Question

plzz solve it...10th one

Answer 1

Step-by-step explanation:

Let the given polynomial be p(x).

Put , x=1 We get p(x)≠0

put ,x=2 we get p(x)≠0

put,x=3 We get p(x)≠0

put,x=4 we get p(x)=0

→4 is a zero of polynomial p(x)

→x-4 is a zero of polynomial p(x)

Divide , p(x) by (x-4)

We get , remainder =0

and , quotient = x^2-8x+7

For , other zeros ,

x^2-8x+7=0

→x=1 or , 7

Thus , zeros of p(x) are

1,4,and ,7

Since , 7-4=4-1 =3

→common difference is same

Hence , the zeros are in AP