Chapter 2.8, Problem 12E

Solution

To Solve:

The polynomial inequality $x^3-4x^2+x+6\le 0$ , by completing the factoring if needed and then using a sign chart.

To Graph:

Support the solution graphically.

  $x\in \left(-\infty ,-1\right]\cup \left[2,3\right]$

Given:

The polynomial inequality $x^3-4x^2+x+6\le 0$ .

Concepts Used:

The Rational Zeroes Theorem: A polynomial of degree $n$ , $a_n{x}^n+a_{n-1}{x}^{n-1}+\dots +a_2{x}^2+a_{1}x+a_0$ , where $a_n,a_0\ne 0$ , has all its rational zeroes among the numbers obtained by the expression $\pm \frac{{\text{Factors of a}}_0}{\text{Factors of }{a}_n}$ .

Factor Theorem. If $a$ is a zero of polynomial $p\left(x\right)$ then $x-a$ is a factor of $p\left(x\right)$ .

Synthetic division of polynomials.

Splitting the linear term to factorize quadratic polynomials.

The graph of $y=f\left(x\right)$ where $f\left(x\right)$ is a polynomial in one variable crosses the $x$ -axis at the zeroes of $f\left(x\right)$ . It changes sign at a particular zero if the multiplicity of that zero is odd, it undergoes no such sign change if the multiplicity of that zero is even.

If ${\left(ax+b\right)}^n$ is a factor of the polynomial $f\left(x\right)$ then $-\frac{b}{a}$ is a zero of $f\left(x\right)$ with multiplicity $n$ .

Polynomials are continuous everywhere in $ℝ$ . So they can change signs only at the points where they cross the $x$ -axis.

The solution to $f\left(x\right)\le 0$is the region in the graph of $y=f\left(x\right)$ where the curve lies on or below the $x$ -axis. This represents the region where $f\left(x\right)$ is not positive.

Calculations:

For the given polynomial $x^3-4x^2+x+6$ , all its rational zeroes are among the numbers obtained by the expression $\pm \frac{{\text{Factors of a}}_0=6}{\text{Factors of }{a}_3=1}$ .

Factors of $6$ are $1,2,3,6$ . Factors of $1$ is $1$ . The possible values of the expression $\pm \frac{{\text{Factors of a}}_0=6}{\text{Factors of }{a}_3=1}$ are $\pm 1,\pm 2,\pm 3\text{, and,}\pm 6$ .

Check the possible rational zeroes $\pm 1,\pm 2,\pm 3\text{, and,}\pm 6$ one by one to determine if one of them gives a factor of $x^3-4x^2+x+6$ .

By synthetic division, check if the possible zero $-1$ gives that $x-\left(-1\right)$ is a factor of $x^3-4x^2+x+6$ .

  $\begin{array}{cc}\overline{)-1}& \begin{array}{rrrr}\hfill 1& \hfill -4& \hfill 1& \hfill 6\\ \hfill & \hfill -1& \hfill 5& \hfill -6\\ \hfill 1& \hfill -5& \hfill 6& \hfill 0\end{array}\end{array}$

The last result is zero. Thus $x+1$ is a factor of $x^3-4x^2+x+6$ . In fact, $x^3-4x^2+x+6=\left(x+1\right)\left(1x^2-5x+6\right)$ , where the coefficients of the quadratic polynomial are obtained from the result of synthetic division.

Factorize the polynomial further by splitting the linear term of the quadratic polynomial:

  $\begin{array}{c}{x}^3-4x^2+x+6=\left(x+1\right)\left(x^2-5x+6\right)\\ =\left(x+1\right)\left(x^2-3x-2x+6\right)\\ =\left(x+1\right)\left(x\left(x-3\right)-2\left(x-3\right)\right)\\ =\left(x+1\right)\left(x-3\right)\left(x-2\right)\end{array}$

The completely factorized expression for the polynomial $x^3-4x^2+x+6$ is $\left(x+1\right)\left(x-3\right)\left(x-2\right)$ .

Let $f\left(x\right)=\left(x+1\right)\left(x-3\right)\left(x-2\right)$ , the zeroes of $f\left(x\right)$ are $-1,2\text{, and, }3$ all with an odd multiplicity of $1$ . Thus, $x=-1,2,3$ will all yield a sign change.

Draw the sign chart based on this analysis:

  

  $x^3-4x^2+x+6$ is not positive when $x\in \left(-\infty ,-1\right]\cup \left[2,3\right]$ .

Graph:

Draw the graph of $y=f\left(x\right)$ and shade the region where it lies on or on or below the $x$ -axis.

  

The shaded area is the same as the analytically obtained solution.

Conclusion:

The solution of $x^3-4x^2+x+6\le 0$ is $x\in \left(-\infty ,-1\right]\cup \left[2,3\right]$ .