factorise $$ x^{2}+3 x+2 x^{2}+3 x+2 $$

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To factorize the given expression:

$$
x^{2} + 3x + 2x^{2} + 3x + 2
$$

**Step 1: Combine Like Terms**

First, combine the like terms:

$$
x^{2} + 2x^{2} = 3x^{2}
$$
$$
3x + 3x = 6x
$$

So, the expression simplifies to:

$$
3x^{2} + 6x + 2
$$

**Step 2: Attempt to Factorize**

We need to factorize the quadratic equation:

$$
3x^{2} + 6x + 2
$$

We look for two numbers $$ a $$ and $$ b $$ such that:

$$
a \times b = 3 \times 2 = 6 \quad \text{and} \quad a + b = 6
$$

The possible pairs $$(a, b)$$ are:

- $$ (1, 6) $$
- $$ (2, 3) $$

However, neither of these pairs adds up to 6 when considering the coefficients, meaning the quadratic doesn't factor neatly with integer coefficients.

**Step 3: Use the Quadratic Formula**

Since simple factoring isn't possible, we can use the quadratic formula to find the roots:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

For the equation $$ 3x^{2} + 6x + 2 $$:

- $$ a = 3 $$
- $$ b = 6 $$
- $$ c = 2 $$

Calculating the discriminant:

$$
\Delta = b^2 - 4ac = 36 - 24 = 12
$$

So, the roots are:

$$
x = \frac{-6 \pm \sqrt{12}}{6} = \frac{-6 \pm 2\sqrt{3}}{6} = \frac{-3 \pm \sqrt{3}}{3}
$$

**Step 4: Express the Factorization**

Using the roots, the factorization of the quadratic is:

$$
3x^{2} + 6x + 2 = 3\left(x + 1 - \frac{\sqrt{3}}{3}\right)\left(x + 1 + \frac{\sqrt{3}}{3}\right)
$$

**Final Answer:**

After simplifying, the factored form is

$$ 3\big(x + 1 - \tfrac{\sqrt{3}}{3}\big)\big(x + 1 + \tfrac{\sqrt{3}}{3}\big) $$
The expression $$ x^{2} + 3x + 2x^{2} + 3x + 2 $$ simplifies to $$ 3x^{2} + 6x + 2 $$. This quadratic does not factor neatly with integer coefficients, so it can be expressed in its factored form using the roots found through the quadratic formula: $$ 3\big(x + 1 - \tfrac{\sqrt{3}}{3}\big)\big(x + 1 + \tfrac{\sqrt{3}}{3}\big) $$

factorise \( x^{2}+3 x+2 x^{2}+3 x+2 \)

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