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Chapter 0: Problem 138

Factor completely. $$ x^{2 n}+6 x^{n}+8 $$

Short Answer

Expert verified
The completely factored form of the equation \( x^{2n} + 6x^n + 8 \) is \( (x^n + 2)(x^n + 4) \).

Step by step solution

01

Recognize the Pattern

In the equation \( x^{2n} + 6x^n + 8 \), we can let \( y = x^n \) and rewrite the polynomial as a standard quadratic equation, i.e. \( y^2 + 6y + 8 \).
02

Factorize the Quadratic Polynomial

The above equation can be factorized into \( (y + 2)(y + 4) \). This is achievable by finding two numbers that multiply to give 8 (the constant term) and add up to give 6 (the coefficient of middle term). The two numbers are 2 and 4.
03

Replace y with \(x^n\)

The last step is to substitute back the value of y into the equation, which is \( x^n \). Therefore, the factorized form of the original equation becomes \( (x^n + 2)(x^n + 4) \).

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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 a fundamental concept in algebra, representing polynomials where the highest exponent of the variable is two. A generic quadratic equation is structured as \( ax^2 + bx + c = 0 \). Each component plays a crucial role.
  • \( a \) is the coefficient of the squared term.
  • \( b \) is the coefficient of the linear term.
  • \( c \) is the constant term.

To solve quadratic equations, you can use various methods such as factoring, completing the square, or the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Factoring is often the preferred choice when the equation can be easily expressed as a product of binomials. In the exercise given, by letting \( y = x^n \), the polynomial is transformed into a more familiar quadratic form, \( y^2 + 6y + 8 \), which can be easily factored.
Exponents
Exponents are a way of expressing repeated multiplication of a number by itself. In algebra, exponents help manage larger numbers and simplify expressions. They follow specific rules that are essential to operations involving polynomials.
  • \( x^m \times x^n = x^{m+n} \)
  • \( (x^m)^n = x^{mn} \)
  • \( x^0 = 1 \) (provided \( x eq 0 \))

In the exercise, the variable \( x^{2n} + 6x^n + 8 \) uses exponents in the terms. The critical insight was to treat \( x^n \) as a single variable to convert the equation into a quadratic form. This abstraction highlights the power of understanding exponent rules in simplifying and solving polynomial expressions.
Polynomial Expressions
Polynomial expressions are mathematical phrases involving variables and coefficients, made up of terms with non-negative integer exponents. A term is generally in the form \( ax^k \), where \( a \) is the coefficient and \( k \) is the exponent.
Polynomials can take different forms like:
  • Monomial 鈥� one term, e.g., \( 3x^2 \).
  • Binomial 鈥� two terms, e.g., \( x^2 + 4x \).
  • Trinomial 鈥� three terms, e.g., \( x^2 + 6x + 8 \).

For the factoring process, recognizing patterns in polynomial expressions is key. The given expression \( x^{2n} + 6x^n + 8 \) is structured as a trinomial. By substituting \( x^n \) with \( y \), the expression resembles a simple quadratic trinomial that can be readily factored into \( (y + 2)(y + 4) \), simplifying the solving process.

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