Problem 59 - Elementary Algebra

Question

Factor completely. Identify any prime polynomials. $$ 2 p^{2}-800 $$

Step-by-Step Solution

Answer

The complete factorization of \(2p^2 - 800\) is \(2(p - 20)(p + 20)\).

Step 1: Identify Common Factor

Look at the given polynomial expression, which is \(2p^2 - 800\). Identify any common factors in the terms of the polynomial. Here, the number 2 is a common factor.

Step 2: Factor Out the Common Factor

Factor the common factor 2 out of the polynomial: \[2(p^2 - 400)\]

Step 3: Recognize and Apply Difference of Squares

Observe the expression inside the parentheses: \(p^2 - 400\). Notice it is a difference of squares where \(400\) can be written as \(20^2\): \[p^2 - 20^2\]. The difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\) can now be applied here.

Step 4: Factor the Difference of Squares

Using the difference of squares formula, factor \(p^2 - 20^2\) into: \[(p - 20)(p + 20)\].

Step 5: Combine All Factors

Include the common factor of 2 that was factored out initially with the factored difference of squares: \[2(p - 20)(p + 20)\]

Key Concepts

Common Factors

Common factors are numbers or variables that are common to all terms in a polynomial.
They can be factored out to simplify the polynomial.
For instance, in the polynomial expression ote{2p^2 - 800}, we notice that the number 2 is a common factor. Here's a simple way to spot common factors and factor them out:

• Look for the greatest common factor (GCF) of all coefficients.
• If there are variables, check if they share any common powers.

Once identified, you factor out the GCF from each term. So in our example,
we factor 2 out to get: ote{2(p^2 - 400)}. This makes the polynomial easier to work with in subsequent steps.
Factoring out common factors is a key first step in polynomial factorization and helps in recognizing further simplifications.

Difference of Squares

The difference of squares is a special form of polynomial that looks like ote{a^2 - b^2}.
This can be factored using the formula: ote{a^2 - b^2 = (a - b)(a + b)}. This formula leverages the idea that:

• The two terms are perfect squares.
• There is a subtraction operation between them.

In our polynomial example, after factoring out the 2, we have ote{p^2 - 400}.
Notice that 400 is a perfect square because ote{400 = 20^2}. We can then rewrite the expression inside the parentheses as a difference of squares: ote{p^2 - 20^2}.
Applying the difference of squares formula, it factors to ote{(p - 20)(p + 20)}.
Understanding the difference of squares is essential for quickly recognizing and factoring such expressions.

Prime Polynomials

Prime polynomials are polynomials that cannot be factored any further over a given set of numbers.
For most exercises, this means they cannot be factored using integers.
For instance, a common quadratic polynomial like ote{x^2 + 1} does not factor further in the set of real numbers.
In our example, after completely factoring the polynomial ote{2p^2 - 800} into ote{2(p - 20)(p + 20)}, none of these factors can be factored further.
Thus, they are as simplified as possible.
To identify prime polynomials, always try to factor them using techniques like finding common factors, checking for patterns (such as difference of squares), and applying factorization formulas.
If none of these methods work, the polynomial is likely prime.
Recognizing when a polynomial is prime is a valuable skill, especially as problems become more complex.