Find zeroes of the polynomial p(x)`x^(3)-9x^(2)+26x-24`, if it is given that the product of its two zeroes is 8.

Answer

N/a

Explanation

Let `alpha,beta,gamma` be zeroes of the given polynomial p(x), such that `alphabeta=8 " " ` ……………(1)
`alpha+beta+gamma=-((-9))/(1)=9 " "`..........(2)
`alphabeta+betagamma +gamma alpha=(26)/(1)=26 " "`...............(3)
`alpha betagamma=-((-24))/(1)=24 " "`................(4)
From (1) and (4)
`8gamma=24 " " implies " "gamma=3`
Put `gamma=3` in (2), we get `alpha+beta=6 " "`..............(5)
Now, `" "(alpha-beta)^(2)=(alpha+beta)^(2)-4alphabeta`
`=(6)^(2)-4(8)`
=36-32
=4
`:. " "alpha-beta=+-2 " "` ...............(6)
Solving (5) and (6), we get
`alpha=4,beta=2 " "or " "alpha=2,beta=4` and `gamma=3`
So, zeroes are 2,3 and 4.`" "` Ans.