Question

please solve the question
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Answer 1

Answer:

Let f(x)=a0xn+a1xn−1+a2xn−2+...........+an, where a0,a1,.......,an∈Z

Using Division Algorithm.

f(x) = x⋅q1(x)+1989 and f(x) =(x−1)⋅q2(x)+9891, Where q1(x) and q2(x)

are polynomial with integer Coefficients.

So

x⋅(q1(x)−q2(x))+q2(x)−7902 = 0

=> x⋅(q1(x)−q2(x)) = q2(x)−7902

Put x=0, we get q2(0)=7902.

Put x=1, we get

=> q1(1)−q2(1)=q2(1)−7902⇒2q2(1)−q1(1)=7902

If m ≡ n(modq), then f(m) ≡ f(n)(modq).

Try q = 2.

A more direct proof would look at:

g(x) = f(x) −1989−7902x

Which must be divisible x(x−1). So g(x) is always even for integer x, and hence f(x)=g(x)+7902x+1989 is always odd form integer x.

Hence proved.

Thus f(x) has no integral roots.

Answer 2

Answer:

$\huge\underline\mathfrak{\green{Answer :-} }\:$

Step-by-step explanation:

Answer:

Let f(x)=a0xn+a1xn−1+a2xn−2+...........+an, where a0,a1,.......,an∈Z

Using Division Algorithm.

f(x) = x⋅q1(x)+1989 and f(x) =(x−1)⋅q2(x)+9891, Where q1(x) and q2(x

x⋅(q1(x)−q2(x))+q2(x)−7902 = 0

=> x⋅(q1(x)−q2(x)) = q2(x)−7902

Put x=0, we get q2(0)=7902.

Put x=1, we get

=> q1(1)−q2(1)=q2(1)−7902⇒2q2(1)−q1(1)=7902

If m ≡ n(modq), then f(m) ≡ f(n)(modq).

Try q = 2.

A more direct proof would look at:

g(x) = f(x) −1989−7902x

Which must be divisible x(x−1). So g(x) is always even for integer x, and hence f(x)=g(x)+7902x+1989 is always odd form integer x.