Problem 96 - Algebra and Trigonometry

Question

Differences of Even Powers (a) Factor the expressions completely: \(A^{4}-B^{4}\) and \(A^{6}-B^{6} .\) (b) Verify that \(18,335=12^{4}-7^{4}\) and that \(2,868,335=12^{6}-7^{6} .\) (c) Use the results of parts (a) and (b) to factor the integers \(18,335\) and \(2,868,335 .\) Show that in both of these factorizations, all the factors are prime numbers.

Step-by-Step Solution

Answer

18,335: 5 x 19 x 193; 2,868,335: 5 x 19 x 11 x 23 x 97 (all factors are prime).

Step 1: Factor the Expression A^4 - B^4

The expression \(A^4 - B^4\) is a difference of squares. We can apply the identity \(a^2 - b^2 = (a - b)(a + b)\). Rewrite \(A^4 - B^4\) as \((A^2)^2 - (B^2)^2\). Factoring this using the difference of squares gives \((A^2 - B^2)(A^2 + B^2)\). Notice, \(A^2 - B^2\) itself is a difference of squares and can be further factored as \((A - B)(A + B)\). Thus, the complete factorization of \(A^4 - B^4\) is \((A - B)(A + B)(A^2 + B^2)\).

Step 2: Factor the Expression A^6 - B^6

The expression \(A^6 - B^6\) is also a difference of powers, specifically of cubes twice. We first factor it as \((A^3)^2 - (B^3)^2\), which is a difference of squares: \((A^3 - B^3)(A^3 + B^3)\). Now, each term \(A^3 - B^3\) and \(A^3 + B^3\) can be factored using the sum and difference of cubes. The difference of cubes \(A^3 - B^3\) is factored as \((A - B)(A^2 + AB + B^2)\), and the sum of cubes \(A^3 + B^3\) is factored as \((A + B)(A^2 - AB + B^2)\). Thus, the complete factorization of \(A^6 - B^6\) is \((A - B)(A + B)(A^2 + AB + B^2)(A^2 - AB + B^2)\).

Step 3: Verify and Calculate 12^4 - 7^4 and 12^6 - 7^6

Calculate \(12^4\) and \(7^4\): \(12^4 = 20,736\) and \(7^4 = 2,401\). Subtract them: \(20,736 - 2,401 = 18,335\). Now, calculate \(12^6\) and \(7^6\): \(12^6 = 2,985,984\) and \(7^6 = 117,649\). Subtract them: \(2,985,984 - 117,649 = 2,868,335\). Both calculations verify the claims in part (b).

Step 4: Factor and Verify Prime Factors of 18,335

Using the factorization from part (a), apply it to \(12^4 - 7^4 = 18,335\). This expression is \((12^2 - 7^2)(12^2 + 7^2)\). Calculate these squares: \(12^2 = 144\) and \(7^2 = 49\). So it becomes \((144 - 49)(144 + 49)\) = \(95 \times 193\). Both 95 and 193 are composite: 95 = 5 x 19, and 193 is prime.

Step 5: Factor and Verify Prime Factors of 2,868,335

Using the factorization from part (a), apply it to \(12^6 - 7^6 = 2,868,335\). This expression is \((12 - 7)(12 + 7)(12^2 + 12\times7+ 7^2)(12^2 - 12\times7+ 7^2)\). Calculate these: \((5)(19)(253)(97)\). Check primality: 5 and 19 are prime; 253 = 11 x 23 (both prime); 97 is prime.

Key Concepts

Factorization

Factorization is a mathematical process where we break down complex expressions into a product of simpler factors. This concept is particularly beneficial for simplifying polynomial expressions, solving equations, and finding roots. By recognizing and applying various factorization techniques, we can decompose expressions into a form that reveals their inherent structures. In the exercise,

• For example, by recognizing that expressions like \(A^4 - B^4\) need simplification, we apply factorization rules like the difference of squares.
• Similarly, \(A^6 - B^6\) can be simplified using a sequence of factorization processes.

Thus, learning how to factor expressions not only makes calculations more manageable but also enhances our understanding of the expression's properties.

Difference of Squares

The difference of squares is a specific form of factorization used for expressions that take the form \(a^2 - b^2\). This form is special because it can always be factored into two linear expressions: \((a - b)(a + b)\). This method allows us to break down even powers of numbers and expressions into their roots.

• In our example with \(A^4 - B^4\), we notice that this can be rewritten as \((A^2)^2 - (B^2)^2\), essentially a difference of squares.

Breaking it down further using the same principle gives you \((A^2 - B^2)(A^2 + B^2)\).
This principle of using a difference of squares is foundational in algebra and aids in simplifying quadratic expressions into linear factors.

Difference of Cubes

The difference of cubes builds on the difference of powers and provides strategies to factor expressions of the form \(a^3 - b^3\). The formula used for this factorization is \((a - b)(a^2 + ab + b^2)\). Using this, we can turn cubic expressions into simpler, more solvable forms.

• For instance, \(A^6 - B^6\) initially looks daunting, but if we think of it in terms of cubes, we realize that we can break it down using the difference of squares first, then factor the resulting pieces further.
• We apply the difference of cubes in the steps to further break down these components: \((A^3 - B^3)(A^3 + B^3)\).

The **difference of cubes** is a powerful method that offers insight into handling higher degree polynomials by simplifying them into a product of more straightforward expressions.

Prime Factorization

Prime factorization involves breaking down a number into its basic building blocks – prime numbers. This process is crucial for fully understanding an expression's composition and for applications like simplifying fractions and finding greatest common divisors or least common multiples.

• Given an integer like 18,335, breaking it down to its prime factors involves recognizing its expression as a product, here verified by calculations: \( 95 \times 193 \).
• Similarly, for 2,868,335, we find a step-by-step lineup of prime factors: \((5)(19)(253)(97)\).

Each non-prime factor, like 95 and 253 in our example, needs further decomposition until all factors in the product are prime. Understanding **prime factorization** enriches our problem-solving toolkit, equipping us to handle both theoretical and practical math problems efficiently.