Problem 56 - Elementary Algebra

Question

Use a pattern to factor. Check. Identify any prime polynomials. $$ 100 j^{2}-k^{12} $$

Step-by-Step Solution

Answer

The factors are (10j + k^{6}) and (10j - k^{6}), which are both prime polynomials.

Step 1: Identify the pattern

Notice that the given polynomial, 100j^{2} - k^{12}, has the form of a difference of squares, a^{2} - b^{2}, because 100j^{2} and k^{12} are both perfect squares.

Step 2: Rewrite as a difference of squares

Express 100j^{2} and k^{12} as squares: 100j^{2} = (10j)^{2} and k^{12} = (k^{6})^{2}. Now the expression can be rewritten as: (10j)^{2} - (k^{6})^{2}.

Step 3: Factor using the difference of squares formula

Recall the difference of squares formula: a^{2} - b^{2} = (a + b)(a - b). Applying this to the expression (10j)^{2} - (k^{6})^{2}, we get: (10j + k^{6})(10j - k^{6}).

Step 4: Check for prime polynomials

Examine the factors (10j + k^{6}) and (10j - k^{6}). Neither of these factors can be factored further, and thus they are considered prime polynomials.

Key Concepts

difference of squares

A difference of squares is a specific type of polynomial that can be factored using the formula: \[a^{2} - b^{2} = (a + b)(a - b)\] This means when you have a polynomial where one square term is subtracted from another, you can rewrite it into a product of two binomials. Let's explore this with the example provided in the exercise: Given polynomial \[100j^{2} - k^{12}\], you first recognize that both \[100j^{2}\] and \[k^{12}\] are perfect squares. You can rewrite \[100j^{2}\] as \[(10j)^{2}\] and \[k^{12}\] as \[(k^{6})^{2}\]. This transforms our polynomial into \[(10j)^{2} - (k^{6})^{2}\]. Now, applying the difference of squares formula: \[(10j)^{2} - (k^{6})^{2} = (10j + k^{6})(10j - k^{6})\]. Remember, this factorization technique only works if you have a 'minus' between the squares. If you have a 'plus,' different techniques are required.

prime polynomials

Prime polynomials are polynomials that cannot be factored further over the set of integers. Think of them as the polynomial equivalent of prime numbers in arithmetic. In the provided example once we've factored \[100j^{2} - k^{12}\] into \[(10j + k^{6})(10j - k^{6})\], we need to check if these factors can be broken down even further.Examine \[10j + k^{6}\] and \[10j - k^{6}\]. There are no further common factors or specific patterns like the difference of squares that apply. Thus, both binomials are considered prime polynomials because they cannot be factored into simpler polynomials with integer coefficients.

perfect squares

A perfect square is a number or term that can be expressed as the square of another number or term. For instance, \[16\] is a perfect square because it equals \[4^{2}\], and \[x^{4}\] is a perfect square because it equals \[(x^{2})^{2}\]. In our exercise example, we had two perfect squares: \[100j^{2}\] and \[k^{12}\]. Here’s how we identify and rewrite them:

• \[100j^{2}\] can be written as \[(10j)^{2}\] because \[10j\] times \[10j\] yields \[100j^{2}\].
• \[k^{12}\] can be written as \[(k^{6})^{2}\] because \[k^{6}\] times \[k^{6}\] produces \[k^{12}\].

Recognizing and rewriting these perfect squares are crucial steps in applying the difference of squares formula efficiently. The transformation makes it straightforward to utilize the formula for factorization.