Question

Answer this question........​

Answer 1

Answer:

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Step-by-step explanation:

ANSWER

Let the zeros of the given polynomial be p,q,r. As the roots are in A.P., then it can be assumed as p−k,p,p+k, where k is the common difference.

p−k+p+p+k=−

a

3b

p=−

a

b

And, (p−k)(p)(p+k)=−

a

d

p(p

2

−k

2

)=−

a

d

−

a

b

(p

2

−k

2

)=−

a

d

p

2

−k

2

=

b

d

And, p(p−k)+p(p+k)+(p−k)(p+k)=

a

3c

2p

2

+(p

2

−k

2

)=

a

3c

a

2

2b

2

+

b

d

=

a

3c

a

2

b

2b

3

+a

2

d

=

a

3c

2b

3

−3abc+a

2

d=0