Chapter 4.1, Problem 34E

To determine

To write:a polynomial equation of least degree for $-2,-0.5,4$ and find the degree as even or odd. Also find the times graph of the related function cross the x-axis.

Answer to Problem 34E

  $2x^3-3x^2-18x-8=0$

The degree of this equation is 3, odd degree

Since all the 3 roots are real, the graph will cross the x-axis thrice

Explanation of Solution

Given information: $-2,-0.5,4$

Calculation:

If $x=-2,$ then

  $x+2\text{ is }$ a factor of the polynomial.

If $x=-0.5,$ then $x=-\frac{1}{2}$

  $2x=-1,$

Then, $2x+1$ is a factor of the polynomial.

Likewise, if $x=4$ then $x-4$ is a factor of the polynomial.

Therefore, the linear factors for the polynomial are

  $\begin{array}{l}\left(x+2\right)\\ \left(2x+1\right)\\ \left(x-4\right)\end{array}$

Now find the product of these factors.

  $\left(x+2\right)\left(2x+1\right)\left(x-4\right)=0$

Multiply the factors

  $\begin{array}{l}\left(x+2\right)\left(x-4\right)\left(2x+1\right)=0\\ \left(x^2-2x-8\right)\left(2x+1\right)=0\\ 2x^3-4x^2-16x+x^2-2x-8=0\\ 2x^3-3x^2-18x-8=0\end{array}$

A polynomial equation with roots $-2,-0.5,4$ is

  $2x^3-3x^2-18x-8=0$

The degree of this equation is 3. Thus the equation has an odddegree since 3 is an odd number.

Since all the 3 roots are real, the graph will cross the x-axis thrice.