Factor the given polynomial. \( x^{2}-3 x+20 \) Select the correct choice below and, if nece A. \( x^{2}-9 x+20=\square \) B. The polynomial is prime.

To factor the polynomial \( x^{2} - 3x + 20 \), let's investigate whether it can be factored into two binomials with integer coefficients. Assume the polynomial can be factored as: \[ (x + a)(x + b) = x^{2} + (a + b)x + ab \] We need to find integers \( a \) and \( b \) such that: \[ a + b = -3 \quad \text{and} \quad ab = 20 \] Let's list the pairs of integers that multiply to 20 and check their sums: - \( 1 \) and \( 20 \): \( 1 + 20 = 21 \) - \( -1 \) and \( -20 \): \( -1 + (-20) = -21 \) - \( 2 \) and \( 10 \): \( 2 + 10 = 12 \) - \( -2 \) and \( -10 \): \( -2 + (-10) = -12 \) - \( 4 \) and \( 5 \): \( 4 + 5 = 9 \) - \( -4 \) and \( -5 \): \( -4 + (-5) = -9 \) None of these pairs add up to \(-3\). Therefore, the polynomial does not factor neatly into binomials with integer coefficients. Additionally, examining the discriminant of the quadratic: \[ \Delta = (-3)^2 - 4(1)(20) = 9 - 80 = -71 \] A negative discriminant indicates that the polynomial has no real roots and cannot be factored over the real numbers. **Conclusion:** The polynomial \( x^{2} - 3x + 20 \) is **prime** (irreducible). **Answer:** **B. The polynomial is prime.**