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Chapter 6: Problem 13

Solve. $$ x(x-3)(x-7)=0 $$

Short Answer

Expert verified
The solutions are \(x=0\), \(x=3\), and \(x=7\).

Step by step solution

01

Set each factor equal to zero

The equation has three factors: \(x\), \(x-3\), and \(x-7\). Set each factor equal to zero to find the values of \(x\).\[ x=0, \quad x-3=0, \quad x-7=0 \]
02

Solve each equation

Solve each equation for \(x\).\[ x=0 \]\[ x-3=0 \implies x=3 \]\[ x-7=0 \implies x=7 \]
03

Write down the solutions

The solutions to the equation are the values of \(x\) that satisfy any of the three equations from Step 2: \(x=0\), \(x=3\), and \(x=7\).

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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 an expression into simpler components, called factors. It is particularly useful in solving quadratic and higher-degree polynomial equations.
To factor an equation like the one given, you look for common factors among the terms. In the exercise, we see that the equation is given in its factored form: \(x(x-3)(x-7) = 0\). Here, the factors are \(x\), \((x-3)\), and \((x-7)\).
This factored form makes it easier to find the solutions by applying the Zero Product Property.
Zero Product Property
The Zero Product Property is a rule in algebra that states if the product of multiple factors equals zero, at least one of the factors must be zero. This property is very useful for solving equations in their factored form.
In the equation \(x(x-3)(x-7)=0\), we use this property to set each factor equal to zero because if any one of them is zero, the entire expression will be zero.
Therefore, we set \(x=0\), \((x-3)=0\), and \((x-7)=0\). This simplifies the problem into three smaller, more manageable equations. Thus, because of the Zero Product Property, we know that one of these factors has to be zero for the product to be zero.
Algebraic Equations
Algebraic equations are expressions that show the equality between two algebraic expressions. Solving these equations often involves finding the values or roots for the variable that satisfy the equation.
In the given exercise, the equation is already nicely factored, so you can apply the Zero Product Property right away.
Often in algebra, we deal with equations that need to be manipulated and factored first. Understanding how to break down and combine terms is essential when working with algebraic equations.
Solving algebraic equations can involve various techniques such as factoring, expanding, and using properties like the distributive property.
Roots of Equations
The roots of an equation are the values of the variable that make the equation true. Simply put, the roots, or solutions, are the values of \(x\) that satisfy the equation.
In our problem \(x(x-3)(x-7)=0\), we found that the roots are \(x=0\), \(x=3\), and \(x=7\).
Finding the roots can be visualized as finding the points where the graph of the equation intersects the x-axis.
Roots are crucial in understanding the behavior of polynomial functions. Each root corresponds to a point where the curve touches or crosses the x-axis.

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Most popular questions from this chapter

Use a pattern to factor. Check. Identify any prime polynomials. $$ h^{3}-27 $$

(a) solve. (b) check. $$ 2 w^{2}+9 w+9=0 $$

(a) factor out the greatest common factor. Identify any prime polynomials. (b) check. $$ 48 v^{3}+56 v^{2}+32 v $$

(a) Describe the mistake in words, or copy down the whole problem and highlight or circle the mistake. (b) Do the problem correctly. Problem: Use a pattern to factor \(25 z^{2}+9\). Incorrect Answer: Since the pattern is \(a^{2}+b^{2}=\) \((a+b)(a+b), a=5 z\) and \(b=3\), and the factored polynomial is \((5 z+3)(5 z+3)\).

The formula for density, \(d\), is \(d=\frac{m}{v}\), where \(m\) is the mass and \(v\) is the volume. The density of a steel sphere is \(7.85 \frac{\mathrm{g}}{\mathrm{cm}^{3}}\), and its mass is \(5 \times 10^{2} \mathrm{~g}\). Solve the formula for \(v\), and find the volume of this sphere. Round to the nearest whole number.
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