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Solve each equation by factoring. [Hint for: First factor out a fractional power.] $$ 5 x^{3}-20 x=0 $$

Short Answer

Expert verified
Solutions are \( x = 0 \), \( x = 2 \), and \( x = -2 \).

Step by step solution

01

Factor Out the Greatest Common Factor

First, identify and factor out the greatest common factor from the equation. The common factor for both terms in the equation \( 5x^3 - 20x = 0 \) is 5x. Thus, we factor it out to get \( 5x(x^2 - 4) = 0 \).
02

Set Each Factor to Zero

Once the factors are identified, set each factor equal to zero to solve for \( x \). We have two factors: \( 5x \) and \( x^2 - 4 \). Set them to zero: \( 5x = 0 \) and \( x^2 - 4 = 0 \).
03

Solve Simple Linear Equation

Start by solving the linear equation \( 5x = 0 \). Divide both sides by 5 to get \( x = 0 \).
04

Solve the Quadratic Equation

Now, solve the quadratic equation \( x^2 - 4 = 0 \). This is a difference of squares, which can be factored into \( (x - 2)(x + 2) = 0 \). Set each factor equal to zero: \( x - 2 = 0 \) and \( x + 2 = 0 \).
05

Solve for Values of x

Solve the simple equations from the previous step: \( x - 2 = 0 \) gives \( x = 2 \), and \( x + 2 = 0 \) gives \( x = -2 \).
06

Compile All Solutions

Combine all solutions from the previous steps. The solutions to the original equation are \( x = 0 \), \( x = 2 \), and \( x = -2 \).

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Key Concepts

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

Factoring
Factoring is a fundamental method in algebra used to simplify polynomial equations, making them easier to solve. The process involves expressing the polynomial as a product of its factors. This technique not only helps in solving equations but also in simplifying expressions. For instance, when asked to solve \(5x^3 - 20x = 0\), the problem can be daunting at first. However, by factoring, the equation can be broken down into manageable parts.

In our example, the factorization of \(5x(x^2 - 4)\) allows us to easily solve for \(x\). The initial step involves identifying the greatest common term that divides each term in the polynomial. Once this common term is factored out, the remaining quadratic can often be further broken down, as seen with the difference of squares method.
Greatest Common Factor (GCF)
The Greatest Common Factor (GCF) is the largest factor that two or more numbers or terms share. Identifying the GCF is critical when factoring polynomials as it simplifies the equation, making further manipulation straightforward. In an equation like \(5x^3 - 20x = 0\), recognizing the GCF involves looking for the highest number and power of \(x\) that can divide each term.

In this particular equation, the GCF is \(5x\). By factoring out \(5x\), the expression simplifies to \(5x(x^2 - 4) = 0\). This operation reduces the polynomial to simpler components, enabling easier resolution of each part. Understanding and identifying the GCF saves time and reduces complexity when solving algebraic expressions.
Quadratic Equations
Quadratic equations are polynomials of degree two, generally expressed in the form \(ax^2 + bx + c = 0\). Solving quadratics is a fundamental part of algebra, as they appear frequently in various math problems. After factoring out the GCF from the equation \(5x^3 - 20x = 0\), one is left with a quadratic portion, \(x^2 - 4\).

Quadratic equations can be solved using several methods, including factoring, completing the square, and using the quadratic formula. The equation \(x^2 - 4\) is particularly straightforward, as it is a difference of squares, allowing it to be easily factored. With proficiency in solving quadratics, you'll find it easier to handle many algebraic challenges.
Difference of Squares
The difference of squares is a special case in polynomial factoring useful for breaking down expressions. The mathematical form is \(a^2 - b^2 = (a-b)(a+b)\), which relies on the identity of squares subtraction. Expressions matching this form can be split into the product of two binomials.

In the polynomial \(x^2 - 4\), you identify it as a difference of squares because it can be rewritten as \((x - 2)(x + 2)\). This form allows the equation to be solved effortlessly, as each factor can be set to zero to solve for the variable \(x\).
  • \(x - 2 = 0\) gives \(x = 2\)
  • \(x + 2 = 0\) gives \(x = -2\)
Recognizing and applying the difference of squares formula can significantly ease the process of solving polynomial equations.

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