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Chapter 5: Problem 2

Choose the letter(s) of the correct response. Which of the following can be solved using the zero-factor property? A. \(25 x^{2}+20 x+4=0\) B. \(3 x^{2}=15 x\) C. \(2 x^{2}-128=0\) D. \(x(4 x-15)-30=0\)

Short Answer

Expert verified
All of the given equations (A, B, C, D) can be solved using the zero-factor property.

Step by step solution

01

- Understand the Zero-Factor Property

The zero-factor property states that if the product of two factors is zero, then at least one of the factors must be zero. Symbolically, if \(a \cdot b = 0\), then \(a = 0\) or \(b = 0\). This property is useful for solving quadratic and other polynomial equations set equal to zero.
02

- Analyze Each Equation

Check each given equation to determine if it can be factored into a product of expressions and set equal to zero:A. \(25x^{2}+20x+4=0\) - This is a quadratic equation that can be factored.B. \(3x^{2}=15x\) - This can be rewritten as \(3x^{2} - 15x = 0\), then factored.C. \(2x^{2}-128=0\) - This can be rewritten as \(2x^{2} = 128\) and then factored.D. \(x(4x-15)-30=0\) - This is a polynomial equation that can be factored.
03

- Apply Zero-Factor Property to Each Equation

Solve each equation by setting the factored expressions equal to zero:A. \(25x^{2}+20x+4=0\) - This can be factored as \((5x + 2)^2 = 0\).B. \(3x^{2} = 15x\) - Factor out common factor: \(3x(x - 5) = 0\), so \(3x = 0\) or \(x - 5 = 0\).C. \(2x^{2} - 128 = 0\) - Factor out the common factor: \(2(x^{2} - 64) = 0\), or factor it further to \(2(x + 8)(x - 8) = 0\).D. \(x(4x - 15) - 30 = 0\) - Distribute and rearrange: \(4x^{2} - 15x - 30 = 0\), then factor.
04

- Identify the Equations Solvable by Zero-Factor Property

Based on the previous steps, identify the equations that can be solved by expressing them in factored form and applying the zero-factor property:A, B, C, and D all can be factored and solved using the zero-factor property.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Quadratic Equations
Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). These equations can often be solved through several methods: graphing, completing the square, using the quadratic formula, or factoring. Factoring is particularly efficient when the quadratic can be nicely decomposed into simpler binomials.
The quadratic formula \(-\frac{b}{2a} \pm \sqrt{\left(\frac{b^2 - 4ac}{2a}\right)}\) can solve any quadratic equation, but factoring is often quicker and simpler for clear-cut factorable quadratics. Always check first if the quadratic can be factored to utilize simpler arithmetic operations.
Factoring Polynomials
Factoring polynomials involves writing the polynomial as a product of its factors. Consider a polynomial expression like \(ax^2 + bx + c\). The goal is to find two binomials \((dx + e)(fx + g)\) that multiply together to give the original polynomial.
Steps to factor a polynomial:
  • Find the product \(ac\) in the quadratic expression \(ax^2 + bx + c\).
  • Look for two numbers that multiply to \(ac\) and add to \(b\).
  • Use these numbers to break the middle term into two terms and factor by grouping.
Factoring makes solving equations easier using properties like the zero-product rule.
Zero-Product Rule
The zero-product rule, also known as the zero-factor property, is a fundamental concept in algebra. If the product of two or more numbers is zero, then at least one of the numbers must be zero. Mathematically, if \(a \cdot b = 0\), then \a = 0\ or \b = 0\.
This rule is crucial when solving polynomial equations because once a polynomial is factored, setting each factor to zero provides the solutions to the equation. For instance, in the factored equation \( (x - 3)(x + 2) = 0 \), solve for \ x \ by setting each factor to zero: \ x - 3 = 0 \ or \ x + 2 = 0 \, yielding the solutions \ x = 3 \ and \ x = -2 \.
Polynomial Equations
Polynomial equations are expressions involving variables raised to whole number powers and coefficients. A polynomial equation, like \(2x^3 - 6x^2 + x - 5 = 0 \, is called so because it involves the terms \2x^3, -6x^2, x,\ and \ -5 \).
To solve polynomial equations, especially those higher than degree two, we often look to factorize these into simpler, solvable components.
Steps to solve polynomial equations:
  • Simplify and rewrite the equation in standard form.
  • Factor the polynomial fully if possible.
  • Apply the zero-product rule to each factor set equal to zero.
This process helps break down complex equations into manageable solutions, making polynomial equations solvable by known methods.

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