7 + (f(7) - 7) = 7 + \frac{3}{6!}(7 - 1)(7 - 2)(7 - ... 13 The polynomial can be rewritten$$ x^7(x-1) + x(x-1) + 15. $$Unless x is between 0 and 1, the first two terms are positive, and so the polynomial is positive. Even if x is between 0 and 1, the first two terms are tiny in magnitude, certainly each individually greater than -1, so that when 15 is added to their sum, the result is positive. ... 9 As Lucian pointed out this follows immediately from the properties of cyclotomic polynomials. We have the factorization (into polynomials irreducible over \Bbb{Q}):$$ x^n-1=\prod_{d\mid n}\Phi_d(x). $$Your observation follows from the following as factorization:$$ a^n-b^n=b^n[\left(\frac ab\right)^n-1]=b^n\prod_{d\mid n}\Phi_d(\frac ab)= \prod_{d\mid ...

8

Let $$f(x)=2x^3-3x^2-12x+1$$ Now, $f(0)>0$ and $f(1)<0 \implies 0< \alpha <1$ $f(-2)<0$ and $f(-1)>0 \implies -2< \beta <-1$ $f(3)<0$ and $f(4)>0 \implies 3< \gamma <4$ $\therefore \lfloor\alpha\rfloor + \lfloor\beta\rfloor + \lfloor\gamma\rfloor = 0+(-2)+3=1$

8

$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$ Set $t=x^2,z=x^2+x+1$. $\Longrightarrow$ \begin{align}10t^2-7tz+z^2&=(2t-z)(5t-z)\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\&=(x^2-x-1)(4x^2-x-1)\end{align} $$\boxed{\color{red}{x_{1,2}=\frac{1}{2}\pm\frac{\sqrt5}{2},\;x_{3,4}=\frac{1}{8}\pm \frac{\sqrt{17}}{8}}}$$

7

Set $A=x^2,B=x^2+x+1$. Then, \begin{align}10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2&=10A^2-7AB+B^2\\&=(2A-B)(5A-B)\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\&=(x^2-x-1)(4x^2-x-1)\end{align}

7

Localise the roots between two consecutive integers and use the Intermediate value theorem. Let $p(x)=2x^3-3x^2-12x+1$. As $p'(x)=6x^2-6x-12=6(x+1)(x-2)$, $p(x)$ has a local maximum at $x=-1$ and a local minimum at $x=2$: $$p(-1)=8,\quad p(2)=-19.$$ Thus we know there are three real roots: $x1<-1$, $-1< x_2 <2$, $x_3>2$. Now a table of the ...

7

For a simple approach consider the function $y=x^4+4x^3-8x^2=x^2\cdot\left((x+2)^2-12\right)$ - the intersections with the line $y=-p$ will give the roots of the original. Since this is just a horizontal line in the normal $x,y$ plane, a quick sketch will show that the number of real roots is governed by the relationship of $p$ to the local minima/maxima of ...

7

The route you take is fruitful. $p\left(x\right)=\left(x-2\right)\left(x-8\right)q\left(x\right)$ leads to: $$\left(x-4\right)q\left(2x\right)=2\left(x-2\right)q\left(x\right)$$ Then $4$ must be a root of $q$, so $q\left(x\right)=\left(x-4\right)r\left(x\right)$ leading to: $$r\left(2x\right)=r\left(x\right)$$ Then $r\left(x\right)$ must be a constant ...

7

Since non-real roots occur in conjugate pairs, number of real roots can either be $1$, $3$ or $5$. Assume, on the contrary, that all roots are real, i.e, $5$ real roots. Let $\{\alpha_{i}\}_{i=1}^5$ be the roots of the equation. By Vieta's Formula, $\displaystyle -a_1= \sum_{i=1}^5 \alpha_{i}$ and $a_2 = \displaystyle \sum_{i < j} ... 6$x\in F[t,t^{-1}]$is a power of$t$iff$x$is invertible (or, equivalently$x\mid 1$) and$t-1\mid x-1$. If$x=t^n$, then$xt^{-n}=1$, and as you already noticed$t-1\mid t^n-1$. For the converse write$x=f(t)/t^n$with$n\ge0$and$f(t)\in F[t]$. Since$x$is invertible there exists$y\in F[t,t^{-1}]$such that$xy=1$. Write$y=g(t)/t^m$with ... 6 It should be clear that on the interval$[-1,1]$you have$|x^8-x^7+x^2-x|\leq |x^8|+|x^7|+|x^2|+|x|\leq 4$and so$x^8-x^7+x^2-x+15\geq 11$Further you should notice that$x^8-x^7>0$when$|x|>1$and that$x^2-x>0$when$|x|>1$, so$x^8-x^7+x^2-x+15\geq 11$for all$x$5 HINT: Let$(x+1)f(x)=1+A\prod_{r=0}^{11}(x-r)$where$A$is an arbitrary constant 5 Usage of the multinomial coefficient$(k_1, k_2, \cdots, k_n)$!: $$\big( 1 + x^5 + x^7\big)^{20} = \sum_{k_1=1}^{20} \sum_{k_2=1}^{20-k_1} (k_1, k_2, 20 - k_1 - k_2)! x^{5k_1} x^{7k_2},$$ where $$(k_1, k_2, \cdots, k_n)! = \frac{ (k_1 + k_2 + \cdots + k_n )! } { k_1! k_2! \cdots k_n!}.$$ So we get$k_1=2$and$k_2=1$, thus $$(2,1,17)! = ... 5 17 can only be obtained by using two 5s and one 7 . These two 5s can be obtained in \binom{20}2 ways which is 190 and the 7 can be got in from one of the remaining 18 brackets. So 190 x 18 = 3420 is the answer. 4 HINT: Let the highest of power of x be n So, (x-8)[a(2x)^n+\cdots]=8(x-1)[ax^n+\cdots] Comparing the coefficients of x^{n+1},$$a2^n=8a\implies n=3$$Let p(x)=(x-2)(x-8)(ax+b) where a,b are arbitrary constants to be determined Hope you take it from here? 4 Apply the Tschirnhaus transformation x=y-\frac{a_1}{5}, which will change your equation to$$y^5+py^3+qy^2+ry+s=0$$Your condition is precisely that p>0. We now apply Descartes' rule of signs to the various cases of the signs of q,r,s. For positive roots, we have the pattern +,q,r,s. For negative roots, we have the pattern -,q,-r,s. Exactly ... 4 If p = 2, then any t will do. Hence, there are 2^n possible values of t in this case. Now assume that p>2. If x^2=t^2-4, then (t-x)(t+x)=t^2-x^2=4. Hence, t-x can be any nonzero element 2u \in \mathbb{F}_{p^n} and t+x=2u^{-1}. Therefore, t=u+u^{-1} for some nonzero u \in \mathbb{F}_{p^n}. Conversely, if t=u+u^{-1} for ... 4 One of the reasons why the methods you mentioned do not work for multiple roots is that, in this situation, the root-finding problem is numerically ill-posed. That is, an arbitrarily small perturbation of the input will change the structure of your solution (just as you recognized, multiple roots will split into clusters). The same holds for the computation ... 4 Yes. Let f\in\mathbb R[X_1,\ldots,X_n] be a polynomial that vanishes on an open subset of \mathbb R^n, wlog. an open neighbourhood of 0. Then for any (a_1,\ldots,a_n)\in\mathbb R^n the polynomial g(T)=f(a_1T,\ldots, a_nT)\in\mathbb R][T] vanishes in a neighbourhood of 0, hence in infinitely many points, hence is the zero polynomial. Now assume ... 4 10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0 Divide by x^4 on both sides, 10-\frac{7x^2(x^2+x+1)}{x^4}+\frac{(x^2+x+1)^2}{x^4}=0 10-7(1+\frac{1}{x}+\frac{1}{x^2})+(1+\frac{1}{x}+\frac{1}{x^2})^2=0 Put (1+\frac{1}{x}+\frac{1}{x^2})=t 10-7t+t^2=0,solving we get t=2,5 when t=2 1+\frac{1}{x}+\frac{1}{x^2}=2 simplify we ... 4 Divide by (x^2 + x + 1)^2, the equation becomes:$$10\frac{x^4}{(x^2 + x + 1)^2} - 7\frac{x^2}{x^2 + x + 1} + 1 = 0$$Let z = \frac{x^2}{x^2 + x + 1}. The equation now is$$10z^2 - 7z + 1 = 0$$Solving it like an ordinary quadratic equation on z you get at most two roots z_{1,2}. Then let \frac{x^2}{x^2 + x + 1} = z_1, or, equivalently,$$x^2 = z_1 ... 3 You have: $$x + y= 2-z\\ xy = 2 + \frac{z^2}{2}.$$ From Vieta's formulas,$x$and$y$are roots of $$t^2 - (2-z)t + 2 + \frac{z^2}{2} = 0;$$ hence $$t = \frac{2 - z \pm \sqrt{(2-z)^2 - 8 - 2z^2}}{2} = \frac{2 - z \pm \sqrt{-(z+2)^2}}{2}.$$ So, we have$z=-2$, and$t=2$; therefore$t=x=y=2$and$z=-2$. Greate thanks to @Shailesh for notes. 3 For a method which is intermediate in cleverness between the one given by the other answers and just solving the system of simultaneous equations, you could use successive differences. We write the following table: $$\begin{array}{ccccccccccccccc} h & & 1 & & 2 & & 3 & & 4 & & 5 & & 6 & & x_6 \\ & ... 3 Solution for lazy people. If f(0) were 0 then the answer would be 7, because fixing the 7 points leads to a unique solution and f(x) = x is obviously a solution. Now, suppose we change f(0) to some value h. One can then argue that the function g(x) = f(x) - x will have the symmetry g(7-x) = g(x). Therefore adding h to f(0) will increase the value of f(7) ... 3 Hint: This is based on @columbus8myhw's hint. Consider the polynomial$$kf(k)-1$$which has degree 99 is zero for all k=1,2,3, \cdots, 99. So you know the polynomial kf(k)-1 has the form$$C(k-1)(k-2)(k-3)\cdots(k-99)$$and you can find C when you put k=0. In general, its a good idea to find a polynomial that has as many zeros as its degree ... 3 Assuming that all the roots \xi_1,\xi_2,\xi_3,\xi_4,\xi_5 of your polynomial are real,$$ (\xi_1+\xi_2+\xi_3+\xi_4+\xi_5)^2 \leq 5(\xi_1^2+\xi_2^2+\xi_3^2+\xi_4^2+\xi_5^2) \tag{1}$$must hold: it is the Cauchy-Schwarz inequality. On the other hand, by Viète's theorem the LHS of (1) is a_1^2, while the RHS of (1) is 5(a_1^2-2a_2). So if all the ... 3 So if you think about$$ (1 + x^5 + x^7)^{20} $$That intuitively is just$$ ((1 + x^5) + x^7) \times ((1 + x^5) + x^7) \times ((1 + x^5) + x^7) ... $$Which can be expanded out term by term. By the Binomial Theorem as$$ (1 + x^5)^{20} (x^7)^0 + \begin{pmatrix} 20 \\ 1\end{pmatrix}(1 + x^5)^{19}x^7 + \begin{pmatrix} 20 \\ 2\end{pmatrix}(1 + ... 3$(1+x^5+x^7)^{20}=\{(1+x^5)+x^7\}^{20}=(1+x^5)^{20}+\binom{20}1(1+x^5)^{20-1}(x^7)^1+\binom{20}2(1+x^5)^{20-2}(x^7)^2+\cdots+(x^7)^{20}$So the required sum will be the coefficient of$x^{17}$in$(1+x^5)^{20}+\binom{20}1\cdot$the coefficient of$x^{17-7}$in$(1+x^5)^{20-1}+\binom{20}2\cdot$the coefficient of$x^{17-7\cdot2}\$ in ...

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