Question

1. \\(\frac{x^2 + 8x + 12}{x^2 + 3x - 18}\\)\
2. \\(\frac{x^2 - 11x + 18}{x^2 + 2x - 8}\\)\
3. \\(\frac{b^2 + 3b - 28}{b^2 - 49}\\)\
4. \\(\frac{v^2 - 3v - 40}{v^2 - 11v + 24}\\)\
5. \\(\frac{4n - 4}{6n - 20}\\)\
6. \\(\frac{v^2 - 5v - 14}{v^2 + 4v + 4}\\)\
7. \\(\frac{6v^3 + 42v^2}{2v^2 + 26v + 84}\\)\
8. \\(\frac{x^3 - x^2 - 42x}{2x^2 - 20x + 42}\\)\
9. \\(\frac{2v^2 + 10v - 48}{8v + 64}\\)\
10. \\(\frac{9x^2 + 81x}{x^3 + 8x^2 - 9x}\\)

Explanation:

Let's solve problem 15 as an example:

Step1: Factor numerator and denominator

Factor \(x^2 + 8x + 12\): We need two numbers that multiply to \(12\) and add to \(8\). Those numbers are \(2\) and \(6\), so \(x^2 + 8x + 12=(x + 2)(x + 6)\).

Factor \(x^2 + 3x - 18\): We need two numbers that multiply to \(-18\) and add to \(3\). Those numbers are \(6\) and \(-3\), so \(x^2 + 3x - 18=(x + 6)(x - 3)\).

So the fraction becomes \(\frac{(x + 2)(x + 6)}{(x + 6)(x - 3)}\).

Step2: Cancel common factors

We can cancel the common factor \((x + 6)\) (assuming \(x
eq - 6\) to avoid division by zero), so we get \(\frac{x + 2}{x - 3}\).

Step1: Factor numerator and denominator

Factor \(b^2 + 3b - 28\): We need two numbers that multiply to \(-28\) and add to \(3\). Those numbers are \(7\) and \(-4\), so \(b^2 + 3b - 28=(b + 7)(b - 4)\).

Factor \(b^2 - 49\): This is a difference of squares, \(b^2 - 49=(b + 7)(b - 7)\).

So the fraction becomes \(\frac{(b + 7)(b - 4)}{(b + 7)(b - 7)}\).

Step2: Cancel common factors

Cancel the common factor \((b + 7)\) (assuming \(b
eq - 7\)), so we get \(\frac{b - 4}{b - 7}\).

Answer:

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

If you want solutions for other problems, you can follow a similar process: factor the numerator and denominator (if they are quadratic or can be factored) and then cancel out the common factors. For example, for problem 17: