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Chapter 10: Problem 33

Use the zero-product property to solve the equation. $$ 7(b-5)^{3}=0 $$

Short Answer

Expert verified
The value of 'b' that satisfies the equation \(7(b-5)^3=0\) is \(b = 5\).

Step by step solution

01

Apply Zero-Product Property

In accordance with the zero product property, set each factor equal to zero, since the product of the factors equals zero. The equation: \(7(b-5)^3=0\), can be broken up into two factors: \(7\) and \((b-5)^3\). Now set each separate factor to zero.
02

Solve for 'b' in (b-5)^3 = 0

Only concentrating on the factor \((b-5)^3=0\), take a cube root on both sides which gives you the equation \(b-5=0\). Solving this equation for 'b', you will find the values of variable 'b'.
03

Final Solution

On solving the equation \(b-5=0\), you obtain \(b = 5\). Since setting \(7 = 0\) does not give us any meaningful output, our solution is \(b = 5\).

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Key Concepts

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

Solving Equations
Solving equations is the process of finding the value of the variable that makes the equation true. In our exercise, we are given the equation \(7(b-5)^3=0\). This equation represents a situation where two factors' product is zero. According to the Zero-Product Property, if the product of two or more factors equals zero, then at least one of these factors must also be zero.

To solve such an equation, follow these steps:
  • First, identify the factors in the equation. For \(7(b-5)^3=0\), the factors are \(7\) and \((b-5)^3\).
  • Since any constant (like 7) will never equal zero, we can ignore it for solving purposes.
  • Focus on the factor \((b-5)^3 = 0\). This tells us that the quantity \((b-5)\) must equal zero for the cube to be zero.
Thus, by setting \((b-5) = 0\), we can solve for \(b\), finding that \(b = 5\). This approach simplifies solving equations by narrowing down to meaningful parts of the expression.
Factoring
Factoring is a critical skill in algebra. It involves breaking down expressions into simpler multipliers or factors that, when multiplied together, return the original expression. In the equation \(7(b-5)^3=0\), the expression already presents factors: \(7\) and \((b-5)^3\).

The key here is to recognize premade factors like \((b-5)^3\). Such forms often result from previous expressions where a single term is repeated multiple times, such as \((b-5)\) being cubed.
  • In this case, we'd write \((b-5)^3\) as \((b-5)(b-5)(b-5)\). Each repetition suggests that if any single instance equals zero, so does the set.
  • It simplifies equations by reducing them to component parts that are easier to handle.
  • Once the factors are recognized, they can be individually equated to zero, as necessary.
Mastering factoring skills paves the way for solving more complex polynomial equations and boosting overall algebraic proficiency.
Algebraic Manipulation
Algebraic manipulation refers to the techniques used to rearrange and simplify algebraic equations to make them solvable. It involves using a variety of operations to change the form of equations without changing their solutions.

In our exercise, algebraic manipulation is employed to solve for \(b\) after recognizing \((b-5)^3 = 0\). By taking the cube root of both sides, we simplify the factor to \(b-5=0\), which is a more straightforward equation. This technique is powerful:
  • It allows us to isolate the variable \(b\), making it easier to solve.
  • Taking roots is a form of inverse operation that helps undo terms raised to powers.
  • Simplification makes equations clearer and highlights the relationships between different terms.
Thus, algebraic manipulation is essential for effectively handling and solving equations, especially those with multiple terms or higher exponents.

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Most popular questions from this chapter

Decide whether the ordered pair is a solution of the inequality. $$y>2 x^{2}-x+7 ;(2,15)$$

$$x^{2}-\frac{5}{3} x+\frac{25}{36}=0$$ $$112 x^{2}-252=0$$

Use a is calculator to evaluate the expression. Round the result to two decimal places when appropriate. $$\left(-6 \cdot 3^{4}\right)^{3}$$

Factor the expression. Tell which special product factoring pattern you used. $$169-x^{2}$$

Solve the equation. \(|x|=3\)
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