Quadratic expressions

The direct answer is B: \((r-s)^2\). Use the identity \(A^2-2AB+B^2=(A-B)^2\). Match A with r and B with s. The first term becomes \(r^2\), the middle term becomes \(-2rs\), and the final term becomes \(s^2\), exactly matching the given trinomial. Therefore \(r^2-2rs+s^2=(r-s)^2\). Expanding the answer gives \(r^2-rs-rs+s^2=r^2-2rs+s^2\). A would give a positive middle term, \(r^2+2rs+s^2\). C is the difference of squares and simplifies to \(r^2-s^2\), so it has no middle term. D expands to \(r^2-rs-2s^2\), which is different. The key check is the sign of the middle term: a negative middle term means the binomial uses subtraction.

The governing concept is the coefficient of a term. A coefficient is the numerical factor multiplying a variable term, while the exponent identifies the power of that variable. In 3x² + 2x − 1, the term containing x to the first power is 2x, so the coefficient of x is 2. Therefore, option B is correct. The number 3 is the coefficient of x², not of x, because it belongs to the quadratic term. The number −1 is the constant term and has no x factor. Although the coefficient of a separately written x is 1, this expression contains 2x, not x alone. Carefully matching the requested power prevents confusing the linear coefficient with the quadratic coefficient or the constant term.