Question

select the correct answer, what is the simplest form of this expression? \\(\frac{2x^2 + 7x + 3}{x^2 - 9}\\) \\(\frac{2x + 1}{x + 3}\\) \\(\frac{2x - 1}{x + 3}\\) \\(\frac{2x - 1}{x - 3}\\) \\(\frac{2x + 1}{x - 3}\\)

Explanation:

Step1: Factor numerator

Factor $2x^2+7x+3$:
Find two numbers that multiply to $2\times3=6$ and add to $7$, which are $6$ and $1$.
Split the middle term: $2x^2+6x+x+3 = 2x(x+3)+1(x+3)=(2x+1)(x+3)$

Step2: Factor denominator

Factor $x^2-9$ using difference of squares: $a^2-b^2=(a-b)(a+b)$
$x^2-9=(x-3)(x+3)$

Step3: Cancel common factors

Substitute factored forms into the expression, then cancel $(x+3)$ (where $x
eq-3$):
$\frac{(2x+1)(x+3)}{(x-3)(x+3)}=\frac{2x+1}{x-3}$

Answer:

$\frac{2x+1}{x-3}$ (corresponding to the fourth option: $\boldsymbol{\frac{2x + 1}{x - 3}}$)