$$p, q, r$$ are given real numbers. Consider the polynomials $$(q-r)(x-q)(x-r) + (r-p)(x-r)(x-p) + (p-q)(x-p)(x-q) + k$$ for all real $$k$$. Then [ ]

The coefficient of $$x^2$$ is $$(q-r) + (r-p) + (p-q) = 0$$, and the coefficient of $$x$$ is $$-\left[(q^2 - r^2) + (r^2 - p^2) + (p^2 - q^2)\right] = 0$$. The constant term is $$qr(q-r) + rp(r-p) + pq(p-q) = -(p-q)(q-r)(r-p)$$, so the polynomial is just the constant $$k - (p-q)(q-r)(r-p)$$. For every $$k \ne (p-q)(q-r)(r-p)$$ its degree is 0, and for $$k = (p-q)(q-r)(r-p)$$ it is the zero polynomial, whose degree is undefined. So degrees 2 and 1 never occur and statements (c) and (d) are false, which leaves (e).