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Chapter 6: Problem 45

$$ \text { For Problems } 41-70 \text {, solve each equation. (Objective } 3 \text { ) } $$ $$ -2 x^{3}+8 x=0 $$

Short Answer

Expert verified
The solutions are \(x = 0, x = 2, x = -2\).

Step by step solution

01

Identify Common Factors

Examine the equation \(-2x^3 + 8x = 0\) and notice that both terms on the left have a common factor of \(2x\). Therefore, we can factor \(2x\) out of the equation.
02

Factor the Equation

Factor \(2x\) out of the equation: \(-2x(x^2 - 4) = 0\). This shows us the outer factor as well as a new quadratic expression within the parenthesis.
03

Use the Zero Product Property

According to the Zero Product Property, if a product of factors is zero, at least one of the factors must be zero. Thus, we can set each factor equal to zero: \(-2x = 0\) and \((x^2 - 4) = 0\).
04

Solve for x from Each Factor

First solve \(-2x = 0\). This simplifies to \(x = 0\). Next, solve \(x^2 - 4 = 0\). This can be factored further into \((x - 2)(x + 2) = 0\).
05

Solve the Quadratic Factors

Solve \((x - 2)(x + 2) = 0\) by setting each factor to zero: \(x - 2 = 0\) which gives \(x = 2\), and \(x + 2 = 0\) which gives \(x = -2\).
06

Determine the Solution

Combine all possible solutions: \(x = 0\), \(x = 2\), and \(x = -2\). These are the values of \(x\) that satisfy the original equation.

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Key Concepts

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

Factoring
Factoring is a technique used to simplify an equation or expression by breaking it down into smaller and more manageable parts known as factors. These factors, when multiplied together, yield the original equation or expression. In the context of the exercise \(-2x^3 + 8x = 0\), factoring plays a crucial role in solving the equation. To begin factoring the equation:
  • Identify the greatest common factor of the terms present. This step helps in simplifying the equation from a complex form into a product of simpler terms.
  • For \(-2x^3 + 8x\), both terms share \(2x\) as a common factor. By factoring it out, you form \(-2x(x^2 - 4)\).
  • This breakdown reveals a quadratic equation inside the parentheses, ready for further analysis.
Factoring is like peeling layers of an onion, where each layer removed makes the problem simpler and often exposes new aspects of the expression to explore further.
Zero Product Property
The Zero Product Property is a fundamental principle in algebra that makes solving equations involving products of factors straightforward. It states that if the product of two or more factors is zero, then at least one of the factors must be zero. When applied to the equation \(-2x(x^2 - 4) = 0\):
  • You separate each factor and set them equal to zero: \(-2x = 0\) and \(x^2 - 4 = 0\). This highlights the potential solutions hidden in the equation.
  • The simplicity of this property lies in it's straightforward application. Once a product equals zero, you can treat each component as a stand-alone equation.
  • In \(-2x = 0\), the solution comes easy: \(x = 0\).
  • In \(x^2 - 4 = 0\), further solving reveals: \((x - 2)(x + 2) = 0\).
The Zero Product Property thus segments the problem into simpler parts that are individually solvable, making it more manageable.
Quadratic Equations
Quadratic equations are polynomial equations of the second degree, generally expressed in the standard form \(ax^2 + bx + c = 0\). Solving such equations is a fundamental skill in algebra. In the given exercise, the equation \(x^2 - 4 = 0\) surfaces after initial factoring.To tackle quadratic equations:
  • Check if it can be factored easily. In this case, \(x^2 - 4\) is a classic difference of squares, which can be factored as \((x - 2)(x + 2)\).
  • Each of these factors can be set to zero individually, giving \(x = 2\) and \(x = -2\) as solutions.
  • Quadratic equations can also be solved using other methods, such as completing the square or utilizing the quadratic formula, especially when factoring is not straightforward.
In this exercise, factoring was sufficient, but it鈥檚 always helpful to be aware of multiple solving strategies. Quadratic equations are central to algebra and appear frequently in various mathematical contexts.

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Most popular questions from this chapter

For Problems \(13-40\), factor each polynomial completely. Indicate any that are not factorable using integers. (Objective 2) $$ 4 x^{2}-28 x y+49 y^{2} $$

What does the expression "not factorable using integers" mean to you?

For Problems \(71-88\), set up an equation and solve each problem. (Objective 4) Suppose that the combined area of two squares is 360 square feet. Each side of the larger square is three times as long as a side of the smaller square. How big is each square?

$$ \text { For Problems } 41-70 \text {, solve each equation. (Objective } 3 \text { ) } $$ $$ 24 x^{2}+17 x-20=0 $$

Suppose that we want to factor \(n^{2}+26 n+168\) so that we can solve the equation \(n^{2}+26 n+168=0\). We need to find two positive integers whose product is 168 and whose sum is 26 . Since the constant term, 168, is rather large, let's look at it in prime factored form: $$ 168=2 \cdot 2 \cdot 2 \cdot 3 \cdot 7 $$ Now we can mentally form two numbers by using all of these factors in different combinations. Using two 2 s and the 3 in one number and the other 2 and the 7 in another number produces \(2 \cdot 2 \cdot 3=12\) and \(2 \cdot 7=14\). Therefore, we can solve the given equation as follows: $$ \begin{aligned} n^{2}+26 n+168 &=0 & \\ (n+12)(n+14)=0 & & \\ n+12=0 & \text { or } & n+14 &=0 \\ n=-12 & \text { or } & n &=-14 \end{aligned} $$ The solution set is \(\\{-14,-12\\}\). Solve each of the following equations. (a) \(n^{2}+30 n+216=0\) (b) \(n^{2}+35 n+294=0\) (c) \(n^{2}-40 n+384=0\) (d) \(n^{2}-40 n+375=0\) (e) \(n^{2}+6 n-432=0\)
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