Chapter 2.8, Problem 11E

Solution

To Solve:

The polynomial inequality $2x^3-3x^2-11x+6\ge 0$ , by completing the factoring if needed and then using a sign chart.

To Graph:

Support the solution graphically.

  $x\in \left[-2,\frac{1}{2}\right]\cup \left[3,\infty \right)$

Given:

The polynomial inequality $2x^3-3x^2-11x+6\ge 0$ .

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.

Factorization of a quadratic polynomial using the identity $a^2-2ab+b^2={(a-b)}^2$ .

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 $f\left(x\right)\ge 0$is the region in the graph of $y=f\left(x\right)$ where the curve lies on or above the $x$ -axis. This represents the region where $f\left(x\right)$ is not negative.

Calculations:

For the given polynomial $2x^3-3x^2-11x+6$ , all its rational zeroes are among the numbers obtained by the expression $\pm \frac{{\text{Factors of a}}_0=6}{\text{Factors of }{a}_3=2}$ .

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

Check the possible rational zeroes $\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2}\text{, and,}\pm \frac{3}{2}$ one by one to determine if one of them gives a factor of $2x^3-3x^2-11x+6$ .

By synthetic division, check if the possible zero $-1$ gives that $x-\left(-1\right)$ is a factor of $2x^3-3x^2-11x+6$ .

  $\begin{array}{cc}\overline{)-1}& \begin{array}{rrrr}\hfill 2& \hfill -3& \hfill -11& \hfill 6\\ \hfill & \hfill -2& \hfill 5& \hfill 6\\ \hfill 2& \hfill -5& \hfill -6& \hfill 12\end{array}\end{array}$

The last result is not zero. Thus $x-\left(-1\right)=x+1$ is not a factor of $2x^3-3x^2-11x+6$ .

By synthetic division, check if the possible zero $1$ gives that $x-1$ is a factor of $2x^3-3x^2-11x+6$ .

  $\begin{array}{cc}\overline{)1}& \begin{array}{rrrr}\hfill 2& \hfill -3& \hfill -11& \hfill 6\\ \hfill & \hfill 2& \hfill -1& \hfill -12\\ \hfill 2& \hfill -1& \hfill -12& \hfill -6\end{array}\end{array}$

The last result is not zero. Thus $x-1$ is not a factor of $2x^3-3x^2-11x+6$ .

By synthetic division, check if the possible zero $-2$ gives that $x+2$ is a factor of $2x^3-3x^2-11x+6$ .

  $\begin{array}{cc}\overline{)-2}& \begin{array}{rrrr}\hfill 2& \hfill -3& \hfill -11& \hfill 6\\ \hfill & \hfill -4& \hfill 14& \hfill -6\\ \hfill 2& \hfill -7& \hfill 3& \hfill 0\end{array}\end{array}$

The last result is zero. Thus $x+2$ is a factor of $2x^3-3x^2-11x+6$ . In fact, $2x^3-3x^2-11x+6=\left(x+2\right)\left(2x^2-7x+3\right)$ , 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:

  $\begin{array}{c}2x^3-3x^2-11x+6=\left(x+2\right)\left(2x^2-7x+3\right)\\ =\left(x+2\right)\left(2x^2-6x-x+3\right)\\ =\left(x+2\right)\left(2x\left(x-3\right)-1\left(x-3\right)\right)\\ =\left(x+2\right)\left(x-3\right)\left(2x-1\right)\end{array}$

The completely factorized expression for the polynomial $2x^3-3x^2-11x+6$ is $\left(x+2\right)\left(x-3\right)\left(2x-1\right)$ .

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

Draw the sign chart based on this analysis:

  

  $2x^3-3x^2-11x+6$ is not negative when $x\in \left[-2,\frac{1}{2}\right]\cup \left[3,\infty \right)$ .

Graph:

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

  

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

Conclusion:

The solution of $2x^3-3x^2-11x+6\ge 0$ is $x\in \left[-2,\frac{1}{2}\right]\cup \left[3,\infty \right)$ .