Question

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

Explanation:

Step1: Recall factoring quadratic

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 \(c\) such that no two integers \(m,n\) satisfy \(m + n=-3\) and \(mn = c\) (except trivial or non - integer, but we consider integer factors for complete factoring in basic cases).

Step2: Test non - factorable \(c\)

If \(c = 2\), then \(x^2-3x + 2=(x - 1)(x - 2)\) (factorable). If \(c = 4\), \(x^2-3x + 4\): we look for two numbers that add to \(- 3\) and multiply to \(4\). The possible pairs for \(4\) are \(1\) and \(4\) (sum \(5\)), \(-1\) and \(-4\) (sum \(-5\)), \(2\) and \(2\) (sum \(4\)), \(-2\) and \(-2\) (sum \(-4\)). None add to \(-3\). So \(x^2-3x + 4\) is prime (completely factored as it can't be factored into linear binomials with integer coefficients). Another example: \(c = 5\), \(x^2-3x + 5\): pairs for \(5\) are \(1\) and \(5\) (sum \(6\)), \(-1\) and \(-5\) (sum \(-6\)), no sum of \(-3\). So any non - factorable constant term (where the quadratic is prime) works. A common example is \(4\) (or \(5\), etc.). Let's take \(c = 4\) as an example.

Answer:

\(4\) (or other non - factorable constants like \(5\), \(7\) etc. depending on the context, but \(4\) is a common choice)