If x^2-5x+6<0 and P=x^2+5x+6, then

1967 AMC 12 Problem 18

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18.

x^2-5x+6\lt0

and

P=x^2+5x+6,

then

P

can take any real value

20\lt P\lt30

0\lt P\lt20

P\lt0

P\gt30

Factor

x^2-5x+6

to locate

x

Check how

x^2+5x+6

varies on that interval

The inequality

(x-2)(x-3)\lt0

gives

2\lt x\lt3.

On this interval

P=x^2+5x+6

is increasing, with endpoint values

P(2)=20

and

P(3)=30.

The endpoints are excluded, so

20\lt P\lt30.

Therefore, the correct answer is B.