Chapter 2.8, Problem 4E

Solution

(a)

To Determine:

The values of $x$ for which the polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ is zero.

  $x=-\frac{3}{5},1$ .

Given:

The polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ .

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.

Calculations:

Determine the zeroes of the factor $\left(5x+3\right)$ of the polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ .

  $\begin{array}{cc}& 5x+3=0\\ \Rightarrow & x=-\frac{3}{5}\end{array}$

Determine the zeroes of the factor $\left(x^2+6\right)$ of the polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ .

  $\begin{array}{cc}& x^2+6=0\\ \Rightarrow & x^2=-6\end{array}$

Note that $\left(x^2+6\right)$ 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 $\left(x-1\right)$ of the polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ .

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

Conclusion:

The polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ of degree $4$ is zero for the two values $x=-\frac{3}{5},1$ .

(b)

To Determine:

The values of $x$ for which the polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ is positive.

  $x\in \left(-\infty ,-\frac{3}{5}\right)\cup \left(1,\infty \right)$ .

Given:

The polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ .

Known from previous part:

The polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ of degree $4$ is zero for the two values $x=-\frac{3}{5},1$ .

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 $x=-\frac{3}{5},1$ of the polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ divide the set of real numbers $ℝ$ into three open intervals, namely, $\left(-\infty ,-\frac{3}{5}\right)$ , $\left(-\frac{3}{5},1\right)$ , and $\left(1,\infty \right)$ . 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 $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ in the interval $\left(-\infty ,-\frac{3}{5}\right)$ . Take a test point $x=-1$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)\\ \Rightarrow & f\left(-1\right)=\left(5\left(-1\right)+3\right)\left({\left(-1\right)}^2+6\right)\left(-1-1\right)\\ \Rightarrow & f\left(-1\right)=\left(-5+3\right)\left(1+6\right)\left(-2\right)\\ \Rightarrow & f\left(-1\right)=-2\times 7\times -2\\ \Rightarrow & f\left(-1\right)=28\\ \Rightarrow & f\left(-1\right)>0\end{array}$

  $f\left(x\right)$ is positive in the interval $\left(-\infty ,-\frac{3}{5}\right)$ .

Determine the sign of $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ in the interval $\left(-\frac{3}{5},1\right)$ . Take a test point $x=0$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)\\ \Rightarrow & f\left(0\right)=\left(5\left(0\right)+3\right)\left({\left(0\right)}^2+6\right)\left(0-1\right)\\ \Rightarrow & f\left(0\right)=\left(0+3\right)\left(0+6\right)\left(-1\right)\\ \Rightarrow & f\left(0\right)=3\times 6\times -1\\ \Rightarrow & f\left(0\right)=-18\\ \Rightarrow & f\left(0\right)<0\end{array}$

  $f\left(x\right)$ is negative in the interval $\left(-\frac{3}{5},1\right)$ .

Determine the sign of $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ in the interval $\left(1,\infty \right)$ . Take a test point $x=2$ in the interval and evaluate $f\left(x\right)$ :

  $\begin{array}{cc}& f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)\\ \Rightarrow & f\left(2\right)=\left(5\left(2\right)+3\right)\left({\left(2\right)}^2+6\right)\left(2-1\right)\\ \Rightarrow & f\left(2\right)=\left(10+3\right)\left(4+6\right)\left(1\right)\\ \Rightarrow & f\left(2\right)=13\times 10\times 1\\ \Rightarrow & f\left(2\right)=130\\ \Rightarrow & f\left(2\right)>0\end{array}$

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

Conclusion:

The polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ is positive in the intervals $\left(-\infty ,-\frac{3}{5}\right)$ and $\left(1,\infty \right)$ .

The polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ is negative in the interval $\left(-\frac{3}{5},1\right)$ .

(c)

To Determine:

The values of $x$ for which the polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ is negative.

  $x\in \left(-\frac{3}{5},1\right)$ .

Given:

The polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ .

Known from previous part:

The polynomial function $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ is negative in the interval $\left(-\frac{3}{5},1\right)$ .

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 $f\left(x\right)=\left(5x+3\right)\left(x^2+6\right)\left(x-1\right)$ is negative in the interval $\left(-\frac{3}{5},1\right)$ .