Question

what is the simplest form of this rational expression?\\(\frac{4x^2 - 14x + 6}{x^2 - 7x + 12}\\)\\(\frac{4x - 1}{x - 3}\\), where \\(x \
eq 2\\) and \\(x \
eq 4\\)\\(4\\), where \\(x \
eq 6\\)\\(\frac{2(2x - 1)}{x - 4}\\), where \\(x \
eq 3\\) and \\(x \
eq 4\\)\\(\frac{2x - 1}{x - 2}\\), where \\(x \
eq 2\\) and \\(x \
eq 3\\)

Explanation:

Step1: Factor numerator

First, factor out a 2 from the numerator, then factor the quadratic:
$4x^2 -14x +6 = 2(2x^2 -7x +3) = 2(2x-1)(x-3)$

Step2: Factor denominator

Factor the quadratic denominator:
$x^2 -7x +12 = (x-3)(x-4)$

Step3: Cancel common factors

Cancel the $(x-3)$ term (note $x
eq 3$ to avoid division by zero):
$\frac{2(2x-1)(x-3)}{(x-3)(x-4)} = \frac{2(2x-1)}{x-4}$

Step4: Identify restricted values

Original denominator is zero when $(x-3)(x-4)=0$, so $x
eq 3$ and $x
eq 4$.

Answer:

$\boldsymbol{\frac{2(2x - 1)}{x - 4}}$, where $\boldsymbol{x
eq 3}$ and $\boldsymbol{x
eq 4}$