Problem 94 - Algebra and Trigonometry with Analytic Geometry

Question

Factor the polynomial. $$ x^{8}-16 $$

Step-by-Step Solution

Answer

The polynomial can be factored as \((x^2 - 2)(x^2 + 2)(x^4 + 4)\).

Step 1: Recognize the Difference of Squares

The given polynomial is in the form of a difference of squares: \(x^8 - 16\). Recognize that \(x^8\) is a perfect square, as it is \((x^4)^2\), and 16 is also a perfect square, equal to \((4)^2\). Therefore, \(x^8 - 16 = (x^4)^2 - 4^2\).

Step 2: Apply Difference of Squares Formula

Use the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\). Let \(a = x^4\) and \(b = 4\). Substitute these into the formula to get \((x^4 - 4)(x^4 + 4)\).

Step 3: Recognize Further Difference of Squares

Notice that the new polynomial \(x^4 - 4\) is again a difference of squares. Here \(x^4\) can be expressed as \((x^2)^2\) and 4 as \((2)^2\), so \(x^4 - 4 = (x^2)^2 - 2^2\).

Step 4: Apply Difference of Squares Again

Apply the difference of squares formula to \(x^4 - 4\). Let \(a = x^2\) and \(b = 2\). Substitute these into the formula to get \((x^2 - 2)(x^2 + 2)\).

Step 5: Combine All Factors

Combine all the factors determined from steps 2 and 4. The original polynomial can be factored into \((x^4 - 4)(x^4 + 4) = (x^2 - 2)(x^2 + 2)(x^4 + 4)\).

Step 6: Confirm Irreducibility of Final Factors

Observe that \(x^4 + 4\) cannot be factored further using real numbers, as it does not satisfy the conditions for any currently applicable simple factoring techniques like the difference of squares.

Key Concepts

Difference of Squares

The difference of squares is a fundamental concept in algebra that simplifies the factoring of polynomial expressions. The formula for the difference of squares is:

• \(a^2 - b^2 = (a - b)(a + b)\).

Here, both \(a^2\) and \(b^2\) are perfect squares. The difference between them can be expressed as the product of two binomials. When applied, this formula quickly breaks down complex expressions into simpler parts.

Take for example our initial polynomial, \(x^8 - 16\). Recognizing \(x^8\) as \((x^4)^2\) and 16 as \(4^2\), we can directly factor it using the difference of squares strategy:

• \((x^4)^2 - 4^2 = (x^4 - 4)(x^4 + 4)\).

Overall, understanding this concept allows us to identify and reduce complex polynomial expressions into more manageable factors.

Polynomial Expressions

Polynomial expressions consist of variables and coefficients, expanded with the operations of addition, subtraction, multiplication, and non-negative integer powers.

Factoring polynomials involves finding expressions that can be multiplied together to obtain the original polynomial. This process is invaluable for simplifying equations and solving for roots.

• The exercise we dealt with here was focused on simplifying \(x^8 - 16\).
• By factoring this polynomial, we reduced it to \((x^2 - 2)(x^2 + 2)(x^4 + 4)\), which is much simpler to handle than the original expanded form.

Understanding polynomial structures and their properties helps in grasping various algebraic techniques, including factorization. It provides the foundation needed for tackling more complex algebraic problems.

Irreducibility

Irreducibility refers to the point where a polynomial cannot be factored further over the set of real numbers using standard algebraic methods. Determining irreducibility helps recognize when to stop factoring, saving time and simplifying calculations.

In our example, we explored the polynomial \(x^8 - 16\) and factored it to obtain \((x^2 - 2)(x^2 + 2)(x^4 + 4)\).

• Here \(x^4 + 4\) is determined to be irreducible over the reals.
• Unlike the differences of squares, no easy formula or factorization strategy applies here, as it doesn't split into simpler real factors.

Thus, recognizing irreducibility means identifying when further factorization is not possible or necessary, allowing one to present the cleanest, simplest form of a polynomial.