Divide `2x^(2) + 3x + 4` by `x+1` and find the quotient.

Options

A. `2x+1`
B. `x-4`
C. `2x+4`
D. `2x-4`

Answer

To divide the polynomial \(2x^2 + 3x + 4\) by \(x + 1\) and find the quotient, we can use polynomial long division. Here are the steps:

### Step 1: Set up the division
We write \(2x^2 + 3x + 4\) (the dividend) under the long division symbol and \(x + 1\) (the divisor) outside.

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

### Step 3: Multiply and subtract
Now, multiply \(2x\) (the result from Step 2) by the entire divisor \(x + 1\):
\[ 2x \cdot (x + 1) = 2x^2 + 2x \]
Next, subtract this result from the original polynomial:
\[ (2x^2 + 3x + 4) - (2x^2 + 2x) = (2x^2 - 2x^2) + (3x - 2x) + 4 = x + 4 \]

### Step 4: Repeat the process
Now, we need to divide the new leading term \(x\) by the leading term of the divisor \(x\):
\[ \frac{x}{x} = 1 \]
Add this to the quotient.

### Step 5: Multiply and subtract again
Multiply \(1\) by the divisor \(x + 1\):
\[ 1 \cdot (x + 1) = x + 1 \]
Now subtract this from the current polynomial:
\[ (x + 4) - (x + 1) = (x - x) + (4 - 1) = 3 \]

### Step 6: Write the final result
At this point, we cannot divide further since \(3\) (the remainder) is a constant and \(x + 1\) is a linear polynomial. Thus, the division gives us:
\[ \text{Quotient} = 2x + 1 \]
\[ \text{Remainder} = 3 \]

### Final Answer
The quotient when dividing \(2x^2 + 3x + 4\) by \(x + 1\) is:
\[ \boxed{2x + 1} \]