Question

1. \\(\frac{x^2 - 2x - 48}{2x^2 - 19x + 24}\\)

Explanation:

Step1: Factor the numerator

We factor \(x^{2}-2x - 48\). We need two numbers that multiply to \(- 48\) and add up to \(-2\). The numbers are \(-8\) and \(6\). So \(x^{2}-2x - 48=(x - 8)(x+6)\)

Step2: Factor the denominator

We factor \(2x^{2}-19x + 24\). We need two numbers that multiply to \(2\times24 = 48\) and add up to \(-19\). The numbers are \(-16\) and \(-3\). Then we rewrite the middle term:
\(2x^{2}-16x-3x + 24=2x(x - 8)-3(x - 8)=(2x - 3)(x - 8)\)

Step3: Simplify the fraction

Now we have \(\frac{(x - 8)(x + 6)}{(2x - 3)(x - 8)}\). We can cancel out the common factor \((x - 8)\) (assuming \(x
eq8\)) and we get \(\frac{x + 6}{2x - 3}\)

Answer:

\(\frac{x + 6}{2x - 3}\) (for \(x
eq8\))