What is the minimum value of the function ( $x^{2}$ - 6x + 8) ( $x^{2}$ - 14x + 48) + 20.

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Solution

The correct option is C 4
Let f(x) = ( $x^{2}$ - 6x + 8) ( $x^{2}$ - 14x + 48) + 20
= (x - 2) (x - 4) (x - 6) (x - 8) + 20
= (x - 2) (x - 8) (x - 4) (x - 6) + 20
= ( $x^{2}$ - 10x + 16) ( $x^{2}$ - 10x + 24) + 20
Put $x^{2}$ - 10x = y = (y + 16)(y + 24) + 20
= $y^{2}$ + 40y + 384 + 20
= $(y + 20)^{2}$ + 4
Minimum value of given function = 4. (At $x^{2}$ - 10x + 20 = 0)

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