Chapter 5, Problem 10CT

Solution

To calculate : The number of complex roots, the possible number of real roots and the possible rational roots.

The given polynomial has one complex root, three real roots, and no rational roots.

Given information :

  $3x^4+5x^3-2x^2+x-9=0$ .

Concept used:

Fundamental Theorem: If the degree of the polynomial is $n$ , which is more than or equal to one, then the polynomial will have exactly $n$ number of roots including real and complex roots.

Rational Root Theorem: For a given polynomial with integer coefficients, the rational roots will be in form of $\frac{p}{q}$ , where $p$ is the integer factor of the constant term and $q$ is the integer factor of the leading coefficient.

Calculation :

To find the possible number of real roots use Descartes' rule.

  $P\left(x\right)=3x^4+5x^3-2x^2+x-9$

The number of positive real zeroes is equal to the number of sign changes in consecutive coefficients of $P\left(x\right)$ or less than that by an even number.

  $\left\{\begin{array}{c}P\left(x\right)=3x^4+5x^3-2x^2+x-9\\ \text{Signs of cefficients}=++-+-\\ \text{Number of sign changes }=2\end{array}\right\}$

So, there are two positive real zeros.

The number of negative real zeroes is equal to the number of sign changes in consecutive coefficients of $P\left(-x\right)$ or less than that by an even number.

  $\left\{\begin{array}{c}P\left(x\right)=3x^4+5x^3-2x^2+x-9\\ P\left(-x\right)=3x^4-5x^3-2x^2-x-9\\ \text{Signs of cefficients}=+----\\ \text{Number of sign changes }=1\end{array}\right\}$

So, there is only one negative real zero.

Use the Fundamental Theorem to find the number of roots including both real and complex.

Here, the degree of polynomial is $4$ , so it will have exactly four roots out of which three roots are real so the possible number of complex root is $1$ .

Use Rational Root Theorem to find rational roots.

In the given polynomial, the constant term is $9$ and the leading coefficient is $3$ .

The factors of $9$ are $\pm 1,\pm 3,\text{ and}\pm 9$ and of $3$ are $\pm 1\text{ and }\pm 3$ .

So, the rational roots in term of $\frac{p}{q}$ will be $\pm 1,\pm 3,\pm 9\text{ and}\pm \frac{1}{3}$ .

Use synthetic division method to test the roots.

  $\begin{array}{l}\left\{\begin{array}{l}\overline{)1}35-21-9\\ 3867\\ \overline{3867\overline{)-2}}\end{array}\right\}\\ \left\{\begin{array}{l}\overline{)-1}35-21-9\\ -3-2\text{ 4 }-5\\ \overline{32-45\overline{)-14}}\end{array}\right\}\end{array}$

As the remainder is not equal to zero, so $\pm 1$ is not a root of the given equation.

  $\begin{array}{l}\left\{\begin{array}{l}\overline{)3}35-21-9\\ 942\text{ 120 }363\\ \overline{31440121\overline{)354}}\end{array}\right\}\\ \left\{\begin{array}{l}\overline{)-3}35-21-9\\ -912-3087\\ \overline{3-410-29\overline{)78}}\end{array}\right\}\end{array}$

As the remainder is not equal to zero, so $\pm 3$ is not a root of the given equation.

  $\begin{array}{l}\left\{\begin{array}{l}\overline{)9}35-21-9\\ 27\text{ 28}8\text{ 2574 }23175\\ \overline{332286\text{ 25}75\overline{)23166}}\end{array}\right\}\\ \left\{\begin{array}{l}\overline{)-9}35-21-9\\ -27198-1\text{764 }15867\\ \overline{3-22196-1763\overline{)15858}}\end{array}\right\}\end{array}$

As the remainder is not equal to zero, so $\pm 9$ is not a root of the given equation.

  $\begin{array}{l}\left\{\begin{array}{l}\overline{)\frac{1}{3}}35-21-9\\ 12\text{ 0 }\frac{1}{3}\\ \overline{3601\overline{)-\frac{26}{3}}}\end{array}\right\}\\ \left\{\begin{array}{l}\overline{)-\frac{1}{3}}35-21-9\\ -1-\frac{4}{3}\frac{10}{9}-\frac{19}{27}\\ \overline{34-\frac{10}{3}\frac{19}{9}\overline{)-\frac{262}{27}}}\end{array}\right\}\end{array}$

As the remainder is not equal to zero, so $\pm \frac{1}{3}$ is not a root of the given equation.

Therefore, the given polynomial has one complex root, three real roots, and no rational roots.