Question

simplify the expression: \\(\frac{3}{4x^2 + 4x + 1} + \frac{x + 1}{4x^2 - 1}\\)\
(1 point)\
\\(\bigcirc\\) \\(\frac{7x + 2}{8x^3 + 4x^2 - 2x - 1}\\)\
\\(\bigcirc\\) \\(\frac{2x^2 + 9x - 2}{8x^3 + 4x^2 - 2x - 1}\\)\
\\(\bigcirc\\) \\(\frac{4 + x}{8x^3 + 4x^2 - 2x - 1}\\)\
\\(\bigcirc\\) \\(\frac{3x + 3}{8x^3 + 4x^2 - 2x - 1}\\)

Explanation:

Step1: Factor denominators

First, factor the denominators of the two fractions.
For \(4x^2 + 4x + 1\), we can use the perfect square formula \((a + b)^2=a^2 + 2ab + b^2\). Here, \(a = 2x\) and \(b=1\), so \(4x^2+4x + 1=(2x + 1)^2\).
For \(4x^2-1\), we use the difference of squares formula \(a^2 - b^2=(a + b)(a - b)\). Here, \(a = 2x\) and \(b = 1\), so \(4x^2-1=(2x + 1)(2x - 1)\).
The least common denominator (LCD) of the two fractions \(\frac{3}{(2x + 1)^2}\) and \(\frac{x + 1}{(2x + 1)(2x - 1)}\) is \((2x + 1)^2(2x - 1)\).

Step2: Rewrite fractions with LCD

Rewrite each fraction with the LCD:

• For \(\frac{3}{(2x + 1)^2}\), multiply numerator and denominator by \((2x - 1)\):

\(\frac{3(2x - 1)}{(2x + 1)^2(2x - 1)}=\frac{6x-3}{(2x + 1)^2(2x - 1)}\)

• For \(\frac{x + 1}{(2x + 1)(2x - 1)}\), multiply numerator and denominator by \((2x + 1)\):

\(\frac{(x + 1)(2x + 1)}{(2x + 1)^2(2x - 1)}=\frac{2x^2+x+2x + 1}{(2x + 1)^2(2x - 1)}=\frac{2x^2+3x + 1}{(2x + 1)^2(2x - 1)}\)

Step3: Add the fractions

Now add the two fractions:
\(\frac{6x-3}{(2x + 1)^2(2x - 1)}+\frac{2x^2+3x + 1}{(2x + 1)^2(2x - 1)}=\frac{(6x - 3)+(2x^2+3x + 1)}{(2x + 1)^2(2x - 1)}\)
Simplify the numerator:
\(6x-3 + 2x^2+3x + 1=2x^2+(6x + 3x)+(-3 + 1)=2x^2+9x - 2\)

Step4: Simplify the denominator

Simplify the denominator \((2x + 1)^2(2x - 1)\):
First, expand \((2x + 1)^2=4x^2+4x + 1\)
Then multiply by \((2x - 1)\):
\((4x^2+4x + 1)(2x - 1)=4x^2(2x - 1)+4x(2x - 1)+1(2x - 1)\)
\(=8x^3-4x^2+8x^2-4x + 2x - 1=8x^3+4x^2-2x - 1\)

Answer:

\(\boldsymbol{\frac{2x^2 + 9x - 2}{8x^3 + 4x^2 - 2x - 1}}\) (corresponding to the second option)