By actual division show that x+2 is a factor of `x^(3)+4x^(2)+3x-2`.

Explanation

To show that \( x + 2 \) is a factor of the polynomial \( x^3 + 4x^2 + 3x - 2 \) using polynomial long division, we will follow these steps:

### Step 1: Set up the division
We will divide the polynomial \( x^3 + 4x^2 + 3x - 2 \) (the dividend) by \( x + 2 \) (the divisor).

### Step 2: Divide the leading terms
Divide the leading term of the dividend \( x^3 \) by the leading term of the divisor \( x \):
\[ \frac{x^3}{x} = x^2 \]
This gives us the first term of the quotient.

### Step 3: Multiply and subtract
Now, multiply \( x^2 \) by the entire divisor \( x + 2 \):
\[ x^2(x + 2) = x^3 + 2x^2 \]
Subtract this from the original polynomial:
\[ (x^3 + 4x^2 + 3x - 2) - (x^3 + 2x^2) = (4x^2 - 2x^2) + 3x - 2 = 2x^2 + 3x - 2 \]

### Step 4: Repeat the process
Now, we will repeat the process with the new polynomial \( 2x^2 + 3x - 2 \).

1. Divide the leading term \( 2x^2 \) by \( x \):
\[ \frac{2x^2}{x} = 2x \]
2. Multiply \( 2x \) by \( x + 2 \):
\[ 2x(x + 2) = 2x^2 + 4x \]
3. Subtract:
\[ (2x^2 + 3x - 2) - (2x^2 + 4x) = 3x - 4x - 2 = -x - 2 \]

### Step 5: Final division
Now, we will divide \( -x - 2 \) by \( x + 2 \).

1. Divide the leading term \( -x \) by \( x \):
\[ \frac{-x}{x} = -1 \]
2. Multiply \( -1 \) by \( x + 2 \):
\[ -1(x + 2) = -x - 2 \]
3. Subtract:
\[ (-x - 2) - (-x - 2) = 0 \]

### Conclusion
Since the remainder is \( 0 \), we conclude that \( x + 2 \) is indeed a factor of \( x^3 + 4x^2 + 3x - 2 \).