- Identify \(p^4 - 16\) as a difference of squares: \((p^2)^2 - 4^2\).
- Apply the difference of squares formula to factor it into \((p^2 - 4)(p^2 + 4)\).
- Further factor \((p^2 - 4)\) into \((p - 2)(p + 2)\).
- Conclude that the completely factored form is \(\boxed{{(p - 2)(p + 2)(p^2 + 4)}}\).
Explanation
- Identify the difference of squares
The expression given is \(p^4 - 16\). This is a polynomial of the form \(a^2 - b^2\), where \(a = p^2\) and \(b = 4\), since \((p^2)^2 = p^4\) and \(4^2 = 16\). The difference of squares formula, \(a^2 - b^2 = (a - b)(a + b)\), allows us to express the original polynomial as a product of two factors:
\[p^4 - 16 = (p^2 - 4)(p^2 + 4)\]
- Factor the remaining expression
We have the expression \((p^2 - 4)(p^2 + 4)\). The first factor, \(p^2 - 4\), is itself a difference of squares because it can be written as \(p^2 - 2^2\). Applying the same difference of squares formula with \(a = p\) and \(b = 2\), we get:
\[p^2 - 4 = (p - 2)(p + 2)\]
The second factor, \(p^2 + 4\), cannot be further factored into polynomials with real coefficients, as there are no real numbers that satisfy \(p^2 = -4\).
- Combine factors and conclude
Combining the factors obtained in the previous steps, we get the complete factorization:
\[p^4 - 16 = (p - 2)(p + 2)(p^2 + 4)\]
Comparing this result to the given options, we find that the last option \((p - 2)(p + 2)(p^2 + 4)\) matches our result perfectly.
Examples
Factoring polynomials is a fundamental skill in algebra that allows us to simplify complex expressions and solve equations efficiently. For instance, in physics, when calculating the trajectory of an object, you might encounter equations that need to be broken down into simpler parts. Understanding how to use the difference of squares, such as \(a^2 - b^2 = (a - b)(a + b)\), simplifies these problems, helping you find roots or critical points of functions, which in turn helps in modeling real-world physical systems.