`p(x)=6x^(2)+7x-2`
`q(x)=2x^(2)+7x-1`
Two polynomial functions p(x) and q(x) are defind above. Which of the following polynomial functions is divisible by 2x+5?

Options

A. `r(x)=p(x)+q(x)`
B. `r(x)+2p(x)+q(x)`
C. `r(x)=p(x)+3q(x)`
D. `r(x)=2p(x)+3q(x)`

Explanation

To determine which polynomial function is divisible by \(2x + 5\), we will analyze the combinations of the given polynomials \(p(x) = 6x^2 + 7x - 2\) and \(q(x) = 2x^2 + 7x - 1\). We will use the property that for a polynomial to be divisible by another polynomial, the constant term of the polynomial must be divisible by the constant term of the divisor.

### Step-by-Step Solution:

1. **Identify the Constant Term of the Divisor**:
The divisor is \(2x + 5\), and its constant term is \(5\).

2. **Evaluate Each Option**:
We will evaluate the combinations of \(p(x)\) and \(q(x)\) given in the options to find their constant terms.

- **Option A: \(p(x) + q(x)\)**:
\[ p(x) + q(x) = (6x^2 + 7x - 2) + (2x^2 + 7x - 1) = 8x^2 + 14x - 3 \]
The constant term is \(-3\). Since \(-3\) is not divisible by \(5\), we can eliminate this option.

- **Option B: \(2p(x) + q(x)\)**:
\[ 2p(x) + q(x) = 2(6x^2 + 7x - 2) + (2x^2 + 7x - 1) = 12x^2 + 14x - 4 + 2x^2 + 7x - 1 = 14x^2 + 21x - 5 \]
The constant term is \(-5\). Since \(-5\) is divisible by \(5\), we will keep this option for further evaluation.

- **Option C: \(p(x) + 3q(x)\)**:
\[ p(x) + 3q(x) = (6x^2 + 7x - 2) + 3(2x^2 + 7x - 1) = 6x^2 + 7x - 2 + 6x^2 + 21x - 3 = 12x^2 + 28x - 5 \]
The constant term is \(-5\). Since \(-5\) is divisible by \(5\), we will keep this option for further evaluation.

- **Option D: \(2p(x) + 3q(x)\)**:
\[ 2p(x) + 3q(x) = 2(6x^2 + 7x - 2) + 3(2x^2 + 7x - 1) = 12x^2 + 14x - 4 + 6x^2 + 21x - 3 = 18x^2 + 35x - 7 \]
The constant term is \(-7\). Since \(-7\) is not divisible by \(5\), we can eliminate this option.

3. **Long Division**:
Now, we need to perform long division for the remaining options (B and C) to check for divisibility by \(2x + 5\).

- **Option B: \(14x^2 + 21x - 5\)**:
- Divide \(14x^2\) by \(2x\) to get \(7x\).
- Multiply \(7x\) by \(2x + 5\) to get \(14x^2 + 35x\).
- Subtract: \(21x - 35x = -14x\).
- Bring down \(-5\): \(-14x - 5\).
- Divide \(-14x\) by \(2x\) to get \(-7\).
- Multiply \(-7\) by \(2x + 5\) to get \(-14x - 35\).
- Subtract: \(-5 + 35 = 30\).
- Since we have a remainder of \(30\), this polynomial is **not divisible** by \(2x + 5\).

- **Option C: \(12x^2 + 28x - 5\)**:
- Divide \(12x^2\) by \(2x\) to get \(6x\).
- Multiply \(6x\) by \(2x + 5\) to get \(12x^2 + 30x\).
- Subtract: \(28x - 30x = -2x\).
- Bring down \(-5\): \(-2x - 5\).
- Divide \(-2x\) by \(2x\) to get \(-1\).
- Multiply \(-1\) by \(2x + 5\) to get \(-2x - 5\).
- Subtract: \(-5 + 5 = 0\).
- Since the remainder is \(0\), this polynomial is **divisible** by \(2x + 5\).

### Conclusion:
The polynomial function that is divisible by \(2x + 5\) is **Option C: \(p(x) + 3q(x)\)**.