Problem 70 - Elementary Algebra

Question

Use any of the factoring methods to factor. Identify any prime polynomials. $$ 36 h^{6}-49 j^{10} $$

Step-by-Step Solution

Answer

\[(6h^3 - 7j^5)(6h^3 + 7j^5)\]are prime factors.

Step 1: Identify Type of Polynomial

Recognize that the given polynomial is a difference of squares since both terms are perfect squares. A difference of squares has the form: \[a^2 - b^2\]

Step 2: Rewrite the Expression

Express each term as a square: \[36h^6 = (6h^3)^2\] \[49j^{10} = (7j^5)^2\]

Step 3: Apply the Difference of Squares Formula

Using the formula \[a^2 - b^2 = (a - b)(a + b)\] where \(a = 6h^3\) and \(b = 7j^5\), rewrite the polynomial: \[(6h^3)^2 - (7j^5)^2 = (6h^3 - 7j^5)(6h^3 + 7j^5)\]

Step 4: Verify Factors

Check that \[(6h^3 - 7j^5)\] and \[(6h^3 + 7j^5)\] are both prime polynomials, meaning they cannot be factored further.

Key Concepts

difference of squares

In algebra, the difference of squares is a special case of polynomial factoring. It appears when you have two perfect squares subtracted from each other. The general formula for the difference of squares is: \[a^2 - b^2 = (a - b)(a + b)\] This type of polynomial is significant because it can always be factored into two binomials. Perfect squares are numbers or expressions that result from squaring a number or expression. For example, \[36 = 6^2\] and \[h^6 = (h^3)^2\], making \[36h^6 = (6h^3)^2\]. Similarly, \[49 = 7^2\] and \[j^{10} = (j^5)^2\], so \[49j^{10} = (7j^5)^2\]. In this exercise, the given polynomial \[36h^6 - 49j^{10}\] can be written as a difference of squares and factored accordingly. For more complex polynomials, it is important to recognize when they can be expressed in this form to simplify the process of factoring.

prime polynomials

A prime polynomial is a polynomial that cannot be factored into the product of two non-constant polynomials over its coefficient field. It is similar to the concept of prime numbers in integers. Identifying prime polynomials is crucial because it lets you know when the factoring process is complete. In the step-by-step solution, after factoring \[36h^6 - 49j^{10}\] into \[(6h^3 - 7j^5)(6h^3 + 7j^5)\], we need to check if these factors can be broken down further. Since neither \[6h^3 - 7j^5\] nor \[6h^3 + 7j^5\] can be factored into simpler polynomials with integer coefficients, they are considered prime polynomials. Remember, a polynomial is prime if no further factoring is possible using rational numbers.

factoring techniques

Factoring polynomials is an essential skill in algebra. It involves expressing a polynomial as a product of simpler polynomials. Several techniques exist for factoring, with each suited for a different kind of polynomial. Common factoring techniques include:

• Factoring out the greatest common factor (GCF)
• Factoring by grouping
• Factoring trinomials
• Special cases like difference of squares, perfect square trinomials, and sum/difference of cubes

In our exercise, we used the difference of squares method because the polynomial was a difference of perfect squares. After identifying the type of polynomial, we expressed each term as a square and applied the appropriate formula. Understanding various factoring techniques helps in breaking down complex polynomials into simpler factors, making algebraic manipulations more manageable.