Problem 8 - Elementary Algebra

Question

Factor each binomial completely. $$ 16 y^{8}-1 $$

Step-by-Step Solution

Answer

Question: Factor the binomial completely: 16y^8 - 1 Answer: (4y^4 + 1)(4y^4 - 1)

Step 1: Rewrite the binomial

Rewrite the given binomial as a difference of squares: $$ 16y^8 - 1 = (4y^4)^2 - (1)^2 $$

Step 2: Apply the difference of squares

Using the difference of squares identity, factor the binomial: $$ (4y^4)^2 - (1)^2 = (4y^4 + 1)(4y^4 - 1) $$

Step 3: Verify if further factoring is possible

Since both factors are sums/differences of prime power terms, no further factoring is possible. The completely factored form of the given binomial is: $$ 16y^8 - 1 = (4y^4 + 1)(4y^4 - 1) $$

Key Concepts

Difference of Squares

The difference of squares is a powerful algebraic tool that helps in factoring certain types of binomials. It involves expressions where one term is squared and then subtracted from another squared term. The general form is given by:
\[ a^2 - b^2 = (a + b)(a - b) \]
This identity is unique because it allows us to break down complex expressions into simpler terms quickly, without needing to solve quadratic equations or apply more complicated methods.

• For example, if you have an expression like \( x^2 - 9 \), it fits the difference of squares format with \( a = x \) and \( b = 3 \).
  By applying the formula, we can factor it into \( (x + 3)(x - 3) \).
• In the original exercise we worked with the expression \( 16y^8 - 1 \), and identified it as a difference of squares with \( a = 4y^4 \) and \( b = 1 \).

Using this method can simplify otherwise tough-looking polynomials, making them easier to handle and solve.

Polynomial Expressions

Polynomial expressions are sums or differences of terms, each of which is composed of a coefficient, a variable raised to a power, and possibly a constant term. They are a fundamental component of algebra and can vary in complexity.
The degree of a polynomial is determined by the highest power of the variable present in the expression. In our example problem, the polynomial expression is \( 16y^8 - 1 \).
Here are a few key points about polynomials:

• Each term of a polynomial consists of a coefficient and a variable raised to a certain power. In \( 16y^8 \), \( 16 \) is the coefficient and \( y^8 \) is the variable term.
• Polynomials can be classified by their number of terms: monomial (one term), binomial (two terms), and trinomial (three terms).
• Understanding the structure of polynomial expressions allows you to apply the appropriate factoring or simplification techniques, such as using algebraic identities like the difference of squares.

Polynomials are versatile; they model a variety of real-world phenomena and possess numerous algebraic properties that can be explored through factorization and operation.

Algebraic Identities

Algebraic identities are equations that hold true for all values of the involved variables. They are used extensively in algebra to simplify calculations, prove other mathematical statements, and solve equations. One important identity relevant to our exercise is the difference of squares identity:
\[ a^2 - b^2 = (a + b)(a - b) \]
This identity allows us to factor calculations rapidly, enabling a more efficient mathematical manipulation of expressions. Here are some insights:

• Algebraic identities help reduce the complexity of polynomial equations and expressions. They are crucial when we need to factor complex expressions or when we are trying to simplify algebraic forms.
• Besides the difference of squares, other notable identities include the square of a sum and the square of a difference, which take the form \( (a + b)^2 = a^2 + 2ab + b^2 \) and \( (a - b)^2 = a^2 - 2ab + b^2 \) respectively.
• Mastering these identities improves conceptual understanding and problem-solving skills in algebra, providing a toolkit for tackling a broad range of mathematical problems.

Using algebraic identities, students can enhance their ability to work through algebraic expressions and equations with greater ease and insight.