Problem 38 - Elementary Algebra

Question

Use a pattern to factor. Check. Identify any prime polynomials. $$ x^{3}+216 $$

Step-by-Step Solution

Answer

\(x^{3} + 216 = (x + 6)(x^{2} - 6x + 36)\). The quadratic \(x^{2} - 6x + 36\) is prime.

Step 1 - Identify the polynomial form

Recognize that the given polynomial, \(x^{3} + 216\), is a sum of cubes. It matches the form \(a^{3} + b^{3}\) where \(a = x\) and \(b = 6\).

Step 2 - Apply the sum of cubes formula

Use the sum of cubes factorization formula: \[a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})\]Substituting \(a = x\) and \(b = 6\), we get:\[x^{3} + 6^{3} = (x + 6)(x^{2} - 6x + 36)\]

Step 3 - Verify the factorization

Multiply the factors \((x + 6)(x^{2} - 6x + 36)\) to ensure they produce the original polynomial:\[(x + 6)(x^{2} - 6x + 36) = x^{3} - 6x^{2} + 36x + 6x^{2} - 36x + 216 = x^{3} + 216\]Thus, the factorization is correct.

Step 4 - Identify prime polynomials

Identify any prime polynomials within the factorized form. The quadratic \(x^{2} - 6x + 36\) does not have real roots and cannot be factored further using real numbers, making it a prime polynomial over the real numbers.

Key Concepts

sum of cubes

In algebra, a sum of cubes is a special type of polynomial. It takes the form \(a^{3} + b^{3}\), where both terms are perfect cubes. For example, \(x^{3} + 216\) can be seen as \(x^{3} + 6^{3}\). To factor such expressions, we use the sum of cubes formula:
\ \ \(a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})\)
This form helps us rewrite the polynomial into simpler terms. For instance, if we have \(x^{3} + 216\), we identify \(a = x\) and \(b = 6\).
Plugging these values into the sum of cubes formula, we get \((x + 6)(x^{2} - 6x + 36)\).
The resulting expression is more manageable and easier to work with in further algebraic manipulations.

polynomial factorization

Polynomial factorization is the process of expressing a polynomial as a product of its factors. This helps us simplify complex expressions and solve equations more efficiently. Consider the example \(x^{3} + 216\). By recognizing it as a sum of cubes, we can use the formula:
\ \ \(a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})\).
Here, we factor it into \((x + 6)(x^{2} - 6x + 36)\).
We should always verify our factorization by expanding the factors to ensure they produce the original polynomial. In this case, multiplying \((x + 6)(x^{2} - 6x + 36)\) gives us back \(x^{3} + 216\), confirming the factorization is correct.
Factorizing polynomials simplifies solving and understanding their behavior.

prime polynomials

A prime polynomial is a polynomial that cannot be factored into products of polynomials with lower degrees using real numbers. After factorizing \(x^{3} + 216\) into \((x + 6)(x^{2} - 6x + 36)\), we need to check if any of the factors can be split further.
In our case, the quadratic polynomial \(x^{2} - 6x + 36\) does not have real roots, making it non-factorable over the real numbers. Thus, we classify it as a prime polynomial.
\ \ Key points to remember about prime polynomials:

• They cannot be expressed as a product of lower-degree polynomials with real coefficients.
• They greatly simplify the process of solving polynomial equations, since they represent the most reduced form of the polynomial components.

Recognizing prime polynomials helps in understanding the structure and properties of polynomial equations more thoroughly.