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Chapter 1: Problem 15

Solve each equation by factoring. $$ z^{2}+z-6=0 $$

Short Answer

Expert verified
The solutions are \( z = -3 \) and \( z = 2 \).

Step by step solution

01

- Identify the quadratic equation

Recognize that the equation is a quadratic in the form of: \[ z^2 + z - 6 = 0 \]
02

- Factor the quadratic equation

Find two numbers that multiply to the constant term (-6) and add to the coefficient of the linear term (1). The numbers 3 and -2 satisfy this condition because: \[ 3 \times (-2) = -6 \] \[ 3 + (-2) = 1 \] Thus, the equation can be factored as: \[ (z + 3)(z - 2) = 0 \]
03

- Set each factor to zero

Use the zero product property, which states that if a product of factors equals zero, then at least one of the factors must be zero: \[ z + 3 = 0 \] \[ z - 2 = 0 \]
04

- Solve each equation

Solve for \( z \) in each equation: For \( z + 3 = 0 \): \[ z = -3 \] For \( z - 2 = 0 \): \[ z = 2 \]

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Key Concepts

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

Factoring
Factoring is a method used to break down a complex expression into simpler elements. This is very useful for solving quadratic equations. In the original exercise, we start with the quadratic equation Recognizing that the quadratic is in the standard form The next step is to find two numbers that multiply to the constant term (-6) and add to the coefficient of the linear term (1). Here, those two numbers are 3 and -2 because: Therefore, the quadratic equation can be factored as , breaking it down into two binomial expressions. This makes it easier to find the values of z that solve the original equation.
Zero Product Property
The Zero Product Property is a fundamental principle in algebra, which states that if So, after factoring the original quadratic equation into , we use the Zero Product Property to solve the problem. Here, it means that each factor can be set to zero: .
Solving Equations
Once we've set each factor equal to zero using the Zero Product Property, the next step is to solve these simple linear equations. By solving , we get: , meaning . Similarly, for , we find that . Therefore, the solutions to the original quadratic equation are and These solutions can be verified by substituting back into the original equation to ensure they satisfy . Solving quadratic equations through factoring, applying the Zero Product Property, and solving linear equations makes the overall process straightforward and efficient.

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Most popular questions from this chapter

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Find the real solutions, if any, of each equation. Use any method. $$ \frac{3 x}{x+2}+\frac{1}{x-1}=\frac{4-7 x}{x^{2}+x-2} $$

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