Solving Inequalities (Tables)

Solving Polynomial Inequalities through tables involves creating a table of intervals and determining the sign of the polynomial within each interval to find where the polynomial is positive or negative.

Steps for Solving Inequalities

1. Move all terms to one side and make sure the polynomial is in Standard Form
2. Identify Critical Points of the polynomial by setting it equal to \(0\). This can be found by factoring the polynomial
3. Identify Intervals by using the Critical Points
4. Create a plus/minus table. This will outline the factors as rows and the intervals as columns
5. Examine the signs of the polynomial in each interval. Identify the intervals where the polynomial is positive or negative
6. Write the solution to the inequality based on the intervals where the polynomial is positive or negative

Example

Consider the polynomial inequality \( P(x) = x^3 - 3x^2 - 4x + 12 > 0 \).

Since the polynomial is already written in descending order of degree, we won't need to set things up any further.

In order to identify Critical Points, we can first set the polynomial equal to \(0\):

\(P(x) = x^3 - 3x^2 - 4x + 12 = 0\)

Next, we can fully factor the polynomial to determine its roots:

\(P(x) = x^2(x - 3) -4(x -3) = 0\)

\(P(x) = (x^2 -4)(x-3) = 0\)

\(P(x) = (x+2)(x-2)(x-3) = 0\)

We can determine the Critical Points as \(x = -2, 2, 3\).

Using the Critical Points, we can determine the intervals as \( (-\infty, -2), (-2,2), (2,3) \) and \( (3,\infty) \).

We can create our plus/minus table as such:

	\( (-\infty, -2)\)	\((-2,2)\)	\((2,3)\)	\( (3,\infty) \)
\(x+2\)
\(x-2\)
\(x-3\)
\(P(x)\)

Then, we need to choose test points for each interval. In this instance we can choose \( -3 \) for \( (-\infty,-2)\), \(0\) for \( (-2,2)\), \(2.5\) for \( (2,3) \), and \( 4 \) for \( (3,\infty) \).

For each factor, we are going to put a \(+\) on the table if it's positive on the interval and \(-\) if its negative.

Here are the results when we evaluate \(x+2\):

	\( (-\infty, -2)\)	\((-2,2)\)	\((2,3)\)	\( (3,\infty) \)
\(x+2\)	\(-\)	\(+\)	\(+\)	\(+\)
\(x-2\)
\(x-3\)
\(P(x)\)

Here are the results when we evaluate \(x-2\):

	\( (-\infty, -2)\)	\((-2,2)\)	\((2,3)\)	\( (3,\infty) \)
\(x+2\)	\(-\)	\(+\)	\(+\)	\(+\)
\(x-2\)	\(-\)	\(-\)	\(+\)	\(+\)
\(x-3\)
\(P(x)\)

Here are the results when we evaluate \(x-3\):

	\( (-\infty, -2)\)	\((-2,2)\)	\((2,3)\)	\( (3,\infty) \)
\(x+2\)	\(-\)	\(+\)	\(+\)	\(+\)
\(x-2\)	\(-\)	\(-\)	\(+\)	\(+\)
\(x-3\)	\(-\)	\(-\)	\(-\)	\(+\)
\(P(x)\)

For the final table, we are going to multiply the signs within each interval to determine the signage of the whole function:

	\( (-\infty, -2)\)	\((-2,2)\)	\((2,3)\)	\( (3,\infty) \)
\(x+2\)	\(-\)	\(+\)	\(+\)	\(+\)
\(x-2\)	\(-\)	\(-\)	\(+\)	\(+\)
\(x-3\)	\(-\)	\(-\)	\(-\)	\(+\)
\(P(x)\)	\(-\)	\(+\)	\(-\)	\(+\)

From looking at the final table, we can identify that the polynomial is positive on the intervals \( (-2,2) \) and \( (3, \infty) \) and negative on the intervals \( (-\infty, -2) \) and \( (2, 3) \).

Therefore, we can determine that the solution to the inequality \( P(x) > 0 \) is \(\boldsymbol{x \in (-2,2) \cup (3,\infty)}\).

\((x-3)(x+4)(x-2) \gt 0\)

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\( (x-4)(x+8)(x-2) \lt 0 \)

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