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Chapter 6: Problem 62

Solve. $$ (z+4)(z-2)=-5 $$

Short Answer

Expert verified
z = -3 or z = 1

Step by step solution

01

Expand the Equation

First, expand the left-hand side of the equation \((z+4)(z-2) = z^2 - 2z + 4z - 8 \). The equation becomes \( z^2 + 2z - 8 \).
02

Set Equation to Zero

Now set the expanded equation equal to -5 from the right-hand side: \( z^2 + 2z - 8 = -5 \). To solve for z, we need the equation in standard form, so add 5 to both sides:\( z^2 + 2z - 8 + 5 = 0 \), which simplifies to \( z^2 + 2z - 3 = 0 \).
03

Factor the Quadratic Equation

Factor the quadratic equation \( z^2 + 2z - 3 \). Find two numbers that multiply to -3 and add to 2. These numbers are 3 and -1, so the equation factors to \( (z+3)(z-1) = 0 \).
04

Solve for z

Set each factor equal to zero and solve for z:\( z+3 = 0 \) gives \( z = -3 \), and \( z-1 = 0 \) gives \( z = 1 \).

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Key Concepts

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

factoring quadratics
Factoring quadratics is a critical skill when dealing with second-degree polynomial equations, often written in the form \( ax^2 + bx + c \). The goal is to break down the polynomial into two binomials. For example, consider the quadratic equation \( z^2 + 2z - 3 \). To factor it, we look for two numbers that multiply to -3 (the constant term) and add up to 2 (the coefficient of the linear term). These numbers are 3 and -1, transforming \( z^2 + 2z - 3 \) into \( (z+3)(z-1) \). This technique makes it easier to solve quadratic equations by applying the zero-product property.
solving equations
Solving equations involves finding the values of variables that make the equation true. For quadratic equations like \( (z+4)(z-2) = -5 \), start by expanding both sides to simplify the expression. Once in standard form, set the equation to zero and factor if possible. After factoring, apply the zero-product property, which states that if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \). For our example, \( (z+3)(z-1) = 0 \) leads to two solutions: \( z + 3 = 0 \), giving \( z = -3 \), and \( z - 1 = 0 \), giving \( z = 1 \).
expanding expressions
Expanding expressions means rewriting a product of binomials as a polynomial. For example, expanding \( (z+4)(z-2) \) involves using the distributive property (FOIL method):
  • First: \( z \times z = z^2 \)
  • Outside: \( z \times -2 = -2z \)
  • Inside: \( 4 \times z = 4z \)
  • Last: \( 4 \times -2 = -8 \)
Combine these results: \( z^2 - 2z + 4z - 8 = z^2 + 2z - 8 \). Expanding binomials does not change the value but converts it into a more manageable form for further algebraic operations.
standard form of quadratic
The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. To solve quadratic equations, convert them to this form. For the equation \( (z+4)(z-2) = -5 \), start by expanding to get \( z^2 + 2z - 8 \). Then set the equation to zero by adding 5 to both sides: \( z^2 + 2z - 8 + 5 = 0 \), resulting in \( z^2 + 2z - 3 = 0 \). This standard form allows us to apply factoring, the quadratic formula, or completing the square to solve for \( z \).

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Most popular questions from this chapter

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