**Explanation:** A polynomial is an algebraic expression that involves only non-negative integer powers of \( x \). \( 3x^2 + 2x + 1 \) is a polynomial, while the other options involve fractional or irrational powers of \( x \) or \( x \) in the exponent, which disqualifies them as polynomials.

**Explanation:** The degree of a polynomial is the highest power of the variable \( x \) in the polynomial. In the polynomial \( 4x^3 – 2x^2 + x – 7 \), the highest power of \( x \) is 3, so the degree is 3.

**Explanation:** To find \( p(2) \), substitute \( x = 2 \) into the polynomial \( p(x) \):
\[ p(2) = 2^2 – 3(2) + 2 = 4 – 6 + 2 = 0 \]

**Explanation:** The polynomial \( x^2 – 5x + 6 \) can be factorized into \( (x – 2)(x – 3) \). This is verified by expanding the factors:
\[ (x – 2)(x – 3) = x^2 – 3x – 2x + 6 = x^2 – 5x + 6 \]

**Explanation:** To solve for \( x \), isolate it on one side of the equation:
\[ 2x + 3 = 7 \]
\[ 2x = 7 – 3 \]
\[ 2x = 4 \]
\[ x = 2 \]

**Explanation:** Given \( a + b = 5 \) and \( ab = 6 \), the quadratic equation is:
\[ x^2 – 5x + 6 = 0 \]
Factoring, we get:
\[ (x – 2)(x – 3) = 0 \]
So, the roots are \( x = 2 \) and \( x = 3 \).

**Explanation:** For a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots is given by \( -b/a \). In this case, \( a = 1 \) and \( b = -4 \), so the sum of the roots is \( -(-4)/1 = 4 \).

**Explanation:** Let the roots of the equation be 3 and \( r \). By Vieta’s formulas, the sum of the roots \( 3 + r = -k \) and the product of the roots \( 3r = 12 \). Solving for \( r \):
\[ 3r = 12 \]
\[ r = 4 \]
Thus, \( 3 + 4 = -k \), so \( k = -7 \).

**Explanation:** The equation \( x^2 – 9 = 0 \) can be factored as:
\[ (x – 3)(x + 3) = 0 \]
So, the roots are \( x = 3 \) and \( x = -3 \).

**Explanation:** To find \( p(-2) \), substitute \( x = -2 \) into the polynomial \( p(x) \):
\[ p(-2) = (-2)^2 + 4(-2) + 4 = 4 – 8 + 4 = 0 \]