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Question : 3 of 160
Marks:
+1,
-0
Solution:
Given that
(x−n)((x2−2nx)2+(2n2−5)(x2−2nx)+(n4−5n2+4))=0
⇒(x−n)(x4−4nx3+4n2x2+2n2x2−4n3x−5x2+10nx+n4−5n2+4)=0
⇒(x−n)(x4−4nx3+6n2x2−4n3x−5x2+10nx+n4−5n2+4)=0
⇒x5−4nx4+6n2x3−4n3x2−5x3+10nx2+n4x
−5n2x+4x−x4n+4n2x3−6n3x2+4n4x+5x2n−10n2x−n5+5n3−4n=0
⇒x5−5nx4+5(2n2−1)x3+5n(3−2n2)x2
+(5n4−15n2+4)x−n(n4−5n2+4)=0
Now, product of roots,
P=−| Constant term |
| Coefficient of x5 |
=− P=n(n2−1)(n2−4) P=n(n−1)(n+1)(n−2)(n+2) Given that,
n>2 So, putting
n=3,
P=3(3−1)(3+1)(3−1)(3+2)=3(2)(4)(2)(5)
P=120 Putting
n=4,
P=4(4−1)(4+1)(4−1)(4+2) =4(3)(5)(3)(6)=1080 Therefore, products of roots is always divisible by
120.
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