Find the zeroes of polynomial f(x)=x3-12x2+39-28 if it is given that zeroes in A.P
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Step-by-step explanation:
f(x)=x
3
−12x
2
+19x−28
Zeroes are in AP.
Let the zeroes be of the form
a-d,a,a+d
sum of roots = 12
3a=12
a = 4
product of roots =−
a
d
a(a
2
−d
2
)=(−28)
a
3
−ad
2
=28
(4)
3
−4d
2
=28
64−4d
2
=28
4d
2
=36
d=±3
Zeroes are 1,4,7 or 7,4,1
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