Question

1. simplify the rational function \\(\frac{3x^2 - 16x + 5}{2x^2 - 9x - 5}\\)

Explanation:

Step1: Factor numerator and denominator

First, factor the quadratic in the numerator \(3x^2 - 16x + 5\). We need two numbers that multiply to \(3\times5 = 15\) and add to \(-16\). Those numbers are \(-15\) and \(-1\). So, we rewrite the middle term:
\(3x^2 - 15x - x + 5 = 3x(x - 5) - 1(x - 5) = (3x - 1)(x - 5)\)

Next, factor the quadratic in the denominator \(2x^2 - 9x - 5\). We need two numbers that multiply to \(2\times(-5) = -10\) and add to \(-9\). Those numbers are \(-10\) and \(1\). Rewrite the middle term:
\(2x^2 - 10x + x - 5 = 2x(x - 5) + 1(x - 5) = (2x + 1)(x - 5)\)

Step2: Cancel common factors

Now, the rational function becomes \(\frac{(3x - 1)(x - 5)}{(2x + 1)(x - 5)}\). We can cancel the common factor \((x - 5)\) (assuming \(x
eq5\) to avoid division by zero), so we get \(\frac{3x - 1}{2x + 1}\)

Answer:

\(\frac{3x - 1}{2x + 1}\) (for \(x
eq5\) and \(x
eq-\frac{1}{2}\))