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Q.
If the minimum value of $y=(x-2)(x-4)(x-6)(x-8)+16, x \in R$ is $m$ then $(m+3)$ equals
Complex Numbers and Quadratic Equations
Solution:
$y=(x-2)(x-6)(x-8)(x-4)+16 $
$=\left(x^2-10 x+16\right)\left(x^2-10 x+24\right)+16$
$=(t+16)(t+24)+16, \text { where } t \in x^2-10 x $
$= t ^2+40 t +400=( x -5)^2-25$
$=( t +20)^2 t \geq-25 $
$\left.y\right|_{\min }=0= m $
$\therefore m +3=3$