Question

If a is rational and x – a divides an integer monic polynomial, prove that a must be an
integer.

Answer 1

Let,A be 1 and x be 2

x-a=2-1

1 so 1 is the answer and it is an integer

Answer 2

Answer with explanation:

It is given that , a is a rational number, which includes integers, whole number as well as natural number.

A Monic polynomial is that polynomial ,and the entire term is composed of single variable, in which coefficient of highest degree polynomial is 1.

→x-a , divides an integer monic polynomial.

→ x-a , divides ,the polynomial, P(x) =x²-a²

Substituting , x=a ,in the polynomial

P(a)=a²-a²=0

It is given that the monic polynomial is a polynomial having roots , only integers.

x²-a²=0

x²=a²

x= a ∧ x= -a,

→a is an integer.

In the beginning it,is given that a is rational , but the monic polynomial has roots equal to integers,

So →, Rational Root ∩ Integral Root = An Integral root

Hence, a is an integer.