1588

${\displaystyle f(x)=(x^2-8x+15)(x^2+bx+c)}$ 关于 ${\displaystyle x=2}$ 对称, 求其最小值.

因为 ${\displaystyle f(x)=(x-3)(x-5)(x^2+bx+c)}$ 关于 ${\displaystyle x=2}$ 对称, 所以零点 ${\displaystyle 3,5}$ 的对称点 ${\displaystyle 1,-1}$ 也应该是 ${\displaystyle f(x)}$ 的零点, 从而$$\begin{array}{rl}f(x)& =(x-3)(x-5)(x-1)(x+1)=(x^2-4x+3)(x^2-4x-5).\end{array}$$ 记 ${\displaystyle x^2-4x=m\ge -4}$, 则 $$f(x)=(m+3)(m-5)=(m-1{)}^2-16\ge -16.$$