$$ 3 x^{2}-7 x-6 \quad $$ 2) $$ 9 x^{4}-13 x^{2}+4 $$ $$ x(3 x $$

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It seems you have provided two polynomial expressions:

1. $$ 3x^2 - 7x - 6 $$
2. $$ 9x^4 - 13x^2 + 4 $$

Additionally, there is a fragment $$ x(3x) $$ which seems incomplete.

Let's proceed with the following steps:

1. **Factor the first polynomial** $$ 3x^2 - 7x - 6 $$.
2. **Factor the second polynomial** $$ 9x^4 - 13x^2 + 4 $$.

I'll start by factoring both polynomials.

### Step 1: Factor $$ 3x^2 - 7x - 6 $$

To factor this quadratic expression, we can look for two numbers that multiply to $$ 3 \times (-6) = -18 $$ and add to $$ -7 $$.

The numbers that satisfy this are $$ -9 $$ and $$ 2 $$.

Now we can rewrite the expression:

$$
3x^2 - 9x + 2x - 6
$$

Next, we can group the terms:

$$
(3x^2 - 9x) + (2x - 6)
$$

Factoring out the common factors in each group:

$$
3x(x - 3) + 2(x - 3)
$$

Now, we can factor out the common binomial factor $$ (x - 3) $$:

$$
(3x + 2)(x - 3)
$$

### Step 2: Factor $$ 9x^4 - 13x^2 + 4 $$

This is a quadratic in terms of $$ x^2 $$. Let's set $$ y = x^2 $$, so we rewrite the expression as:

$$
9y^2 - 13y + 4
$$

Now we need to find two numbers that multiply to $$ 9 \times 4 = 36 $$ and add to $$ -13 $$. The numbers that satisfy this are $$ -9 $$ and $$ -4 $$.

Now we can rewrite the expression:

$$
9y^2 - 9y - 4y + 4
$$

Next, we can group the terms:

$$
(9y^2 - 9y) + (-4y + 4)
$$

Factoring out the common factors in each group:

$$
9y(y - 1) - 4(y - 1)
$$

Now, we can factor out the common binomial factor $$ (y - 1) $$:

$$
(9y - 4)(y - 1)
$$

Substituting back $$ y = x^2 $$:

$$
(9x^2 - 4)(x^2 - 1)
$$

The term $$ (x^2 - 1) $$ can be factored further as $$ (x - 1)(x + 1) $$.

Thus, the complete factorization of $$ 9x^4 - 13x^2 + 4 $$ is:

$$
(9x^2 - 4)(x - 1)(x + 1)
$$

### Summary of Results

1. The factorization of $$ 3x^2 - 7x - 6 $$ is:
   $$
   (3x + 2)(x - 3)
   $$

2. The factorization of $$ 9x^4 - 13x^2 + 4 $$ is:
   $$
   (9x^2 - 4)(x - 1)(x + 1)
   $$

If you need further assistance or have additional questions, feel free to ask!
The first polynomial $$ 3x^2 - 7x - 6 $$ factors to $$ (3x + 2)(x - 3) $$. The second polynomial $$ 9x^4 - 13x^2 + 4 $$ factors to $$ (9x^2 - 4)(x - 1)(x + 1) $$.

\( 3 x^{2}-7 x-6 \quad \) 2) \( 9 x^{4}-13 x^{2}+4 \) \( x(3 x \)

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