Problem 12 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

simplify the expression. \(\frac{8 r^{3}-s^{3}}{2 r^{2}+r s-s^{2}}\)

Step-by-Step Solution

Answer

\[\frac{4r^2 + 2rs + s^2}{(2r + s)(r - s)}\]

Step 1: Factorize the Numerator (Difference of Cubes)

The numerator is a difference of cubes, so we can factorize it using the difference of cubes formula: \[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\] Applying this to our numerator, we get: \[8r^3 - s^3 = (2r - s)((2r)^2 + (2r)(s) + s^2)\] \[8r^3 - s^3 = (2r - s)(4r^2 + 2rs + s^2)\]

Step 2: Factorize the Denominator (Factor by Grouping)

Now, let's factorize the denominator using the factor by grouping method. Group the first two and the last two terms separately: \[2r^2 + rs - s^2 = 2r^2 + rs \: - (s^2)\] Next, factor out the common terms from each group: \[2r^2 + rs - s^2 = r(2r + s) - s(2r + s)\] Now, factor out the common binomial factor (2r + s): \[2r^2 + rs - s^2 = (2r + s)(r - s)\]

Step 3: Simplify the Expression

Now that we have factored both the numerator and the denominator, we can substitute these factored forms back into the expression and simplify: \[\frac{8r^3 - s^3}{2r^2 + rs - s^2} = \frac{(2r - s)(4r^2 + 2rs + s^2)}{(2r + s)(r - s)}\] We can see that the term (2r - s) is common in both the numerator and the denominator, so we can cancel it out: \[\frac{(2r - s)(4r^2 + 2rs + s^2)}{(2r + s)(r - s)} = \frac{4r^2 + 2rs + s^2}{(2r + s)(r - s)}\] Our simplified expression is: \[\frac{4r^2 + 2rs + s^2}{(2r + s)(r - s)}\]

Key Concepts

Difference of Cubes

The difference of cubes is a mathematical concept used to simplify expressions involving perfect cube terms. To understand this, it's essential to recognize that a cube is a number or a variable raised to the third power, like \(a^3\) or \(b^3\). The difference of cubes formula is given by \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). This tells us that when we subtract one cube from another, we can factor it as a product of a binomial and a trinomial.

To apply this to an example, suppose we have the expression \(8r^3 - s^3\). Identifying \(8r^3\) as \((2r)^3\) and \(s^3\) as \((s)^3\), we can use the difference of cubes formula:

• The binomial is \((2r - s)\).
• The trinomial is given by \((2r)^2 + (2r)(s) + s^2\), which simplifies to \(4r^2 + 2rs + s^2\).

You can now replace the original expression with its factored form, which can simplify further steps in algebraic operations.

Factor by Grouping

Factor by grouping is a useful technique in algebra for factoring polynomials that have four or more terms. It involves rearranging and grouping terms in pairs, then factoring out the greatest common factor from each pair. This helps in breaking down complex polynomials into simpler factors.

Consider the expression \(2r^2 + rs - s^2\). To factor by grouping:

• First, group the terms: \(2r^2 + rs\) and \(-s^2\).
• From the first group, the common factor is \(r\), resulting in \(r(2r + s)\).
• From the second term, factor out \(-s\), yielding \(-s(2r + s)\).

Notice that \((2r + s)\) is a common factor in both groups. This allows us to factor the entire expression as \((2r + s)(r - s)\). Using factor by grouping makes complex expressions more manageable and can lead to further simplifications.

Simplification of Expressions

Simplification of expressions involves reducing an algebraic expression to its simplest form. This means removing common factors, reducing fractions, and simplifying any remaining terms. It helps in making calculations easier and reveals the core components of the expression.

In the context of our factorization example, after applying the difference of cubes to the numerator and factor by grouping to the denominator, we obtained:

• Numerator: \((2r - s)(4r^2 + 2rs + s^2)\)
• Denominator: \((2r + s)(r - s)\)

The next step in simplification is to cancel any common factors between the numerator and denominator. In this case, there isn't a direct cancelation possible immediately from the new expression, unlike what was previously mentioned in the original simplification steps. So, the expression remains factored unless further simplifying circumstances arise in specific contexts. Always re-check for common factors, as oversights can lead to errors. The goal is to express complex algebraic functions in their simplest possible forms.