Problem 48 - Elementary Algebra

Question

Factor completely. Identify any prime polynomials. $$ 80 p^{3}-180 p v^{2} $$

Step-by-Step Solution

Answer

The factored form is \(20p(2p + 3v)(2p - 3v)\).

Step 1: Identify the common factor

The first step is to find the greatest common factor (GCF) of the terms in the expression. For the terms \(80p^3\) and \(-180pv^2\), the GCF is \(10p\).

Step 2: Factor out the GCF

Factor \(10p\) out of each term: \[ 80p^3 - 180pv^2 = 10p (8p^2 - 18v^2) \]

Step 3: Simplify the expression inside the parentheses

Simplify \(8p^2 - 18v^2\) recognizing any common numerical factor: \[ 8p^2 - 18v^2 = 2(4p^2 - 9v^2) \]

Step 4: Combine factored terms

Combine the factors \(10p\) and \(2\): \[ 10p \times 2(4p^2 - 9v^2) = 20p(4p^2 - 9v^2) \]

Step 5: Recognize a difference of squares

'\(4p^2 - 9v^2\)' is a difference of squares, which can be factored as \((2p)^2 - (3v)^2\): \[ 20p(4p^2 - 9v^2) = 20p( (2p)^2 - (3v)^2 ) \]

Step 6: Apply the difference of squares formula

Use the difference of squares formula \(a^2 - b^2 = (a + b)(a - b)\) to factor completely: \[ 20p((2p)^2 - (3v)^2) = 20p(2p + 3v)(2p - 3v) \]

Key Concepts

Greatest Common Factor

When factoring polynomials, the first step is to find the Greatest Common Factor (GCF). The GCF is the highest number that can evenly divide all terms in an expression.
For example, in the expression \(80p^3 - 180pv^2\), we identify the GCF by breaking down the coefficients and variables:

• The coefficients 80 and 180 share a GCF of 10.
• The variable part common to both terms is \(p\).

Thus, the GCF of the polynomial is \(10p\). Factoring it out simplifies the expression significantly.

Difference of Squares

The difference of squares is a specific factoring technique applied to polynomials of the form \(a^2 - b^2\). It can be factored into \((a + b)(a - b)\).
In our problem, after factoring out the GCF, the expression inside the parentheses \(8p^2 - 18v^2\) can be recognized and simplified like so:

• Identify the common factor within \(8p^2 - 18v^2\), which is 2.
• Resulting in: \(2(4p^2 - 9v^2)\).
• Next, observe that \(4p^2 - 9v^2\) is a classic difference of squares with \((2p)^2 - (3v)^2\).

Finally, applying the difference of squares formula: \( (2p + 3v)(2p - 3v)\), we can factor completely.

Factoring Techniques

There are several techniques used for factoring polynomials:

• Greatest Common Factor (GCF): We factor out the GCF common to all terms, simplifying the expression.
• Grouping: Used when polynomials have four or more terms, grouped in pairs.
• Difference of Squares: Applied when we have expressions like \(a^2 - b^2\).
• Trinomials: Factored into the product of two binomials.

In our exercise, we mainly used the GCF and Difference of Squares techniques. First, we factored out \(10p\) from the polynomial. Then, we recognized \(4p^2 - 9v^2\) as a difference of squares and applied its specific factoring formula.

Prime Polynomials

A polynomial is prime if it cannot be factored into the product of two non-constant polynomials over the integers. Determining the primality of a polynomial involves checking if there are no integer factors other than 1 and itself.
Let's look at whether \(20p(2p + 3v)(2p - 3v)\) contains any prime polynomials:

• \(2p + 3v\) cannot be factored further using integer coefficients, so it's prime.
• Similarly, \(2p - 3v\) is also a prime polynomial for the same reason.

In conclusion, although the overall expression \(20p(2p + 3v)(2p - 3v)\) is fully factored, the polynomials \(2p + 3v\) and \(2p - 3v\) remain prime.