Problem 51 - College Algebra with Modeling and Visualization

Question

Solve the polynomial inequality. $$ 2 x^{4}+2 x^{3} \leq 12 x^{2} $$

Step-by-Step Solution

Answer

The solution is \([-3, 2]\).

Step 1: Move all terms to one side

Begin by bringing the right-hand side terms to the left side of the inequality. Subtract \(12x^2\) from both sides to set the inequality to zero: \[2x^4 + 2x^3 - 12x^2 \leq 0\]

Step 2: Factor out the greatest common factor

Factor out the greatest common factor from the left side of the inequality. The common factor here is \(2x^2\): \[2x^2(x^2 + x - 6) \leq 0\]

Step 3: Factor the quadratic expression

Next, factor the quadratic \(x^2 + x - 6\). Look for two numbers that multiply to \(-6\) and add to \(1\). These are \(3\) and \(-2\). So the quadratic factors to: \[(x + 3)(x - 2)\]

Step 4: Write the factored inequality

Substitute the factored form back into the inequality: \[2x^2(x + 3)(x - 2) \leq 0\]

Step 5: Determine critical points

Identify the critical points by setting each factor equal to zero. Solve for \(x\): \[2x^2 = 0 \Rightarrow x = 0\]\[x + 3 = 0 \Rightarrow x = -3\]\[x - 2 = 0 \Rightarrow x = 2\]So the critical points are \(x = -3\), \(x = 0\), and \(x = 2\).

Step 6: Test intervals around critical points

Test the intervals determined by the critical points in the factored inequality:- Test interval \((-\infty, -3)\): Choose \(x = -4\)- Test interval \((-3, 0)\): Choose \(x = -1\)- Test interval \((0, 2)\): Choose \(x = 1\)- Test interval \((2, \infty)\): Choose \(x = 3\) Substitute these values into \(2x^2(x + 3)(x - 2)\) and check which intervals satisfy \(\leq 0\).

Step 7: Analyze each interval

Carry out substitution for chosen test points:- For \(x = -4\): \(2(-4)^2(-4+3)(-4-2) = 2 \times 16 \times (-1) \times (-6) = 192 > 0\)- For \(x = -1\): \(2(-1)^2(-1+3)(-1-2) = 2 \times 1 \times 2 \times (-3) = -12 < 0\)- For \(x = 1\): \(2(1)^2(1+3)(1-2) = 2 \times 1 \times 4 \times (-1) = -8 < 0\)- For \(x = 3\): \(2(3)^2(3+3)(3-2) = 2 \times 9 \times 6 \times 1 = 108 > 0\)Intervals \((-3, 0)\) and \((0, 2)\) satisfy the inequality.

Step 8: Include critical points

Check the inequality at the critical points:- For \(x = -3\), substitute into the original equation, we get \(2(-3)^4 + 2(-3)^3 - 12(-3)^2 = 0\).- For \(x = 0\), substitute into the original equation, we get \(2(0)^4 + 2(0)^3 - 12(0)^2 = 0\).- For \(x = 2\), substitute into the original equation, we get \(2(2)^4 + 2(2)^3 - 12(2)^2 = 0\).All critical points satisfy \(\leq 0\), so they are included in the solution.

Key Concepts

Factoring Polynomials

Factoring polynomials is a critical step in solving polynomial inequalities. Essentially, this means rewriting the polynomial as a product of its factors. The simplification makes it much easier to identify the critical points, which are vital for testing intervals.

In our exercise, we initially had the inequality \(2x^4 + 2x^3 \leq 12x^2\). The first step required moving all terms to one side of the inequality to form:

• \(2x^4 + 2x^3 - 12x^2 \leq 0\)

Finding the greatest common factor is the next step in factoring. For our equation, the common factor was \(2x^2\), and factoring it out simplified the expression to:

• \(2x^2(x^2 + x - 6) \leq 0\)

This step reduces the complexity of the polynomial and paves the way for further breakdown.

Next, focus shifts to the quadratic part, \(x^2 + x - 6\). Identifying the roots of the polynomial through factoring not only assists in better comprehension but also directly ties into finding critical points—key to solving inequalities successfully.

Quadratic Expressions

Quadratic expressions are integral to solving polynomial inequalities. A quadratic expression is of the form \(ax^2 + bx + c\), and understanding how to factor and solve these is crucial.

In the context of the exercise, we considered the quadratic expression \(x^2 + x - 6\):

• This needs to be factored into its component products.
• We aimed to find numbers that multiply to \(-6\) and add to \(1\).

The numbers \(3\) and \(-2\) were good candidates, allowing us to rewrite the quadratic as:

• \((x + 3)(x - 2)\)

This step is fundamental because it converts complex expressions into simpler ones, facilitating easier solution of the inequality.

Quadratic factoring not only breaks down terms but also helps visualize solutions by highlighting crucial intercept points on the graph, essential for analyzing polynomial behaviors.

Critical Points

Critical points in polynomial inequalities are values of \(x\) where the expression equals zero. These are pivotal, as they provide boundaries for interval testing, determining where an inequality holds true.

After we factored the polynomial into \(2x^2(x + 3)(x - 2) \leq 0\), the next move was to identify its critical points:

• Setting each factor equal to zero gives us \(x = 0\), \(x = -3\), and \(x = 2\).

These values are critical, acting as breakpoints dividing the number line into intervals. Intervals are tested to determine where the original inequality holds.

Testing around these points involved:

• Choosing test values from each interval.
• Substituting them in to evaluate the inequality's truth.

Comprehending critical points not only aids in solving but also strengthens understanding by pinpointing where shifts in inequality validity occur, ultimately leading to correct solutions.