Question

19
select the correct answer.
what is this expression in simplest form?
\\(\frac{x + 2}{4x^2 + 5x + 1} \cdot \frac{4x + 1}{x^2 - 4}\\)
a. \\(\frac{1}{(x + 1)(x - 2)}
\\)b. \\(\frac{x}{(x - 2)}
\\)c. \\(\frac{4x + 1}{(x + 1)(x - 2)}
\\)d. \\(\frac{4x + 1}{x - 2}\\)

Explanation:

Step1: Factor the denominators

First, factor \(4x^{2}+5x + 1\). We need two numbers that multiply to \(4\times1 = 4\) and add to \(5\). The numbers are \(4\) and \(1\). So, \(4x^{2}+5x + 1=4x^{2}+4x+x + 1 = 4x(x + 1)+1(x + 1)=(4x + 1)(x + 1)\).

Next, factor \(x^{2}-4\) using the difference of squares formula \(a^{2}-b^{2}=(a + b)(a - b)\). Here, \(a=x\) and \(b = 2\), so \(x^{2}-4=(x + 2)(x - 2)\).

The original expression becomes \(\frac{x + 2}{(4x + 1)(x + 1)}\cdot\frac{4x + 1}{(x + 2)(x - 2)}\).

Step2: Cancel out common factors

We can cancel out the common factors \((x + 2)\) and \((4x + 1)\) from the numerator and the denominator.

After canceling, we have \(\frac{1}{(x + 1)(x - 2)}\) for the first fraction's remaining part and \(\frac{1}{(x - 2)}\) is not correct. Wait, let's do it step by step.

Multiplying the numerators: \((x + 2)(4x + 1)\)

Multiplying the denominators: \((4x^{2}+5x + 1)(x^{2}-4)=(4x + 1)(x + 1)(x + 2)(x - 2)\)

Now, cancel \((x + 2)\) from numerator and denominator, and cancel \((4x + 1)\) from numerator and denominator.

So we get \(\frac{1}{(x + 1)(x - 2)}\)? Wait, no, wait the numerator after canceling \((x + 2)\) and \((4x + 1)\) is \(1\times1 = 1\)? Wait no, wait the first fraction's numerator is \(x + 2\), the second's numerator is \(4x + 1\). The first denominator is \((4x + 1)(x + 1)\), the second denominator is \((x + 2)(x - 2)\). So when we multiply, it's \(\frac{(x + 2)(4x + 1)}{(4x + 1)(x + 1)(x + 2)(x - 2)}\). Then cancel \((x + 2)\) and \((4x + 1)\) from numerator and denominator, we get \(\frac{1}{(x + 1)(x - 2)}\)? But wait, let's check the options. Option A is \(\frac{1}{(x + 1)(x - 2)}\), but wait maybe I made a mistake. Wait no, wait the original expression:

Wait the first fraction is \(\frac{x + 2}{4x^{2}+5x + 1}\), the second is \(\frac{4x + 1}{x^{2}-4}\).

Factor \(4x^{2}+5x + 1=(4x + 1)(x + 1)\), \(x^{2}-4=(x + 2)(x - 2)\).

So the expression is \(\frac{x + 2}{(4x + 1)(x + 1)}\times\frac{4x + 1}{(x + 2)(x - 2)}\).

Now, cancel \((x + 2)\) (numerator of first and denominator of second), cancel \((4x + 1)\) (denominator of first and numerator of second). So we have \(\frac{1}{(x + 1)}\times\frac{1}{(x - 2)}=\frac{1}{(x + 1)(x - 2)}\), which is option A? Wait but let's check the options again. Wait option A is \(\frac{1}{(x + 1)(x - 2)}\), option C is \(\frac{4x + 1}{(x + 1)(x - 2)}\), wait no, maybe I messed up the factoring.

Wait, wait the first fraction's numerator is \(x + 2\), denominator is \(4x^{2}+5x + 1=(4x + 1)(x + 1)\). The second fraction's numerator is \(4x + 1\), denominator is \(x^{2}-4=(x + 2)(x - 2)\). So when we multiply, it's \(\frac{(x + 2)(4x + 1)}{(4x + 1)(x + 1)(x + 2)(x - 2)}\). Now, cancel \((x + 2)\) and \((4x + 1)\), so we get \(\frac{1}{(x + 1)(x - 2)}\), which is option A. Wait but let me check again.

Wait, maybe I made a mistake in the factoring of \(4x^{2}+5x + 1\). Let's check: \((4x + 1)(x + 1)=4x^{2}+4x+x + 1=4x^{2}+5x + 1\), that's correct. And \(x^{2}-4=(x + 2)(x - 2)\), correct. So canceling \((x + 2)\) and \((4x + 1)\), we get \(\frac{1}{(x + 1)(x - 2)}\), which is option A.

Wait but let's check the options again. Option A is \(\frac{1}{(x + 1)(x - 2)}\), option C is \(\frac{4x + 1}{(x + 1)(x - 2)}\). Wait, maybe I messed up the numerator. Wait the first numerator is \(x + 2\), second is \(4x + 1\). So numerator is \((x + 2)(4x + 1)\), denominator is \((4x + 1)(x + 1)(x + 2)(x - 2)\). So cancel \((x + 2)\) and \((4x + 1)\), so numerator becomes \(1\times1 = 1\), denominator becomes \((x + 1)(x - 2)\). So the result is \(\frac{1}{(x + 1)(x…

Answer:

A. \(\frac{1}{(x + 1)(x - 2)}\)