Factorise (by group $$ x^{2}+4 x+3 x+20 $$

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To factorize the expression $$ x^{2} + 4x + 3x + 20 $$ by grouping, let's follow the steps:

1. **Combine Like Terms:**
   $$
   x^{2} + 4x + 3x + 20 = x^{2} + 7x + 20
   $$

2. **Attempt to Factor by Grouping:**
   To factor by grouping, we need to find two numbers that multiply to the constant term (20) and add up to the coefficient of the $$ x $$ term (7).

Let's list the pairs of factors of 20:
   - $$ 1 \times 20 = 20 $$ (Sum: 21)
   - $$ 2 \times 10 = 20 $$ (Sum: 12)
   - $$ 4 \times 5 = 20 $$ (Sum: 9)

None of these pairs of factors add up to 7.

3. **Conclusion:**
   Since there are no integer pairs that satisfy the conditions needed for factoring by grouping, the quadratic expression $$ x^{2} + 7x + 20 $$ **cannot be factored** using integer coefficients.

If you need to express the factors, you can use the quadratic formula to find the roots and write the expression in its factored form with complex numbers:

$$
   x = \frac{-7 \pm \sqrt{49 - 80}}{2} = \frac{-7 \pm \sqrt{-31}}{2} = \frac{-7 \pm i\sqrt{31}}{2}
   $$

Therefore, the factored form over the complex numbers is:

$$
   \left( x + \frac{7 + i\sqrt{31}}{2} \right)\left( x + \frac{7 - i\sqrt{31}}{2} \right)
   $$

However, for most practical purposes, especially if you're working within the realm of real numbers, the expression remains unfactorable.

**Final Answer:**

The expression cannot be factored by grouping using integer coefficients.
The expression $$ x^{2} + 4x + 3x + 20 $$ cannot be factored by grouping using integer coefficients.

Factorise (by group \( x^{2}+4 x+3 x+20 \)

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