(Q5)If two zeros of the polynomial x4-6x3-26x2+138x-35 are 2±√3 find other zeros.

Answer

Let the other, two zeros are α,β

Then the sum of the zeros of given polynomial = 2+√3+2-√3+α+β

=
-b
a
 =
-(-)
1
 =

4+α+β=

α+β=----------(1)

Now product of the zeros is

(2+√3)(2-√3)(α)(β) =
e
a
 =
-
1

(4-3)(αβ)= -

αβ= ------------(2)

Now (α-β)2= (α+β)2 - 4αβ

(2)2 - 4(-35)=

(α-β)=±---------(3)

Now solving (1) & (3) we get

α + β =
α - β =
2α =

α=; then α + β= + β=2

β= -

The remaining zeros are α,β = , -

So total zeros of given polynomial are

2+√3, 2-√3, , -