Problem 30 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In Exercises, factor the polynomial. If the polynomial is prime, state it. $$ 6 u v^{2}+9 u v-6 v $$

Step-by-Step Solution

Answer

The factored form of the given polynomial is: \(3v(2uv+3u-2)\)

Step 1: Identify common factors

In the given polynomial \(6uv^2 + 9uv - 6v\), we can see that each term has both a numerical and a variable factor in common. In terms of numbers, each term has a factor of 3. In terms of variables, each term has a factor of \(v\).

Step 2: Factor out the common factors

Now we factor out the common factors found in the previous step. We found that 3 and \(v\) are common factors in all the terms. So we factor out \(3v\) from the polynomial: \(3v(2uv + 3u - 2)\)

Step 3: Check for further factoring

Now we need to check if the remaining expression inside the parentheses can be further factored: \(2uv + 3u - 2\) There are no common factors between these three terms, and it's not a quadratic trinomial that we can factor into two binomials. Therefore, this expression is prime, and we have reached the factored form of the original polynomial. #Final Answer# The factored form of the given polynomial is: \(3v(2uv+3u-2)\)

Key Concepts

Common Factor Identification

When you're given a polynomial expression, the first step in factorization is to identify any common factor across all terms. This can dramatically simplify the expression early in the process.
A common factor could be a:

• Numerical Factor: This refers to the greatest common divisor (GCD) of the coefficients of the terms.
• Variable Factor: If each term in the polynomial includes the same variable raised to a power, that variable might be a common factor.

In our exercise with the polynomial \(6uv^2 + 9uv - 6v\), you identify that each term contains the number 3 and the variable \(v\) as common factors. By recognizing these, you can factor these elements out, simplifying the polynomial into a format where further factorization methods can be applied.

Factoring Trinomials

After factoring out the common factors, we often deal with an expression that has three terms or what is called a trinomial. These are typically of the form \(ax^2 + bx + c\). However, in our example, we have the expression \(2uv + 3u - 2\).
Here, we need to check if the trinomial can be further broken down into the product of two simpler binomials. To factor a trinomial:

• Check for any pattern or identities that might simplify factorization.
• Look for potential pairs whose product equals \(a \times c\) and whose sum equals \(b\).

In our example, there is no straightforward way to split \(2uv + 3u - 2\) into binomials using common factored terms or other known identities, so it's determined as non-factorable using this approach at this stage.

Prime Expressions

Expressions that cannot be factored further using integer coefficients are often called prime expressions. This is in a similar context to how prime numbers are numbers that are only divisible by 1 and themselves.
When you factor a polynomial and check the remaining expression for possible reductions, you may find that it is irreducible. This tells you it's "prime." Your determination often arises when:

• No common factors exist across the terms.
• No known factorization identities apply.

In the polynomial task, the expression \(2uv + 3u - 2\) was found to be prime. This makes \(3v(2uv+3u-2)\) the simplest representation of the initial polynomial expression. In such cases, identifying the limitations for further factorization is crucial to accurately representing the polynomial in its factored form.