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Solve by factoring. $$x^{3}-x^{2}-12 x=0$$

Short Answer

Expert verified
The solutions are \(x = 0\), \(x = 4\), and \(x = -3\).

Step by step solution

01

Identify Common Factor

Identify and factor out the greatest common factor from the polynomial. The given equation is: \(x^3 - x^2 - 12x = 0\). Notice that \(x\) is common in all terms so we can factor it out: \(x(x^2 - x - 12) = 0\).
02

Factor the Quadratic

Now, focus on factoring the quadratic equation inside the parentheses: \(x^2 - x - 12\). We need to find two numbers that multiply to \(-12\) (the constant term) and add up to \(-1\) (the coefficient of \(x\)). These numbers are \(-4\) and \(3\). So, the quadratic can be factored as \((x - 4)(x + 3)\).
03

Rewrite the Factored Equation

Substitute the factors of the quadratic back into the original equation to get: \(x(x - 4)(x + 3) = 0\).
04

Solve Each Factor for Zero

Set each factor in the equation \(x(x - 4)(x + 3) = 0\) equal to zero and solve for \(x\). This gives us the equations: \(x = 0\), \(x - 4 = 0\) and \(x + 3 = 0\). Solving these, we get \(x = 0\), \(x = 4\), and \(x = -3\).
05

List the Solutions

The solutions to the original equation \(x^3 - x^2 - 12x = 0\) are \(x = 0\), \(x = 4\), and \(x = -3\).

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Key Concepts

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

Quadratic Equations
Quadratic equations are expressions of the form \( ax^2 + bx + c = 0 \). In these equations, \( x \) represents a variable, and \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). These equations are significant because they often appear in various mathematical and real-world contexts.
To solve a quadratic equation, many methods are available:
  • Factoring: This involves finding two numbers that multiply to \( ac \) and add to \( b \). For example, in the quadratic \( x^2 - x - 12 \), the numbers are \(-4\) and \(3\).
  • Completing the square: A method to convert the equation into a perfect square trinomial, which can then be solved easily.
  • Quadratic formula: Given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), it's a universal method that works for any quadratic equation.
In our original exercise, we used factoring to break down the quadratic part of the polynomial. Factoring is one of the simplest methods when the quadratic can be easily factored.
Roots of Equations
The roots of an equation are the values of \( x \) that satisfy the equation. In simple terms, "roots" are another way of saying "solutions." Finding roots is like finding the numbers that make the equation true.
For equations that are set equal to zero, like \( x(x - 4)(x + 3) = 0 \), the roots are the values of \( x \) that make each factor zero.
In the step-by-step solution, we set each factor equal to zero:
  • For \( x = 0 \), the root is straightforward as it makes the equation zero directly.
  • For \( x - 4 = 0 \), solving yields \( x = 4 \).
  • For \( x + 3 = 0 \), solving gives \( x = -3 \).
Thus, the roots of the polynomial are \( x = 0, 4, \) and \(-3\). These roots are points where the polynomial graph intersects the \( x \)-axis.
Polynomial Equations
Polynomial equations are expressions that involve sums of powers of variables with constant coefficients. These powers are always whole numbers. A polynomial can have terms like \( x^3, x^2, x \), each with a coefficient. The original equation \(x^3 - x^2 - 12x = 0\) is a polynomial equation of degree 3 because its highest power of \( x \) is 3.

Polynomials are ubiquitous in algebra and can represent a wide array of functions and models in different fields. Solving these equations involves finding values for \( x \) that make the equation zero.
Different techniques for solving polynomial equations include:
  • Factoring: Identifying and separating factors makes solving for roots simpler.
  • Synthetic or long division: Useful for dividing polynomials.
  • Using the quadratic formula: Works well when dealing with quadratic polynomials, which was part of our solution.
Factoring is often the initial approach taken, as seen in our exercise, leading us to find the roots of the polynomial.

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