Problem 52 - Algebra 1: Concepts and Skills

Question

Solve \(6(x-3)(x+5)(x-9)=0\). A) \(6,3,5,\) and 9 B) \(3,-5,\) and 9 C) \(6,3,-5,\) and 9 D) \(6,3,5,\) and \(-9\)

Step-by-Step Solution

Answer

The solutions for x are \(3,-5,\) and 9. Hence, the correct answer is B: \(3,-5,\) and 9.

Step 1: Apply Zero-Product Property

Set each of the factors in the equation \(6(x-3)(x+5)(x-9)=0\) equal to zero.

Step 2: Solve the first equation

Solving \(x-3=0\) gives \(x=3\).

Step 3: Solve the second equation

Solving \(x+5=0\) gives \(x=-5\).

Step 4: Solve the third equation

Solving \(x-9=0\) gives \(x=9\).

Key Concepts

Solving Polynomial Equations

Understanding how to solve polynomial equations is a foundational skill in algebra. These equations can often seem daunting due to their multiple terms and variables, but the process can be simplified using various strategies. One such strategy is the Zero-Product Property, which states that if the product of several factors equals zero, then at least one of the factors must be zero.

For example, if you have an equation like the textbook problem, where the product of factors equals zero: \(6(x-3)(x+5)(x-9)=0\), you can apply the Zero-Product Property. To utilize this property, you individually set each factor that contains the variable equal to zero, as shown in the step by step solution: \(x-3=0\), \(x+5=0\), and \(x-9=0\). Solving each of these simple equations will yield the roots of the original polynomial equation. These roots, \(x=3\), \(x=-5\), and \(x=9\), are the solutions because they make the original equation true when substituted back into it.

Factoring Polynomials

Factoring polynomials is a critical operation used in solving polynomial equations. It involves breaking down a complex expression into simpler parts, or 'factors', that when multiplied together give back the original polynomial. This is an effective method to simplify polynomial equations, making it easier to identify solutions.

To factor a polynomial, look for the Greatest Common Factor (GCF) first, and then apply different techniques like grouping, using special product formulas, or long division. The problem provided is already factored, demonstrating the multiplicative nature of polynomials. However, when you encounter a non-factored polynomial equation, your first step should be to express it in its factored form, if possible, which will then allow you to apply the Zero-Product Property to find the solutions.

Quadratic Equations

Quadratic equations, a specific type of polynomial equation with a degree of 2, are ubiquitous in algebra and have the general form \(ax^2+bx+c=0\). Such equations are commonly solved through factoring, completing the square, or using the quadratic formula. Quadratics often appear as one or more factors in more complex polynomial equations.

For example, in the exercise, if one of the factors \((x-3)(x+5)(x-9)=0\) had been a quadratic, like \(x^2-4\), which factors to \((x+2)(x-2)\), you would still use the Zero-Product Property to find its roots. It's also important to note that when solving quadratic equations, there may be two real roots, one real root, or no real roots when considering the real number system. Understanding quadratic equations is key, as they form the basis of polynomial equations and their factorization.