Chapter 2.8, Problem 5E

Solution

(a)

To Determine:

The values of $x$ for which the polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ is zero.

  $x=-1,8$ .

Given:

The polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ .

Concepts Used:

The zeroes of a polynomial function are the set of zeroes of all its factors.

The degree of a polynomial of one variable is the sum of the degrees of all its factors.

A polynomial of degree $n$ has may have as many as $n$ zeroes.

Calculations:

Determine the zeroes of the factor $\left(2x^2+5\right)$ of the polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ .

  $\begin{array}{cc}& 2x^2+5=0\\ \Rightarrow & 2x^2=-5\\ \Rightarrow & x^2=-\frac{5}{2}\end{array}$

Note that $\left(2x^2+5\right)$ cannot be zero for any real number $x$ because there is no real number whose square is a negative number.

Determine the zeroes of the factor $\left(x-8\right)$ of the polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ .

  $\begin{array}{cc}& x-8=0\\ \Rightarrow & x=8\end{array}$

Determine the zeroes of the factor $\left(x+1\right)$ of the polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ .

  $\begin{array}{cc}& x+1=0\\ \Rightarrow & x=-1\end{array}$

Conclusion:

The polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ of degree $7$ is zero for the two values $x=-1,8$ .

(b)

To Determine:

The values of $x$ for which the polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ is positive.

  $x\in \left(-1,8\right)\cup \left(8,\infty \right)$ .

Given:

The polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ .

Known from previous part:

The polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ of degree $7$ is zero for the two values $x=-1,8$ .

Concepts Used:

The zeroes of a polynomial function are the set of zeroes of all its factors.

The degree of a polynomial of one variable is the sum of the degrees of all its factors.

A polynomial of degree $n$ has may have as many as $n$ zeroes.

The sign of a polynomial function $f\left(x\right)$ in open interval $\left(a,b\right)$ may change only if there exists at least one $c\in \left(a,b\right)$ such that $f\left(c\right)=0$ . In other words, the value of a polynomial function may not change sign unless it passes one of its zeros on the $x$ -axis.

Calculations:

The zeroes $x=-1,8$ of the polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ divide the set of real numbers $ℝ$ into three open intervals, namely, $\left(-\infty ,-1\right)$ , $\left(-1,8\right)$ , and $\left(8,\infty \right)$ . Since none of these intervals include any zero of $f\left(x\right)$ , it can be said that $f\left(x\right)$ cannot change sign in any of these intervals.

Determine the sign of $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ in the interval $\left(-\infty ,-1\right)$ . Take a test point $x=-2$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3\\ \Rightarrow & f\left(-2\right)=\left(2{\left(-2\right)}^2+5\right){\left(-1-8\right)}^2{\left(-2+1\right)}^3\\ \Rightarrow & f\left(-2\right)=\left(8+5\right){\left(-9\right)}^2{\left(-1\right)}^3\\ \Rightarrow & f\left(-2\right)=13\times 81\times -1\\ \Rightarrow & f\left(-2\right)=-1053\\ \Rightarrow & f\left(-2\right)<0\end{array}$

  $f\left(x\right)$ is negative in the interval $\left(-\infty ,-1\right)$ .

Determine the sign of $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ in the interval $\left(-1,8\right)$ . Take a test point $x=0$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3\\ \Rightarrow & f\left(0\right)=\left(2{\left(0\right)}^2+5\right){\left(0-8\right)}^2{\left(0+1\right)}^3\\ \Rightarrow & f\left(0\right)=\left(0+5\right){\left(-8\right)}^2{\left(1\right)}^3\\ \Rightarrow & f\left(0\right)=5\times 62\times 1\\ \Rightarrow & f\left(0\right)=310\\ \Rightarrow & f\left(0\right)>0\end{array}$

  $f\left(x\right)$ is positive in the interval $\left(-1,8\right)$ .

Determine the sign of $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ in the interval $\left(8,\infty \right)$ . Take a test point $x=9$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3\\ \Rightarrow & f\left(9\right)=\left(2{\left(9\right)}^2+5\right){\left(9-8\right)}^2{\left(9+1\right)}^3\\ \Rightarrow & f\left(9\right)=\left(36+5\right){\left(1\right)}^2{\left(10\right)}^3\\ \Rightarrow & f\left(9\right)=41\times 1\times 1000\\ \Rightarrow & f\left(9\right)=41000\\ \Rightarrow & f\left(9\right)>0\end{array}$

  $f\left(x\right)$ is positive in the interval $\left(8,\infty \right)$ .

Conclusion:

The polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ is positive in the intervals $\left(-1,8\right)$ and $\left(8,\infty \right)$ .

The polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ is negative in the interval $\left(-\infty ,-1\right)$ .

(c)

To Determine:

The values of $x$ for which the polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ is negative.

  $x\in \left(-\infty ,-1\right)$ .

Given:

The polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ .

Known from previous part:

The polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ is negative in the interval $\left(-\infty ,-1\right)$ .

Concepts Used:

The zeroes of a polynomial function are the set of zeroes of all its factors.

The degree of a polynomial of one variable is the sum of the degrees of all its factors.

A polynomial of degree $n$ has may have as many as $n$ zeroes.

The sign of a polynomial function $f\left(x\right)$ in open interval $\left(a,b\right)$ may change only if there exists at least one $c\in \left(a,b\right)$ such that $f\left(c\right)=0$ . In other words, the value of a polynomial function may not change sign unless it passes one of its zeros on the $x$ -axis.

Conclusion:

The polynomial function $f\left(x\right)=\left(2x^2+5\right){\left(x-8\right)}^2{\left(x+1\right)}^3$ is negative in the interval $\left(-\infty ,-1\right)$ .