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Solve each equation. $$ 3 x^{3}-27 x=0 $$

Short Answer

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

Step by step solution

01

Factor out the greatest common factor

The equation given is \( 3x^3 - 27x = 0 \). Notice that both terms \(3x^3\) and \(-27x\) have a common factor of \(3x\). So, factor \(3x\) from the equation: \( 3x(x^2 - 9) = 0 \).
02

Set each factor equal to zero

Once the equation is factored, set each factor equal to zero to find the solutions. 1. \( 3x = 0 \)2. \( x^2 - 9 = 0 \)
03

Solve \(3x = 0\)

Divide both sides by 3 to isolate \(x\): \[ x = 0 \]
04

Solve \(x^2 - 9 = 0\)

This is a difference of squares. It can be factored as: \( (x - 3)(x + 3) = 0 \)Now, set each factor equal to zero and solve for \(x\): 1. \( x - 3 = 0 \) 2. \( x + 3 = 0 \)
05

Solve \(x - 3 = 0 \)

Add 3 to both sides to solve for \(x\): \[ x = 3 \]
06

Solve \(x + 3 = 0 \)

Subtract 3 from both sides to solve for \(x\): \[ x = -3 \]
07

Combine all solutions

Combine the solutions from all previous steps. Therefore, the solutions are \( x = 0 \), \( x = 3 \), and \( x = -3 \).

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Key Concepts

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

Polynomial Equations
Polynomial equations involve expressions that include terms with variables raised to various powers. In the equation given in the exercise, you see terms like \(3x^3\) and \(-27x\). These are typical components of a polynomial equation because they contain variables with exponents.

Polynomial equations can be simple, involving just one or two terms, or they can be complex with many terms and high degree exponents. The degree of a polynomial is based on the highest power of the variable present. In this particular equation, \(3x^3 - 27x = 0\), the highest power is 3, making it a cubic polynomial. Solving polynomial equations involves finding the values of the variable that make the equation true, also known as the roots or solutions of the equation.
  • The given polynomial can be decomposed into simpler parts which can be solved simultaneously.
  • Recognizing the structure of a polynomial allows us to employ strategies like factoring to simplify the problem.
Difference of Squares
The concept of the difference of squares is a useful algebraic tool for factoring polynomials. Recognizing a difference of squares can simplify the process of finding solutions to an equation. A difference of squares is a specific kind of binomial that takes the form \(a^2 - b^2\).

In the factorized version of the original equation, \(x^2 - 9\) is a classic example of a difference of squares, where \(a^2 = x^2\) and \(b^2 = 9\). The difference of squares can be factored into two binomials: \((x-3)(x+3)\). This transformation is vital because it reduces the quadratic part of the problem into simpler linear equations.
  • The formula for factoring a difference of squares is \(a^2 - b^2 = (a - b)(a + b)\).
  • It offers a direct method to factor and resolve equations quickly.
Common Factor
Finding a common factor is one of the first steps in solving polynomial equations, especially when they involve more than one term. A common factor is a shared number or expression, present in each term of the polynomial.

In the original equation \(3x^3 - 27x = 0\), both terms \(3x^3\) and \(-27x\) share the factor \(3x\), making \(3x\) the greatest common factor (GCF). By factoring out the GCF, the equation simplifies significantly. Factoring creates an equation like \(3x(x^2 - 9) = 0\), which allows for easier manipulation and solving.
  • Always begin by identifying any common factor present in the terms of the polynomial.
  • Factor the GCF from the expression to simplify the equation and isolate variable terms.
Understanding how to extract and use the common factor helps intercept complex equations early in the process, making later steps more straightforward.

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