Problem 12 - Precalculus : Building Concepts and Connections

Question

Solve the polynomial inequality. $$x^{3}+2 x^{2}+x<0$$

Step-by-Step Solution

Answer

The solution to the inequality \(x^{3} + 2 x^{2} + x < 0\) is \(x \in (-\infty, -1) \cup (0, +\infty)\).

Step 1: Factorisation

To factorise the polynomial inequality, recognize that each term contains an x, hence you can factor out x: \(x(x^{2} + 2x + 1)< 0\).

Step 2: Factor further

Considering \(x^{2} + 2x + 1\), we notice that there are two terms that can be squared to give the equation. Therefore, using the perfect square formula, the whole expression can be factorised further to get: \(x(x + 1)^{2} < 0\).

Step 3: Find critical points

Next, we find the critical points of the inequality by setting the factors equal to zero and solving for x: For \(x=0\) and \(x = -1\), we get the critical points. These points divide the number line into distinct intervals.

Step 4: Test the intervals

Test any number in each interval in the inequality: The intervals are \(-\infty \leq x < -1\), \(-1 < x < 0\) and \( 0 < x < +\infty\). If we pick -2, -0.5 and 1 and substitute these into the original inequality, we will find that the inequality holds for the intervals \(-\infty \leq x < -1\) and \( 0 < x < +\infty\).

Step 5: Write the solution

The solution to the inequality \(x^{3} + 2 x^{2} + x < 0\) based on the intervals that satisfy the inequality, is \(x \in (-\infty, -1) \cup (0, +\infty)\).

Key Concepts

Factorisation of Polynomials

Factorisation is a powerful tool in algebra that breaks down complex polynomial expressions into simpler, more manageable factors. It is essential when solving polynomial inequalities because it reveals the structure of the polynomial and helps identify where the expression can switch signs.

In the given problem, the polynomial inequality is: \( x^3 + 2x^2 + x < 0 \). The first step was to factor out the greatest common factor, which is x, giving us \( x(x^2 + 2x + 1) < 0 \). Upon closer inspection, the remaining quadratic expression can be recognized as a perfect square trinomial. A perfect square trinomial is an expression that can be written as \( (a + b)^2 \), where \( a^2 + 2ab + b^2 = (a + b)^2 \). In our problem, this trinomial is \( x^2 + 2x + 1 \), which can be factored further to \( (x + 1)^2 \).

Factorisation not only simplifies the given expression but also provides crucial insights for the next steps in solving the inequality. It transforms the inequality into a product of factors, making it easier to determine when the product will be less than zero - that is, when the inequality holds.

Critical Points in Inequalities

Critical points are values that separate regions on the number line where the polynomial function changes its sign from positive to negative or vice versa. These points are pivotal when solving inequalities because they are the boundaries that help define where the solution lies.

When factoring a polynomial inequality, like the one we have \( x(x + 1)^2 < 0 \), we aim to find these critical points by setting each factor equal to zero and solving for x. By doing this, we essentially determine where the graph of the polynomial touches or crosses the x-axis. In this case, the critical points are x = 0 and x = -1, discovered by solving the equations \( x = 0 \) and \( (x + 1)^2 = 0 \).

These critical points divide the real number line into intervals. Understanding where these points lie and how the sign of the polynomial changes at these points is crucial for identifying where the inequality is satisfied. If a factor appears more than once, as \( (x + 1)^2 \) does in our example, the curve of the polynomial does not cross the x-axis at that point, which influences the testing of intervals and ultimately, the solution.

Testing Intervals in Inequalities

Once you have your polynomial factored and the critical points established, the next step to solve an inequality is to determine the sign of the polynomial in each interval created by the critical points. This step is called 'testing intervals', and it involves picking a test point from each interval and substituting this point into the factored inequality to see if the result is positive or negative.

For our inequality \( x(x + 1)^2 < 0 \), the critical points divided the number line into three intervals: \( (-\infty, -1) \) , \( (-1, 0) \) , and \( (0, +\infty) \). By choosing test points from each interval like -2, -0.5, and 1, and substituting them back into the factored inequality, we can determine which intervals satisfy the original inequality.

In this case, the intervals where the inequality holds true are \( (-\infty, -1) \) and \( (0, +\infty) \), which means when x is between negative infinity and -1 or between 0 and positive infinity, the original polynomial inequality is satisfied. These intervals provide the solution set for the inequality, ensuring a comprehensive understanding of where the polynomial expression is less than zero.