Chapter 2.4, Problem 77E

Solution

(a)

To find: The zeroes of f(x) related to zeroes of g(x).

The explanation is given.

Given information

The function $f(x)=\frac{1}{3}{x}^3+x^2+2x-3$and $g(x)=x^3+3x^2+6x-9$

In f(x) leading term coefficient is $q=\frac{1}{3}$ all the possible factors of it are $\pm 1,\pm \frac{1}{3}$ and coefficient of trailing term is $p=3$ all possible factor of 3 are $\pm 3,\pm 1$ .now all the possible rational zeroes are

  $\frac{p}{q}=\frac{\pm 3}{\pm 1},\frac{\pm 3}{\frac{\pm 1}{3}}=\pm 9$ .

The leading coefficient of g(x) is $q=1$ and trailing coefficient is $p=9$ .all possible factors of 9 are $\pm 1,\pm 3,\pm 9$ .Hence all the rational number of g(x) are $\frac{p}{q}=\frac{\pm 3}{\pm 1},\frac{\pm 9}{\pm 1},$ It is clear that all the rational roots of f(x) and g(x) are equal.

(b)

To find: The zeroes of f(x) .

The rational zeroes of f(x) are $\frac{p}{q}=\frac{\pm 7}{2}$

Given information

Function $f(x)=x^3-\frac{7}{6}{x}^2-\frac{20}{3}x+\frac{7}{2}$

In f(x) leading term coefficient is $q=1$ all the possible factors of it are $\pm 1$ and coefficient of trailing term is $p=\frac{7}{2}$ all possible factor of $\frac{7}{2}$ are $\pm \frac{7}{2},\pm 1$ .now all the possible rational zeroes are

  $\frac{p}{q}=\frac{\pm 7}{2}$ .

(c)

To find: The zeroes of f(x) .

The rational zeroes of f(x) are $\frac{p}{q}=\frac{\pm 5}{2}$

Given information$f(x)=x^3-\frac{5}{2}{x}^2-\frac{37}{12}x+\frac{5}{2}$

In f(x) leading term coefficient is $q=1$ all the possible factors of it are $\pm 1$ and coefficient of trailing term is $p=\frac{5}{2}$ all possible factor of $\frac{5}{2}$ are $\pm \frac{5}{2},\pm 1$ .now all the possible rational zeroes are

  $\frac{p}{q}=\frac{\pm 5}{2}$ .