Chapter 8.6, Problem 6ST

a.

To determine

To write the possibilities of numbers of positive and negative real roots using Descartes’ rule of signs.

Answer to Problem 6ST

Since $P\left(x\right)=0$ has three roots in all, there are only one possibility.

Number of Positive real roots	Number of negative real roots
3	1
1	1

Explanation of Solution

Given information:

The given equation is $2x^4-9x^3+13x^2-x-5=0$

Formula used:

Descartes’ Rule of Signs

Let $P\left(x\right)$ be a simplified polynomial with real coefficients and terms arranged in decreasing degree of x.

1. The number of positive real roots of $P\left(x\right)=0$ equals to the number of variations of sign of $P\left(x\right)$ or is fewer than this number by an even integer.
2. The number of negative real roots of $P\left(x\right)=0$ equals to the number of variations of sign of $P\left(-x\right)$ or is fewer than this number by an even integer.

Calculation :

First write the polynomial

  $P\left(x\right)=2x^4-9x^3+13x^2-x-5=0$

1. $P\left(x\right)$ Has three variation in sign, so by part (1) of Descartes’ rule the number of positive real roots of $P\left(x\right)=0$is 3 or 1.
2. $P\left(-x\right)=2x^4+9x^3+13x^2+x-5=0$

  $P\left(-x\right)$ Has only one variations in sign, so by part (2) of Descartes’ rule the number of negative real roots of $P\left(x\right)=0$ is 1.

3. Since $P\left(x\right)=0$has fourroots in all, there are only two possibility.

Number of Positive real roots	Number of negative real roots
3	1
1	1

b.

To determine

To solve the equation.

Answer to Problem 6ST

The roots of the equation is $2+i,2-i,1,\text{ and }\frac{-1}{2}$

Explanation of Solution

Given information:

The equation is $2x^4-9x^3+13x^2-x-5=0$ and its one root is $2+i$

Calculation:

Since $2+i$ is a root, $2-i$ must also be a root.

Therefore the factors are $x-\left(2+i\right),x-\left(2-i\right)$

The multiplication of the factors is

  $\left(x-\left(2+i\right)\right)\left(x-\left(2-i\right)\right)=x^2-4x+5$

To get other factor of the equation divide the equation by this product.

  $\begin{array}{l}\text{ 2}{x}^2-x-1\\ x^2-4x+5)\overline{2x^4-9x^3+13x^2-x-5}\\ -2x^4+8x^3-10x^2\\ \overline{\begin{array}{l}-x^3\text{+3}{x}^2-x\\ x^3-4x^2-5x\\ \overline{\begin{array}{l}-x^2+4x-5\\ x^2-4x+5\\ \overline{\text{ 0 }}\end{array}}\end{array}}\end{array}$

The quotient is $\text{2}{x}^2-x-1$ , factorise the quotient.

  $\text{2}{x}^2-x-1=\left(2x+1\right)\left(x-1\right)\left[\text{factorise}\right]$

Equate the factors to zero to get the roots

  $\begin{array}{l}2x+1=0\\ 2x=-1\\ x=-\frac{1}{2}\\ or,x-1=0\\ x=1\end{array}$

The roots of the equation is $2+i,2-i,1,\text{ and }\frac{-1}{2}$