Chapter 2.8, Problem 1E

Solution

(a)

To Determine:

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

  $x=-2,-1,5$ .

Given:

The polynomial function $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ .

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.

Calculations:

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

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

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

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

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

  $\begin{array}{cc}& x-5=0\\ \Rightarrow & x=5\end{array}$

Conclusion:

The polynomial function $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ of degree $3$ is zero for the three values $x=-2,-1,5$ .

(b)

To Determine:

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

  $x\in \left(-2,-1\right)\cup \left(5,\infty \right)$ .

Given:

The polynomial function $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ .

Known from previous part:

The polynomial function $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ of degree $3$ is zero for the three values $x=-2,-1,5$ .

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 $x=-2,-1,5$ of the polynomial function $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ divide the set of real numbers $ℝ$ into four open intervals, namely, $\left(-\infty ,-2\right)$ , $\left(-2,-1\right)$ , $\left(-1,5\right)$ , and $\left(5,\infty \right)$ . 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 $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ in the interval $\left(-\infty ,-2\right)$ . Take a test point $x=-3$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)\\ \Rightarrow & f\left(-3\right)=\left(-3+2\right)\left(-3+1\right)\left(-3-5\right)\\ \Rightarrow & f\left(-3\right)=\left(-1\right)\left(-2\right)\left(-8\right)\\ \Rightarrow & f\left(-3\right)=-16\\ \Rightarrow & f\left(-3\right)<0\end{array}$

  $f\left(x\right)$ is negative in the interval $\left(-\infty ,-2\right)$ .

Determine the sign of $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ in the interval $\left(-2,-1\right)$ . Take a test point $x=-1.5$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)\\ \Rightarrow & f\left(-1.5\right)=\left(-1.5+2\right)\left(-1.5+1\right)\left(-1.5-5\right)\\ \Rightarrow & f\left(-1.5\right)=\left(0.5\right)\left(-0.5\right)\left(-6.5\right)\\ \Rightarrow & f\left(-1.5\right)=1.625\\ \Rightarrow & f\left(-1.5\right)>0\end{array}$

  $f\left(x\right)$ is positive in the interval $\left(-2,-1\right)$ .

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

  $\begin{array}{cc}& f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)\\ \Rightarrow & f\left(0\right)=\left(0+2\right)\left(0+1\right)\left(0-5\right)\\ \Rightarrow & f\left(0\right)=\left(2\right)\left(1\right)\left(-5\right)\\ \Rightarrow & f\left(0\right)=-10\\ \Rightarrow & f\left(0\right)<0\end{array}$

  $f\left(x\right)$ is negative in the interval $\left(-1,5\right)$ .

Determine the sign of $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ in the interval $\left(5,\infty \right)$ . Take a test point $x=6$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)\\ \Rightarrow & f\left(6\right)=\left(6+2\right)\left(6+1\right)\left(6-5\right)\\ \Rightarrow & f\left(6\right)=\left(8\right)\left(7\right)\left(1\right)\\ \Rightarrow & f\left(6\right)=56\\ \Rightarrow & f\left(6\right)>0\end{array}$

  $f\left(x\right)$ is positive in the interval $\left(5,\infty \right)$ .

Conclusion:

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

The polynomial function $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ is negative in the intervals $\left(-\infty ,-2\right)$ and $\left(-1,5\right)$ .

(c)

To Determine:

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

  $x\in \left(-\infty ,-2\right)\cup \left(-1,5\right)$ .

Given:

The polynomial function $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ .

Known from previous part:

The polynomial function $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ is negative in the intervals $\left(-\infty ,-2\right)$ and $\left(-1,5\right)$ .

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 $f\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x-5\right)$ is negative in the intervals $\left(-\infty ,-2\right)$ and $\left(-1,5\right)$ .