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Chapter 5: Problem 26

Solve each equation. $$n^{3}=16 n$$

Short Answer

Expert verified
The solutions are \( n = 0, 4, -4 \).

Step by step solution

01

Translate the Equation

The given equation is \( n^3 = 16n \). We need to rearrange this equation so that it equals zero. This can be done by subtracting \( 16n \) from both sides of the equation to get \( n^3 - 16n = 0 \).
02

Factor the Equation

The equation \( n^3 - 16n = 0 \) can be factored by taking out the greatest common factor, which is \( n \). The equation thus becomes \( n(n^2 - 16) = 0 \).
03

Solve the Factored Equation

Set each factor of the equation \( n(n^2 - 16) = 0 \) equal to zero: \( n = 0 \) or \( n^2 - 16 = 0 \). These are the potential solutions to the equation.
04

Solve the Quadratic Equation

We're left with \( n^2 - 16 = 0 \). To solve this, add 16 to both sides to get \( n^2 = 16 \). Then, taking the square root of both sides gives \( n = \pm 4 \).
05

Combine the Solutions

The solutions from the previous steps are \( n = 0 \), \( n = 4 \), and \( n = -4 \). Therefore, the complete solution set for the equation \( n^3 = 16n \) is \( n = 0, 4, -4 \).

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Key Concepts

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

Factoring
Factoring is an essential skill when dealing with polynomial equations. It involves breaking down an equation into simpler pieces, known as factors, which when multiplied together result in the original equation. Consider the equation from the exercise, \( n^3 - 16n = 0 \). A key step here is identifying the greatest common factor (GCF).
For this equation, the GCF is \( n \), since both terms, \( n^3 \) and \( 16n \), include an \( n \) that can be factored out. Factoring the equation gives us \( n(n^2 - 16) = 0 \).
  • The first factor \( n \) suggests that one solution to the equation is \( n = 0 \), because when \( n = 0 \), the product becomes zero.
  • The second factor, \( n^2 - 16 \), is further analyzed to find additional solutions.
Mastering factoring not only simplifies polynomial equations but also paves the way for more advanced algebraic manipulations.
Quadratic Equations
Quadratic equations are a form of polynomial equations where the highest degree of the unknown variable is 2. Generally, these take the form \( ax^2 + bx + c = 0 \). In our exercise, after factoring out \( n \), we are left with the quadratic equation \( n^2 - 16 = 0 \).
This equation can be rewritten using a difference of squares, a common technique for solving quadratics: \( n^2 - 16 \) is \((n + 4)(n - 4) = 0\) due to the identity \( a^2 - b^2 = (a-b)(a+b) \).
  • To solve, set each factor to zero, yielding \( n + 4 = 0 \) or \( n - 4 = 0 \).
  • Solving these gives \( n = -4 \) and \( n = 4 \).
Understanding quadratic equations, and the methods to solve them, is crucial in finding and verifying solutions for these and more complex equations.
Solving Equations
Solving equations involves finding the value of the unknown variable that makes the equation true. In our problem, this process begins once the equation is set to zero and factored, \( n(n^2 - 16) = 0 \).
  • First, look at \( n = 0 \) from the factor \( n \) directly, which offers one solution.
  • Next, the quadratic factor \( n^2 - 16 \) is converted into \((n + 4)(n - 4) = 0\) to get \( n = 4 \) or \( n = -4 \).
These steps lead to potential solutions: \( n = 0 \), \( n = 4 \), and \( n = -4 \). A crucial part of solving equations is verifying each potential solution by substituting them back into the original equation to ensure they work鈥攁 step that reinforces understanding and accuracy.
Solving equations requires careful manipulation and reasoning, helping students see how each step paves the way for the next and how solutions fit within the context of the original problem.

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Most popular questions from this chapter

Recall that a graphing calculator may be used to check addition, subtraction, and multiplication of polynomials. In the same manner, a graphing calculator may be used to check factoring of polynomials in one variable. For example, to see that $$ 2 x^{3}-9 x^{2}-5 x=x(2 x+1)(x-5) $$ graph \(\mathrm{Y}_{1}=2 x^{3}-9 x^{2}-5 x\) and \(\mathrm{Y}_{2}=x(2 x+1)(x-5) .\) Then trace along both graphs to see that they coincide. Factor the following and use this method to check your results. $$ x^{4}+6 x^{3}+5 x^{2} $$

Find the value of \(c\) that makes each trinomial a perfect square trinomial. $$ x^{2}+c x+16 $$

Factor. Assume that variables used as exponents represent positive integers. $$ x^{2 n}-36 $$

Find the value of \(c\) that makes each trinomial a perfect square trinomial. Factor \(x^{6}-1\) completely, using the following methods from this chapter. A. Factor the expression by treating it as the difference of two squares, \(\left(x^{3}\right)^{2}-1^{2}\) B. Factor the expression by treating it as the difference of two squares, \(\left(x^{3}\right)^{2}-1^{2}\) C. Factor the expression by treating it as the difference of two squares, \(\left(x^{3}\right)^{2}-1^{2}\)

Recall that a graphing calculator may be used to check addition, subtraction, and multiplication of polynomials. In the same manner, a graphing calculator may be used to check factoring of polynomials in one variable. For example, to see that $$ 2 x^{3}-9 x^{2}-5 x=x(2 x+1)(x-5) $$ graph \(\mathrm{Y}_{1}=2 x^{3}-9 x^{2}-5 x\) and \(\mathrm{Y}_{2}=x(2 x+1)(x-5) .\) Then trace along both graphs to see that they coincide. Factor the following and use this method to check your results. $$ x^{3}+6 x^{2}+8 x $$
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