Problem 34 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In Exercises, factor the polynomial. If the polynomial is prime, state it. $$ 16 u^{4} v-9 v^{3} $$

Step-by-Step Solution

Answer

The factored polynomial is \(16u^4v - 9v^3 = v(4u^2 + 3v)(4u^2 - 3v)\).

Step 1: Identify common factors

In the given polynomial \(16u^4v - 9v^3\), the common factor between the two terms is \(v\). So, we can factor this out.

Step 2: Factor out the common factor

Now, we will factor \(v\) out of the polynomial: \(v(16u^4 - 9v^2)\).

Step 3: Check for difference of squares

Next, we see if the remaining part of the polynomial, \(16u^4 - 9v^2\), can be factored using the difference of squares method. Since \(16u^4\) can be written as \((4u^2)^2\) and \(9v^2\) can be written as \((3v)^2\), the polynomial is indeed a difference of squares.

Step 4: Factor the difference of squares

Using the difference of squares method, we can factor the polynomial as follows: \(16u^4 - 9v^2 = (4u^2 + 3v)(4u^2 - 3v)\).

Step 5: Combine the factored terms

Finally, we combine the extracted common factor and the factored difference of squares to obtain the fully factored polynomial: \(16u^4v - 9v^3 = v(4u^2 + 3v)(4u^2 - 3v)\).

Key Concepts

Difference of Squares

Factoring polynomials often involves recognizing patterns that simplify the process, and one such pattern is the difference of squares. This occurs when a polynomial can be expressed as the subtraction of two perfect squares. Symbolically, it's written as \(a^2 - b^2\), which can be factored into \((a + b)(a - b)\).

When facing a polynomial like \(16u^4 - 9v^2\), we can clearly see it fits this pattern because \(16u^4\) is the square of \(4u^2\) and \(9v^2\) is the square of \(3v\). So, applying the difference of squares rule, we can factor this polynomial into \((4u^2 + 3v)(4u^2 - 3v)\). This method is particularly useful because it allows us to factor some quadratic expressions quickly and elegantly, providing a significant simplification in our algebraic journey.

Common Factor

Finding a common factor among the terms of a polynomial is like uncovering the thread that holds a fabric together. It's an initial step we take before diving into more complex factoring methods. In our example \(16u^4v - 9v^3\), the common factor is \(v\), which appears in each term of the polynomial.

To factor it out, we simply divide each term by \(v\) and place it in front of a set of parentheses that contain the remaining expression: \(v(16u^4 - 9v^2)\). This is crucial because it simplifies the polynomial and may reveal further factoring opportunities, such as the difference of squares in the remaining expression.

Polynomial Factoring Steps

Factoring polynomials can feel like solving a puzzle – it involves several methodical steps to reduce an expression into its simplest factors. Here's a condensed roadmap:

• Scan the polynomial for any common factors among all the terms.
• Factor out any common factors found.
• Review the remaining polynomial for patterns such as difference of squares, perfect square trinomials, or other identifiable multiplication patterns.
• If applicable, use special factoring formulas or techniques like grouping.
• Always double-check by expanding the factors to ensure they multiply back to the original polynomial.

Using our polynomial as an example, the process involved factoring out the common factor \(v\) and then recognizing and factoring the difference of squares, leading to the fully factored form \(v(4u^2 + 3v)(4u^2 - 3v)\).

Prime Polynomials

Just as prime numbers are numbers only divisible by 1 and themselves, prime polynomials cannot be factored into the product of two non-constant polynomials with integer coefficients. Recognizing prime polynomials saves effort and frustration; if a polynomial is prime, no amount of factoring will break it down any further.

To determine if a polynomial is prime, look for the absence of common factors and the inability to apply special factoring formulas. If neither appears presents, and all coefficients are simplified, you've likely got a prime polynomial on your hands. However, if you do find a way to factor the polynomial, it's not prime. For instance, if we couldn't factor \(16u^4 - 9v^2\) at all, we would then conclude it's a prime polynomial. Fortunately, in this case, it was factorable, revealing that it is not prime.