Step 1: Write down the condition using sum equals product.
For a quadratic polynomial $p(x) = ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$:
\[ \alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a} \]
We are told sum of zeroes equals product of zeroes, so:
\[ -\frac{b}{a} = \frac{c}{a} \]
Since $a \neq 0$, multiply both sides by $a$:
\[ -b = c \quad \text{or equivalently} \quad b + c = 0 \]
Step 2: Narrow down the options using a quick sign check before testing exactly.
For $b + c = 0$, $b$ and $c$ must be equal in size but opposite in sign (or both zero). Look at each option's $b$ and $c$ values:
Option (A) $3x^2 + 2x + 2$: here $b = 2$ and $c = 2$, both positive, so they cannot cancel to zero. This option is ruled out right away.
Option (B) $3x^2 - 2x - 3$: here $b = -2$ and $c = -3$, both negative, so again they cannot cancel to zero. Ruled out.
Option (D) $x^2 - 3x + 2$: here $b = -3$ and $c = 2$, opposite signs but different sizes ($3 \neq 2$), so $b + c = -1 \neq 0$. Ruled out.
This leaves only option (C) as a candidate.
Step 3: Confirm option (C) satisfies the condition.
Option (C) is $3x^2 - 2x + 2$, so $a = 3$, $b = -2$, $c = 2$.
\[ b + c = -2 + 2 = 0 \]
This confirms the condition holds.
Step 4: Verify directly with sum and product.
\[ \text{Sum} = -\frac{b}{a} = -\frac{-2}{3} = \frac{2}{3} \]
\[ \text{Product} = \frac{c}{a} = \frac{2}{3} \]
Both equal $\frac{2}{3}$, confirming the answer.
Final Answer:
The polynomial with sum of zeroes equal to their product is $3x^2 - 2x + 2$, so option (C) is correct.
\[ \boxed{3x^2 - 2x + 2} \]