`f(x)=2x^(3)+6x^(2)+4x`
`g(x)x^(2)+3x+2`
The polynomials f (x ) and g (x ) are defined above. Which of the following polynomials is divisible by 2x+3 ?

Options

A. `h(x)=f(x)+g(x)`
B. `p(x)=f(x)+3g(x)`
C. `r(x)=2f(x)+3g(x)`
D. `s(x)=3f(x)+2g(x)`

Explanation

To determine which polynomial is divisible by \(2x + 3\), we will analyze the given polynomials \(f(x)\) and \(g(x)\) and their combinations.

1. **Define the Polynomials**:
\[ f(x) = 2x^3 + 6x^2 + 4x \]
\[ g(x) = x^2 + 3x + 2 \]

2. **Option A: \(h(x) = f(x) + g(x)\)**:
\[ h(x) = (2x^3 + 6x^2 + 4x) + (x^2 + 3x + 2) \]
Combine like terms:
\[ h(x) = 2x^3 + (6x^2 + x^2) + (4x + 3x) + 2 = 2x^3 + 7x^2 + 7x + 2 \]

3. **Check Divisibility of \(h(x)\) by \(2x + 3\)**:
We will perform polynomial long division of \(h(x)\) by \(2x + 3\).

- Divide the leading term: \(2x^3 \div 2x = x^2\)
- Multiply \(x^2\) by \(2x + 3\):
\[ 2x^3 + 3x^2 \]
- Subtract:
\[ (2x^3 + 7x^2 + 7x + 2) - (2x^3 + 3x^2) = 4x^2 + 7x + 2 \]
- Next, divide \(4x^2\) by \(2x\):
\[ 4x^2 \div 2x = 2x \]
- Multiply \(2x\) by \(2x + 3\):
\[ 4x^2 + 6x \]
- Subtract:
\[ (4x^2 + 7x + 2) - (4x^2 + 6x) = x + 2 \]
- Finally, divide \(x\) by \(2x\):
\[ x \div 2x = \frac{1}{2} \]
- Multiply \(\frac{1}{2}\) by \(2x + 3\):
\[ x + \frac{3}{2} \]
- Subtract:
\[ (x + 2) - (x + \frac{3}{2}) = 2 - \frac{3}{2} = \frac{1}{2} \]

Since the remainder is \(\frac{1}{2}\), \(h(x)\) is **not divisible** by \(2x + 3\).

4. **Option B: \(p(x) = f(x) + 3g(x)\)**:
\[ p(x) = (2x^3 + 6x^2 + 4x) + 3(x^2 + 3x + 2) \]
Simplifying:
\[ p(x) = 2x^3 + 6x^2 + 4x + 3x^2 + 9x + 6 = 2x^3 + 9x^2 + 13x + 6 \]

5. **Check Divisibility of \(p(x)\) by \(2x + 3\)**:
Perform polynomial long division of \(p(x)\) by \(2x + 3\).

- Divide the leading term: \(2x^3 \div 2x = x^2\)
- Multiply \(x^2\) by \(2x + 3\):
\[ 2x^3 + 3x^2 \]
- Subtract:
\[ (2x^3 + 9x^2 + 13x + 6) - (2x^3 + 3x^2) = 6x^2 + 13x + 6 \]
- Next, divide \(6x^2\) by \(2x\):
\[ 6x^2 \div 2x = 3x \]
- Multiply \(3x\) by \(2x + 3\):
\[ 6x^2 + 9x \]
- Subtract:
\[ (6x^2 + 13x + 6) - (6x^2 + 9x) = 4x + 6 \]
- Finally, divide \(4x\) by \(2x\):
\[ 4x \div 2x = 2 \]
- Multiply \(2\) by \(2x + 3\):
\[ 4x + 6 \]
- Subtract:
\[ (4x + 6) - (4x + 6) = 0 \]

Since the remainder is \(0\), \(p(x)\) is **divisible** by \(2x + 3\).

### Conclusion:
The polynomial \(p(x) = f(x) + 3g(x)\) is divisible by \(2x + 3\).