Answered

What is the product?



$\frac{3 k}{k+1} \cdot \frac{k^2-1}{3 k^3}$


A$\frac{k+1}{k^2}$
B$\frac{k-1}{k^2}$
C$\frac{-1}{k}$
D$\frac{1}{k}$

Verified Answer :

  • Multiply the numerators and denominators to combine the expression into one fraction.
  • Factor the difference of squares in the numerator as \(k^2-1 = (k-1)(k+1)\).
  • Cancel the common factors \(3\), \(k+1\), and \(k\) from the numerator and denominator.
  • State the final simplified expression: \(\boxed{\frac{k-1}{k^2}}\).

Explanation

  1. Analysis of the Rational Expression

We are tasked with simplifying the algebraic expression:
\[\frac{3k}{k+1} \cdot \frac{k^2-1}{3k^3}\]
To simplify this product of rational expressions, we must first analyze the components. The numerator of the first fraction is \(3k\) and the denominator is \(k+1\). The numerator of the second fraction is a quadratic expression \(k^2-1\), which can be factored as a difference of squares: \((k-1)(k+1)\). The denominator of the second fraction is \(3k^3\). For this expression to be defined, we must assume \(k eq 0\) and \(k eq -1\).

  1. Setting up the Product

We write the product as a single fraction by multiplying the numerators and the denominators:
\[\frac{3k(k^2-1)}{(k+1)(3k^3)}\]
Next, we substitute the factored form of the difference of squares \(k^2-1 = (k-1)(k+1)\) into the numerator:
\[\frac{3k(k-1)(k+1)}{(k+1)(3k^3)}\]

  1. Simplifying the Expression

Now, we identify and cancel common factors in the numerator and denominator:

  • The constant factor \(3\) exists in both the numerator and denominator, so it cancels out.
  • The factor \((k+1)\) exists in both the numerator and denominator, so it cancels out (given \(k eq -1\)).
  • The factor \(k\) in the numerator and \(k^3\) in the denominator leaves \(k^2\) in the denominator (given \(k eq 0\)).

Performing these cancellations, the expression simplifies to:
\[\frac{(k-1)}{k^2}\]
Thus, the final simplified form is \(\frac{k-1}{k^2}\). Comparing this to the given options, it matches the second option.

Examples

In engineering, rational expressions often appear when calculating transfer functions or resistance in complex electrical circuits. For instance, when analyzing how components connected in series and parallel affect overall current, you might end up with a bulky formula representing the relationship between input and output. Simplifying such expressions allows engineers to understand the system's behavior more clearly and perform calculations more efficiently, just as we simplified the algebraic expression here to make it easier to work with.

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