Chapter 2.8, Problem 3E

Solution

(a)

To Determine:

The values of $x$ for which the polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ is zero.

  $x=-7,-4,6$ .

Given:

The polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ .

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 $\left(x+7\right)$ of the polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ .

  $\begin{array}{cc}& x+7=0\\ \Rightarrow & x=-7\end{array}$

Determine the zeroes of the factor $\left(x+4\right)$ of the polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ .

  $\begin{array}{cc}& x+4=0\\ \Rightarrow & x=-4\end{array}$

Determine the zeroes of the factor $\left(x-6\right)$ of the polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ .

  $\begin{array}{cc}& x-6=0\\ \Rightarrow & x=6\end{array}$

Conclusion:

The polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ of degree $4$ is zero for the three values $x=-7,-4,6$ .

(b)

To Determine:

The values of $x$ for which the polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ is positive.

  $x\in \left(-\infty ,-7\right)\cup \left(-4,6\right)\cup \left(6,\infty \right)$ .

Given:

The polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ .

Known from previous part:

The polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ of degree $3$ is zero for the three values $x=-7,-4,6$ .

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 $x=-7,-4,6$ of the polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ divide the set of real numbers $ℝ$ into four open intervals, namely, $\left(-\infty ,-7\right)$ , $\left(-7,-4\right)$ , $\left(-4,6\right)$ , and $\left(6,\infty \right)$ . 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 $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ in the interval $\left(-\infty ,-7\right)$ . Take a test point $x=-10$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2\\ \Rightarrow & f\left(-10\right)=\left(-10+7\right)\left(-10+4\right){\left(-10-6\right)}^2\\ \Rightarrow & f\left(-10\right)=\left(-3\right)\left(-6\right){\left(-16\right)}^2\\ \Rightarrow & f\left(-10\right)=-3\times -6\times 256\\ \Rightarrow & f\left(-10\right)=4608\\ \Rightarrow & f\left(-10\right)>0\end{array}$

  $f\left(x\right)$ is positive in the interval $\left(-\infty ,-7\right)$ .

Determine the sign of $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ in the interval $\left(-7,-4\right)$ . Take a test point $x=-5$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2\\ \Rightarrow & f\left(-5\right)=\left(-5+7\right)\left(-5+4\right){\left(-5-6\right)}^2\\ \Rightarrow & f\left(-5\right)=\left(2\right)\left(-1\right){\left(-11\right)}^2\\ \Rightarrow & f\left(-5\right)=2\times -1\times 121\\ \Rightarrow & f\left(-5\right)=-242\\ \Rightarrow & f\left(-5\right)<0\end{array}$

  $f\left(x\right)$ is negative in the interval $\left(-7,-4\right)$ .

Determine the sign of $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ in the interval $\left(-4,6\right)$ . Take a test point $x=0$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2\\ \Rightarrow & f\left(0\right)=\left(0+7\right)\left(0+4\right){\left(0-6\right)}^2\\ \Rightarrow & f\left(0\right)=\left(7\right)\left(4\right){\left(-5\right)}^2\\ \Rightarrow & f\left(0\right)=7\times 4\times 25\\ \Rightarrow & f\left(0\right)=700\\ \Rightarrow & f\left(0\right)>0\end{array}$

  $f\left(x\right)$ is positive in the interval $\left(-4,6\right)$ .

Determine the sign of $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ in the interval $\left(6,\infty \right)$ . Take a test point $x=10$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2\\ \Rightarrow & f\left(10\right)=\left(10+7\right)\left(10+4\right){\left(10-6\right)}^2\\ \Rightarrow & f\left(10\right)=\left(17\right)\left(14\right){\left(4\right)}^2\\ \Rightarrow & f\left(10\right)=17\times 14\times 16\\ \Rightarrow & f\left(10\right)=3808\\ \Rightarrow & f\left(10\right)>0\end{array}$

  $f\left(x\right)$ is positive in the interval $\left(6,\infty \right)$ .

Conclusion:

The polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ is positive in the intervals $\left(-\infty ,-7\right)$ , $\left(-4,6\right)$ and $\left(6,\infty \right)$ .

The polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ is negative in the interval $\left(-7,-4\right)$ .

(c)

To Determine:

The values of $x$ for which the polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ is negative.

  $x\in \left(-\infty ,-4\right)\cup \left(-\frac{1}{3},7\right)$ .

Given:

The polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ .

Known from previous part:

The polynomial function $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ is negative in the intervals $\left(-\infty ,-4\right)$ and $\left(-\frac{1}{3},7\right)$ .

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 $f\left(x\right)=\left(x+7\right)\left(x+4\right){\left(x-6\right)}^2$ is negative in the intervals $\left(-\infty ,-4\right)$ and $\left(-\frac{1}{3},7\right)$ .