Problem 9 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ 4 x^{4}+10 x^{3}=6 x^{2}+15 x $$

Step-by-Step Solution

Answer

The solutions are \(x = 0, \pm \sqrt{\frac{3}{2}}, -\frac{5}{2}\).

Step 1: Move All Terms to One Side

Start by moving all terms to the left side of the equation. Subtract \(6x^2\) and \(15x\) from both sides:\[4x^4 + 10x^3 - 6x^2 - 15x = 0\]

Step 2: Factor by Grouping

Look for common factors in pairs of terms. For the first two terms, \(4x^4 + 10x^3\), you can factor out \(2x^3\). For the last two terms, \(-6x^2 - 15x\), you can factor out \(-3x\):\[2x^3(2x + 5) - 3x(2x + 5) = 0\]

Step 3: Factor the Expression Completely

Notice that \((2x + 5)\) is a common factor. Factor \((2x + 5)\) out:\[(2x^3 - 3x)(2x + 5) = 0\]Now, factor out \(x\) from \(2x^3 - 3x\):\[x(2x^2 - 3)(2x + 5) = 0\]

Step 4: Solve for x

Set each factor equal to zero and solve for \(x\):1. \(x = 0\)2. \(2x^2 - 3 = 0\) leads to \(2x^2 = 3\), then \(x^2 = \frac{3}{2}\), so \(x = \pm \sqrt{\frac{3}{2}}\)3. \(2x + 5 = 0\) leads to \(x = -\frac{5}{2}\)

Step 5: List All Solutions

List all solutions found from the previous step:\[x = 0, \pm \sqrt{\frac{3}{2}}, -\frac{5}{2}\]

Key Concepts

Factoring by Grouping

Factoring by grouping is a useful method for breaking down more complex polynomial equations into simpler components that can be solved more easily. In the given problem, we start by rearranging all terms on one side of the equation:

\[4x^4 + 10x^3 - 6x^2 - 15x = 0\]
Next, we look at pairs of terms to find common factors. For example:

• The first pair, \(4x^4 + 10x^3\), has a common factor of \(2x^3\).
• The second pair, \(-6x^2 - 15x\), has a common factor of \(-3x\).

Grouping these terms, we factor them separately, yielding:

\[2x^3(2x + 5) - 3x(2x + 5) = 0\]
Here, we notice that \((2x + 5)\) appears in both groups, allowing us to factor it out, which simplifies our equation further.

Quadratic Solutions

Once the equation is simplified, quadratic solutions can be applied to find the unknown values of \(x\). The equation transforms into:
\[(2x^3 - 3x)(2x + 5) = 0\]
Next, we can factor \(x\) from the term \(2x^3 - 3x\), resulting in:
\[x(2x^2 - 3)(2x + 5) = 0\]
Now, consider each factor separately:

• For \(2x^2 - 3 = 0\), solve by adding 3 to both sides and then dividing by 2 to isolate \(x^2\): \[2x^2 = 3 \rightarrow x^2 = \frac{3}{2}\]
• Then solve for \(x\) by taking the square root of both sides, giving you two solutions: \[x = \pm \sqrt{\frac{3}{2}}\]

These are examples of solving quadratic equations to find various solutions for \(x\). Quadratic solutions often provide multiple answers due to the squared variable.

Zero Product Property

The zero product property is a fundamental concept in algebra that states if the product of multiple factors equals zero, at least one of the factors must be zero. We use this property to solve polynomial equations by setting each factor equal to zero.
For our equation:
\[x(2x^2 - 3)(2x + 5) = 0\]
We apply the zero product property as follows:

• \(x = 0\), since there is a factor \(x\), which immediately leads to one solution.
• From \(2x^2 - 3 = 0\), we've previously calculated solutions \(x = \pm \sqrt{\frac{3}{2}}\).
• From \(2x + 5 = 0\), solve for \(x\) by isolating it: \[2x = -5 \rightarrow x = -\frac{5}{2}\]

Each factor set to zero provides a pathway to finding valid solutions for the variable \(x\). Understanding the zero product property simplifies the process of solving complex polynomial equations like the one provided here.