Answer to Question #307675 in Algebra for Ankit

Question

Consider the polynomial f(x) = x

5 − 34x

3 + 29x

2 + 212x − 300.

(a) Find upperbounds and lowerbounds for the real roots of f(x).

(b) What are the possible integral roots of f(x)?

(c) Illustrate synthetic division by showing that one of the integers obtained in (b)

is/isn’t a root of f(x).

(d) Illustrate Newton’s method by showing that one of the integers obtained in (b)

is/isn’t a root of f(x)

Answer 1

$f(x)=x^5-34x^3+29x^2+212x-300\\ (a) upperbounds\\ N=1+\sqrt[2]{\frac{300}{1}}\approx18\\ \begin{matrix} &1&0&-34&29&212&-300 \\ 17 & 1&17&255&>0&>0&>0\\ 10&1&10&66&>0&>0&>0\\ 6&1&6&2&41&458&>0\\ 5&1&5-9<0 \end{matrix}\\ upperbounds f(x)=5$

$lowerbounds\\ f(-x)=-x^5+34x^3+29x^2-212x-300\\ f_1(x)=-f(-x)=\\ =x^5-34x^3-29x^2+212x+300\\ N=1+\sqrt[2]{\frac{34}{1}}\approx6\\ \begin{matrix} &1&0&-34&-29&212&300 \\ 5 & 1&5&-9<0 \end{matrix}\\ upperbounds f_1(x)=6\\ lowerbounds f(x)=-6\\ x\in(-6,5)$

$(b)\\ 300:\pm1,\pm2,\pm3,\pm4,\pm5,\pm6,\pm10,\pm12,\pm15,\pm20,\\\pm25,\pm30,\pm50,\pm60,\pm75,\pm100,\pm150,\pm300\\ x\in(-6,5)\\ \alpha:\pm1,\pm2,\pm3,\pm4,\pm5\\ f(1)=1-34+29+212-300=-92\\ f(-1)=-1+34+29-212-300=-450\\ \frac{f(1)}{\alpha-1}\in Z, \frac{f(-1)}{\alpha+1}\in Z\\ -\frac{92}{\alpha-1}\in Z, -\frac{450}{\alpha-1}\in Z\\ \alpha=2,-3\\ \begin{matrix} &1&0&-34&29&212&-300 \\ 2 & 1&2&-30&-31&-150&0\\ -3&1&-1&-27&50&0 \end{matrix}\\ x_1=-3, x_2=2 - roots f(x)$

$(c)\\ f(x):x+3\\ \begin{matrix} x^5-34x^3+29x^2+212x-300 & |x+3 \\ x^5+3x^4& x^4-3x^3-25x^2-104x-100\\ -------\\ -3x^4-34x^3+29x^2+212x-300\\ -3x^4-9x^3\\ -------\\ -25x^3+29x^2+212x-300\\ -25x^3-75x^2\\ -------\\ 104x^2+212x-300\\ 104x^2+312x\\ -------\\ -100x-300\\ -100x-300\\ -------\\0 \end{matrix}$

$(d)\\ f(x)= x^5-34x^3+29x^2+212x-300\\ f'(x)=5x^3-102x^2+58x+212\\ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\\ x_n=-3\\ f(-3)=(-3)^5-34(-3)^3+29(-3)^2+\\+212(-3)-300=0\\ f'(-3)=5(-3)^4-102(-3)^2+58(-3)+212=-475\\ x_{n+1}=-3-\frac{0}{-475}=-3$