- Factor the quadratic \(4x^2+5x+1\) into \((4x+1)(x+1)\).
- Factor the difference of squares \(x^2-4\) into \((x-2)(x+2)\).
- Rewrite the expression as \(\frac{x+2}{(4x+1)(x+1)} \cdot \frac{4x+1}{(x-2)(x+2)}\).
- Cancel the common factors \((x+2)\) and \((4x+1)\) to obtain the final simplified expression \(\boxed{\frac{1}{(x+1)(x-2)}}\).
Explanation
- Factoring the expressions
To simplify the expression \(\frac{x+2}{4x^2+5x+1} \cdot \frac{4x+1}{x^2-4}\), we first factor all the polynomials involved in the numerator and the denominator.
- The denominator \(4x^2 + 5x + 1\) can be factored by finding two numbers that multiply to \(4 \times 1 = 4\) and add to 5. These numbers are 4 and 1. So, \(4x^2 + 4x + x + 1 = 4x(x+1) + 1(x+1) = (4x+1)(x+1)\).
- The denominator \(x^2 - 4\) is a difference of squares: \(x^2 - 2^2 = (x-2)(x+2)\).
The original expression becomes:
\[\frac{x+2}{(4x+1)(x+1)} \cdot \frac{4x+1}{(x-2)(x+2)}\]
- Simplifying the expression
Now that we have the factored form, we can simplify the expression by canceling the common factors in the numerator and the denominator:
\[\frac{x+2}{(4x+1)(x+1)} \cdot \frac{4x+1}{(x-2)(x+2)} = \frac{(x+2)(4x+1)}{(4x+1)(x+1)(x-2)(x+2)}\]
Canceling \((x+2)\) and \((4x+1)\) from the numerator and denominator:
\[\frac{1}{(x+1)(x-2)}\]
This expression is now in its simplest form. Comparing this to the given options, it matches the first option.
- Conclusion
The simplification is complete, and the resulting expression is \(\frac{1}{(x+1)(x-2)}\). This confirms that the first choice is the correct one. Therefore, the simplest form is \(\boxed{\frac{1}{(x+1)(x-2)}}\).
Examples
Simplifying rational expressions is essential in engineering and physics, where complex formulas involving ratios of variables often need to be reduced to their simplest form to be easier to work with. For instance, when designing circuits, analyzing signal transfer functions, or calculating physical properties like velocity or force, you will frequently encounter rational expressions. Reducing these expressions helps you understand the underlying relationship between variables more clearly and makes further calculations, like finding roots or limits, much simpler.