Question

1. the simplified rational expression of \\(\frac{2x^2 - 7x - 4}{5x^2 - 16x + 3} \cdot \frac{2x^2 - 7x + 3}{5x^2 - 21x + 4}\\) is

a. \\(\frac{4x}{5x - 1}, x \
eq \frac{1}{5}, 3, 4\\)
b. \\(\frac{4x^2 - 1}{5x - 1}, x \
eq \frac{1}{5}, 3, 4\\)
c. \\(\frac{4x^2 - 1}{25x^2 - 10x + 1}, x \
eq \frac{1}{5}, 3, 4\\)
d. \\(\frac{4x^2 - 1}{25x^2 - 1}, x \
eq \frac{1}{5}, 3, 4\\)

Explanation:

Step1: Factor numerators and denominators

Factor \(2x^2 - 7x - 4\): \(2x^2 - 7x - 4=(2x + 1)(x - 4)\)
Factor \(2x^2 - 7x + 3\): \(2x^2 - 7x + 3=(2x - 1)(x - 3)\)
Factor \(5x^2 - 16x + 3\): \(5x^2 - 16x + 3=(5x - 1)(x - 3)\)
Factor \(5x^2 - 21x + 4\): \(5x^2 - 21x + 4=(5x - 1)(x - 4)\)

Step2: Substitute factored forms into the expression

The original expression \(\frac{2x^2 - 7x - 4}{5x^2 - 16x + 3}\cdot\frac{2x^2 - 7x + 3}{5x^2 - 21x + 4}\) becomes:
\(\frac{(2x + 1)(x - 4)}{(5x - 1)(x - 3)}\cdot\frac{(2x - 1)(x - 3)}{(5x - 1)(x - 4)}\)

Step3: Cancel common factors

Cancel \((x - 4)\) and \((x - 3)\) from numerator and denominator:
\(\frac{(2x + 1)\cancel{(x - 4)}}{(5x - 1)\cancel{(x - 3)}}\cdot\frac{(2x - 1)\cancel{(x - 3)}}{(5x - 1)\cancel{(x - 4)}}=\frac{(2x + 1)(2x - 1)}{(5x - 1)^2}\)

Step4: Simplify the numerator and denominator

Simplify numerator: \((2x + 1)(2x - 1)=4x^2 - 1\)
Simplify denominator: \((5x - 1)^2 = 25x^2 - 10x + 1\)
Also, find the values of \(x\) for which the original expression is undefined (denominators zero):
For \(5x^2 - 16x + 3 = 0\), solutions are \(x=\frac{1}{5},3\); for \(5x^2 - 21x + 4 = 0\), solutions are \(x=\frac{1}{5},4\). So \(x
eq\frac{1}{5},3,4\)

Answer:

C. \(\frac{4x^2 - 1}{25x^2 - 10x + 1}, x
eq\frac{1}{5}, 3, 4\)