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Let r be a root of the equation x^(2)+2x...

Let r be a root of the equation `x^(2)+2x+6=0.` The value of `(r+2)(r+3)(r+4)(r+5)` is equal to-

A

51

B

-51

C

-126

D

126

Text Solution

Verified by Experts

The correct Answer is:
C
r is root, therefore `r^(2)+2r+6=0`
`(r+2)(r+3)(r+4)(r+5)=(r^(2)+2r+6+3r)(r^(2)+4r+5r+20)`
`=(3r)(7r+14)`
`21(r^(2)+2r)=21(-6)=-126`
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