Problem 67 - Beginning and Intermediate Algebra

Question

The following equations are not quadratic but can be solved by factoring and applying the zero product rule. Solve each equation. $$5 w^{2}+36 w=w^{3}$$

Step-by-Step Solution

Answer

The short answer for solving the equation \(5w^2 + 36w = w^3\) by factoring and applying the zero product rule is: First, rewrite the equation in standard form, which gives us \(-w(w-9)(w+4)=0\). Then apply the zero product rule, resulting in three solutions for \(w\): \(w = 0\), \(w = 9\), and \(w = -4\).

Step 1: Rewrite the equation in standard form

To rewrite the given equation in standard form, we need to set it equal to 0. To do that, we subtract the cubic term on the right side from both sides of the equation: $$5w^2 + 36w - w^3 = 0$$

Step 2: Factor the polynomial

Next, we factor the polynomial as much as possible. We can start by factoring out a common factor, which is "w" in this case: $$w(5w^2 + 36w - w^3) = 0$$ Now, we need to factor the quadratic expression inside the parentheses. Since the coefficients don't have any common factors, we'll try to factor by grouping: Rearrange the terms: $$w(-w^3 + 5w^2 + 36w) = 0$$ Factor by grouping: $$w(-w^3 + 5w^2 + 36w) = w(-w(w^2 - 5w - 36)) = 0$$ Now we can factor the quadratic expression inside the parentheses: $$-w(w^2 - 5w - 36) = -w(w-9)(w+4) = 0$$

Step 3: Apply the Zero Product Rule

Now we can apply the zero product rule. According to the zero product rule, if the product of factors is equal to 0, then at least one of the factors must be equal to 0. So we have: 1. \(-w = 0\) 2. \((w-9) = 0\) 3. \((w+4) = 0\)

Step 4: Solve for w

Now, we can solve each of the equations from the previous step to get the solutions for w: 1. \(-w = 0 \Rightarrow w = 0\) 2. \((w-9) = 0 \Rightarrow w = 9\) 3. \((w+4) = 0 \Rightarrow w = -4\) So the solutions for the equation \(5w^2 + 36w = w^3\) are \(w = 0\), \(w = 9\), and \(w = -4\).

Key Concepts

Zero Product Rule

The Zero Product Rule is a fundamental principle in algebra that is used to solve equations involving a product of factors. It states that if the product of two or more terms equals zero, at least one of the terms must be zero. This is expressed as:

• If \( a \times b = 0 \), then \( a = 0 \) or \( b = 0 \).

In the context of solving algebraic equations, this rule is extremely helpful for finding solutions after a polynomial has been factored completely. Once we have a product of factors set equal to zero, we can individually set each factor to zero and solve for the variable.
For instance, if you have factored an equation into \( w(-w)(w-9)(w+4) = 0 \), applying the Zero Product Rule gives you the individual equations: \(-w = 0\), \(w-9 = 0\), and \(w+4 = 0\). Solving these gives the solutions for the variable \( w \). This rule simplifies solving polynomials significantly and is an integral part of algebraic problem-solving.

Cubic Equations

Cubic equations involve polynomial terms where the highest degree is three. The standard form of a cubic equation is generally expressed as \( ax^3 + bx^2 + cx + d = 0 \). These equations can have up to three real roots, or solutions, and are more complex than quadratic equations.
To solve cubic equations, factoring is often employed, especially if the equation can be simplified into form that allows the application of the Zero Product Rule. For example, the equation \( 5w^2 + 36w = w^3 \) can be rearranged into \( 5w^2 + 36w - w^3 = 0 \), making it easier to proceed with factoring.
Once factored, a cubic equation might have several linear (or even quadratic) factors that are easier to solve. While some cubic equations might require specific techniques like synthetic division or the Rational Root Theorem, many can be handled by straightforward factoring if the polynomial terms can be easily grouped and reduced.

Algebraic Solutions

Algebraic solutions refer to solving equations using algebraic techniques, which involve manipulating algebraic expressions using various properties and rules. To determine the solutions of an algebraic equation, especially polynomials, factoring is a key method. It involves breaking down complex expressions into simpler components that are easy to solve.
When solving for \( w \) in the equation \( 5w^2 + 36w - w^3 = 0 \), algebraic techniques such as rearrangement and factoring by substitution are used. Begin by rewriting and simplifying the polynomial expression to set the equation to zero. Then factor out common elements for simplification.

• First factor out the greatest common factor, like \( w \), and then analyze the quadratic-like terms remaining.
• Continue factoring until you reach irreducible linear expressions if possible.

Finally, apply the Zero Product Rule to find each possible value of \( w \). This process not only finds solutions but also equips you with valuable problem-solving skills in algebra. Understanding and applying these strategies are crucial for solving a range of algebraic equations effectively.