Question #307675

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)


1
Expert's answer
2022-03-20T14:39:51-0400

f(x)=x534x3+29x2+212x300(a)upperboundsN=1+300121810342921230017117255>0>0>01011066>0>0>0616241458>05159<0upperboundsf(x)=5f(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

lowerboundsf(x)=x5+34x3+29x2212x300f1(x)=f(x)==x534x329x2+212x+300N=1+341261034292123005159<0upperboundsf1(x)=6lowerboundsf(x)=6x(6,5)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:±1,±2,±3,±4,±5,±6,±10,±12,±15,±20,±25,±30,±50,±60,±75,±100,±150,±300x(6,5)α:±1,±2,±3,±4,±5f(1)=134+29+212300=92f(1)=1+34+29212300=450f(1)α1Z,f(1)α+1Z92α1Z,450α1Zα=2,31034292123002123031150031127500x1=3,x2=2rootsf(x)(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+3x534x3+29x2+212x300x+3x5+3x4x43x325x2104x1003x434x3+29x2+212x3003x49x325x3+29x2+212x30025x375x2104x2+212x300104x2+312x100x300100x3000(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)=x534x3+29x2+212x300f(x)=5x3102x2+58x+212xn+1=xnf(xn)f(xn)xn=3f(3)=(3)534(3)3+29(3)2++212(3)300=0f(3)=5(3)4102(3)2+58(3)+212=475xn+1=30475=3(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


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