Problem 14 - Applied Calculus

Question

Solve each equation by factoring. [Hint for: First factor out a fractional power.] $$ 2 x^{5}-50 x^{3}=0 $$

Step-by-Step Solution

Answer

\(x = 0, -5, \text{ and } 5\).

Step 1: Identify Common Factors

Look for any common factors in the terms \(2x^5\) and \(-50x^3\). Notice that both terms have a factor of \(x^3\). Factor \(x^3\) out from the equation: \[ x^3(2x^2 - 50) = 0 \]

Step 2: Simplify the Factored Expression

Now that you've factored out \(x^3\), focus on simplifying the expression within the parentheses: \(2x^2 - 50\). Notice that 2 is a common factor: \[ x^3(2(x^2 - 25)) = 0 \]

Step 3: Further Factor the Quadratic Expression

The quadratic expression \(x^2 - 25\) is a difference of squares. We can factor it as: \[ x^3(2(x+5)(x-5)) = 0 \]

Step 4: Solve for the Roots

Each factor in the product can be set equal to zero to solve for \(x\). We have three equations: \(x^3 = 0\), \(x+5 = 0\), and \(x-5 = 0\). Solve each equation: - From \(x^3 = 0\), \(x = 0\). - From \(x + 5 = 0\), \(x = -5\).- From \(x - 5 = 0\), \(x = 5\).

Step 5: Write the Complete Solution

The solutions to the equation are the roots found in Step 4. Therefore, the values of \(x\) that satisfy the equation are: \[ x = 0, x = -5, \text{ and } x = 5 \]

Key Concepts

Polynomial Equations

Polynomial equations are mathematical expressions that involve variables raised to different powers and coefficients. They are characterized by the highest power of the variable, known as the degree. For instance, in the equation \(2x^5 - 50x^3 = 0\), the highest power is 5, making it a fifth-degree polynomial equation.
Understanding the structure of polynomial equations is essential to solve them effectively. Each term in a polynomial is a product of a constant coefficient and a variable raised to a power. For example:

• The term \(2x^5\) includes the coefficient 2 and the variable \(x\) raised to the power of 5.
• The term \(-50x^3\) includes the coefficient -50 and the variable \(x\) to the third power.

To solve polynomial equations like this, you'll often need to factor them into simpler expressions or forms. This method may include finding common factors, which helps make the equation easier to work with.

Difference of Squares

The difference of squares is a specific type of polynomial expression where two squared terms are subtracted. A classic formula used to identify and factor these expressions is \(a^2 - b^2 = (a+b)(a-b)\). Recognizing a difference of squares can simplify an equation considerably.
In the example \(x^2 - 25\), notice it fits the pattern \(a^2 - b^2\):

• Here, \(x^2\) is \(a^2\) (where \(a = x\)).
• The number 25 is \(b^2\) (where \(b = 5\)).

Using the difference of squares formula, the expression factors to \((x+5)(x-5)\). This simplification allows for easier solving of each factor set to zero in further steps, streamlining the equation initially given.

Root Solving

Root solving, also known as finding the "zeros," involves determining the solutions for the variable \(x\) when the polynomial is equal to zero. Once an equation is fully factored, each factor can be addressed independently.
For the equation \(x^3(2(x+5)(x-5)) = 0\), you solve by setting each factor to zero:

• \(x^3 = 0\) yields the solution \(x = 0\). Here, \(x\) is zero because \(x\) cannot be anything else to satisfy \(x^3 = 0\).
• \(x + 5 = 0\) gives \(x = -5\). Subtract 5 from both sides to find this root.
• \(x - 5 = 0\) results in \(x = 5\). Add 5 to both sides to solve this equation.

Each solution or "root" provides a value of \(x\) that makes the overall polynomial equal to zero. Thus, the roots of the equation \(2x^5 - 50x^3 = 0\) are \(x = 0, -5,\) and \(5\). These represent the points where the graph of the polynomial intersects the \(x\)-axis.