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Chapter 1: Problem 75

Find the real solutions of each equation by factoring. $$ x^{3}-9 x=0 $$

Short Answer

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

Step by step solution

01

Factor out the common term

First, observe the equation and look for a common factor. In this case, both terms on the left side have a factor of \(x\). Factor \(x\) out of the equation: \[ x(x^2 - 9) = 0 \]
02

Factor the quadratic expression

Next, factor the quadratic expression \(x^2 - 9\). Notice that it is a difference of squares, which can be factored as: \[ x(x - 3)(x + 3) = 0 \]
03

Set each factor equal to zero

Set each factor of the equation to zero and solve for \(x\): For the first factor: \[ x = 0 \]For the second factor: \[ x - 3 = 0 \ \ x = 3 \]For the third factor: \[ x + 3 = 0 \ \ x = -3 \]
04

Collect the solutions

Combine all the solutions obtained from solving each factor: 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.

Difference of Squares
The difference of squares is a vital concept in factoring polynomials. It applies to expressions of the form \[a^2 - b^2 = (a + b)(a - b)\]. This pattern arises because of special rules in algebra. For example, consider the quadratic expression in the solution: \(x^2 - 9\). Here, \(x^2\) is a square term, and \(9\) is also a square term (\(3^2\)).
We rewrite 9 as \(3^2\) to get \(x^2 - 3^2\).
According to the difference of squares formula, it factors into:
  • \(x^2 - 3^2 = (x + 3)(x - 3)\)
. This makes it easier to solve the equation, as factors simplify the polynomial into simpler components.
Quadratic Expression
A quadratic expression is a polynomial of degree 2, generally written as \(ax^2 + bx + c\). Quadratics can appear in various forms, and learning to recognize and factor them is essential. In our example, the expression \(x^2 - 9\) is a quadratic without the middle term (b = 0).
Key things to remember:
  • Standard form: \(ax^2 + bx + c\)
  • Finding common factors
  • Applying special factoring techniques, such as the difference of squares
. Factoring quadratics often leads to simple, solvable equations. Every quadratic expression can be solved using these methods, providing solutions in a straightforward manner.
Solving Equations
Solving polynomial equations by factoring involves breaking down the expression into product form. This converts a complex problem into simpler equations. In our example, after factoring we get: \(x(x - 3)(x + 3) = 0\).
The next step is setting each factor to zero:
  • \(x = 0\)
  • \(x - 3 = 0\) \( \rightarrow x = 3\)
  • \(x + 3 = 0\) \( \rightarrow x = -3\)
This approach leverages the Zero Product Property, which states if \(ab = 0\), then \(a = 0\) or \(b = 0\).
After solving each equation, we collect all the solutions to get the final answer. For polynomial equations, verifying each solution ensures accuracy.

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