Question

identify prime polynomials which constant term would mean that the expression is completely factored? ( x^2 - 3x + ) -10 0 10

Explanation:

Step1: Recall factoring quadratic \(x^2 + bx + c\)

A quadratic \(x^2 + bx + c\) factors as \((x + m)(x + n)\) where \(m + n = b\) and \(mn = c\). For \(x^2 - 3x + c\), we need \(m + n=-3\) and \(mn = c\). A prime polynomial (completely factored here means it can't be factored over integers) occurs when no integer pairs \(m,n\) satisfy \(m + n=-3\) and \(mn = c\).

Step2: Check \(c = - 10\)

Find \(m,n\) with \(m + n=-3\) and \(mn=-10\). Solve: \(m,n\) are roots of \(t^2+3t - 10=0\). Factoring: \((t + 5)(t - 2)=0\), so \(m = - 5,n = 2\) (since \(-5+2=-3\) and \(-5\times2=-10\)). So \(x^2-3x - 10=(x - 5)(x + 2)\), not prime.

Step3: Check \(c = 0\)

\(x^2-3x+0=x(x - 3)\), factored, not prime.

Step4: Check \(c = 10\)

Find \(m,n\) with \(m + n=-3\) and \(mn = 10\). The quadratic \(t^2+3t + 10 = 0\) has discriminant \(D=9-40=-31<0\), so no real (let alone integer) roots. Thus \(x^2-3x + 10\) can't be factored over integers, so it's completely factored (prime in integer factorization).

Answer:

10