Question

1. \\(\frac{x - 3}{x^2 - 6x + 9}\\)\
2. \\(\frac{x^2 - x - 12}{5x^2 + 14x - 3}\\)

Explanation:

Problem 2: Simplify \(\boldsymbol{\frac{x - 3}{x^2 - 6x + 9}}\) (assuming the denominator is \(x^2 - 6x + 9\) as \(x^2 - 6x - 9\) doesn't factor nicely, likely a typo)

Step 1: Factor the denominator

The denominator \(x^2 - 6x + 9\) is a perfect square trinomial. Using the formula \(a^2 - 2ab + b^2=(a - b)^2\), where \(a = x\) and \(b = 3\), we get \(x^2 - 6x + 9=(x - 3)^2\).
So the expression becomes \(\frac{x - 3}{(x - 3)^2}\).

Step 2: Cancel common factors

We can cancel one \((x - 3)\) from the numerator and the denominator (assuming \(x
eq3\) to avoid division by zero).
\(\frac{x - 3}{(x - 3)^2}=\frac{1}{x - 3}\) (for \(x
eq3\))

Problem 5: Simplify \(\boldsymbol{\frac{x^2 - x - 12}{5x^2 + 14x - 3}}\)

Step 1: Factor the numerator

For the numerator \(x^2 - x - 12\), we need two numbers that multiply to \(- 12\) and add to \(-1\). The numbers are \(-4\) and \(3\).
So \(x^2 - x - 12=(x - 4)(x + 3)\).

Step 2: Factor the denominator

For the denominator \(5x^2 + 14x - 3\), we use the AC method. Multiply \(5\) and \(-3\) to get \(-15\). We need two numbers that multiply to \(-15\) and add to \(14\). The numbers are \(15\) and \(-1\).
Rewrite the middle term: \(5x^2+15x - x - 3\)
Group the terms: \((5x^2 + 15x)+(-x - 3)\)
Factor out the GCF from each group: \(5x(x + 3)-1(x + 3)\)
Factor out \((x + 3)\): \((5x - 1)(x + 3)\)

Step 3: Cancel common factors

The expression is \(\frac{(x - 4)(x + 3)}{(5x - 1)(x + 3)}\). Cancel out the common factor \((x + 3)\) (assuming \(x
eq - 3\) to avoid division by zero).
So \(\frac{(x - 4)(x + 3)}{(5x - 1)(x + 3)}=\frac{x - 4}{5x - 1}\) (for \(x
eq - 3\))

Answer:

s:

1. \(\frac{1}{x - 3}\) (for \(x

eq3\))

1. \(\frac{x - 4}{5x - 1}\) (for \(x

eq - 3\))