Chapter 5.8, Problem 42PPS

a.

To determine

To calculate: The zeros of the function $f\left(x\right)=2x^3+7x^2+2x-3\text{ and }g\left(x\right)=2x^3-7x^2+2x+3$ .

Answer to Problem 42PPS

The zeros of the function $f\left(x\right)=2x^3+7x^2+2x-3$ are $-1,-3\text{ and }\frac{1}{2}$ and $g\left(x\right)=2x^3-7x^2+2x+3$ are $1,3\text{ and }-\frac{1}{2}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=2x^3+7x^2+2x-3\text{ and }g\left(x\right)=2x^3-7x^2+2x+3$ .

Formula used:

A polynomial of n degree has n zeros, which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

Calculation:

Consider the function $f\left(x\right)=2x^3+7x^2+2x-3$ .

Observe that degree of polynomial is 3, so the functions has 3 zeros which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(x\right)=2x^3+7x^2+2x-3$

There is 1 positive real zeros.

Now,

  $f\left(-x\right)=-2x^3+7x^2-2x-3$

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(-x\right)=-2x^3+7x^2-2x-3$ .

There are 2 or 0 negative real zero.

Next, construct a table with possible combinations of real and imaginary zeros.

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 1& 0& 2& 1+0+2=3\\ 1& 2& 0& 1+2+0=3\end{array}$

Recall that the Rational zero theorem states that provided a polynomial $P\left(x\right)$ with integral coefficients then every rational zero is of the form $\frac{p}{q}$ that is a rational number in simplest form. Here, p is a factor of the constant term and q is a factor of leading coefficient.

For the provided function leading coefficient is 2 and constant term is 3 Therefore, p is a factor of 3 and q is a factor of 2.

The possible combinations of $\frac{p}{q}$ in simplest form are,

  $\frac{p}{q}=\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}$

Next, construct a table with help of synthetic substitution to compute the value of $f\left(x\right)$ for real values of $x$ .

  $\begin{array}{ccccc}x& 2& 7& 2& -3\\ -1& 2& 5& -3& 0\\ 1& 2& 9& 11& 8\\ -3& 2& 1& -1& 0\\ 3& 2& 13& 41& 120\\ -\frac{1}{2}& 1& 3& -\frac{1}{2}& -\frac{5}{2}\\ \frac{1}{2}& 1& 4& 3& 0\\ -\frac{3}{2}& 1& 2& -2& 3\\ \frac{3}{2}& 1& 5& \frac{17}{2}& \frac{45}{2}\end{array}$

As observed three zeros are resulted at $x=-1,x=-3\text{ and }x=\frac{1}{2}$ .

Thus, the zeros of the function $f\left(x\right)=2x^3+7x^2+2x-3$ are $-1,-3\text{ and }\frac{1}{2}$ .

Consider the function $g\left(x\right)=2x^3-7x^2+2x+3$ .

Observe that degree of polynomial is 3, so the functions has 3 zeros which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $g\left(x\right)=2x^3-7x^2+2x+3$

There are 2 or 0 positive real zeros.

Now,

  $g\left(-x\right)=-2x^3-7x^2-2x+3$

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $g\left(-x\right)=-2x^3-7x^2-2x+3$ .

There is 1 negative real zero.

Next, construct a table with possible combinations of real and imaginary zeros.

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 0& 1& 2& 0+1+2=3\\ 2& 1& 0& 2+1+0=3\end{array}$

Recall that the Rational zero theorem states that provided a polynomial $P\left(x\right)$ with integral coefficients then every rational zero is of the form $\frac{p}{q}$ that is a rational number in simplest form. Here, p is a factor of the constant term and q is a factor of leading coefficient.

For the provided function leading coefficient is 2 and constant term is 3 Therefore, p is a factor of 3 and q is a factor of 2.

The possible combinations of $\frac{p}{q}$ in simplest form are,

  $\frac{p}{q}=\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}$

Next, construct a table with help of synthetic substitution to compute the value of $f\left(x\right)$ for real values of $x$ .

  $\begin{array}{ccccc}x& 2& -7& 2& 3\\ 1& 2& -5& -3& 0\\ -1& 2& -9& 11& -8\\ 3& 2& -1& -1& 0\\ -3& 2& -13& 41& -120\\ \frac{1}{2}& 1& -3& -\frac{1}{2}& \frac{5}{2}\\ -\frac{1}{2}& 1& -4& 3& 0\\ \frac{3}{2}& 1& -2& -2& 3\\ -\frac{3}{2}& 1& -5& \frac{17}{2}& -\frac{45}{2}\end{array}$

As observed three zeros are resulted at $x=1,x=3\text{ and }x=-\frac{1}{2}$ .

Thus, the zeros of the function $g\left(x\right)=2x^3-7x^2+2x+3$ are $1,3\text{ and }-\frac{1}{2}$ .

b.

To determine

To describe: The function among $f\left(x\right)=2x^3+7x^2+2x-3\text{ and }g\left(x\right)=2x^3-7x^2+2x+3$ depicted by the graph.

  

Answer to Problem 42PPS

The function represented by graph is $g\left(x\right)=2x^3-7x^2+2x+3$

Explanation of Solution

Given information:

The functions $f\left(x\right)=2x^3+7x^2+2x-3\text{ and }g\left(x\right)=2x^3-7x^2+2x+3$ .

Consider the graph provided below,

  

Observe the points at which the graph intersects the x -axis.

Such points are $x=1,x=3\text{ and }x=-\frac{1}{2}$ .

This means the zeroes of the function depicted by the graph are $x=1,x=3\text{ and }x=-\frac{1}{2}$ .

From the above calculations the function $g\left(x\right)=2x^3-7x^2+2x+3$ has zeroes as $1,3\text{ and }-\frac{1}{2}$ .

Therefore, the graph of the function represented by graph is $g\left(x\right)=2x^3-7x^2+2x+3$ .