Question

If a is rational and x-a divides an integer monic polynomial prove that a must be an integer

Answer 1

Answer:

The proof is explained step-wise below :

Step-by-step explanation:

Division by ( x - a ) in the ring Z[x] is defined as follows:  

(x - a) divides monic polynomial p(x) in Z[x] if and only if (x - a) is in Z[x] and there is a member q(x) in Z[x] such that p(x) = (x - a) × q(x).  

Therefore, if (x - a) divides the monic polynomial p(x) in Z[x] then (x - a) is a member of Z[x],

Also given x is a rational number which implies that a must be an integer.

Thus, a ∈ Z

Hence, a must be an integer.

Hence Proved

Answer 2

Answer with explanation:

It is given that a, is a Rational Number.

→x-a, divides an integer Monic polynomial.

A polynomial is said to be monic, if the leading coefficient of highest degree variable is equal to 1.

Consider three polynomials,

1.

→2 x²+5 x+3=0,which is not a Monic polynomial,as Coefficient of highest degree variable is equal to 2.

Factorizing the above Quadratic

→2 x² +3 x+2 x+3=0

→ x(2 x+3)+1(2 x+3)=0

→(x+1)×(2 x+3)=0

→x+1=0  ∧ 2 x+3=0

x= -1 ∧ $x=\frac{-3}{2}$

are two Zeroes of Non Monic Polynomial,one of which is rational ,other is non rational that is an integer.

→we can conclude that non monic polynomial can't have all roots equal to integer.

2.

Consider ,the Monic polynomial ,f(x)= x²-3 x+2

which can be factorized as

→x²-2 x-x+2

=x(x-2)-1(x-2)

=(x-1)(x-2)

Finding the Zeroes

x-1= 0 ∧ x-2=0

→x=1 ∧ x=2

So,the Integral monic polynomial has two real integral root.

3.

Now,consider a monic polynomial having irrational root

x²+3 x +1

Zeroes can be obtained by

$x=\frac{-3\pm\sqrt{3^2-4\times 1\times 1}}{2}\\\\x=\frac{-3\pm \sqrt{5}}{2}$

both Zeroes are irrational , as irrational root occur in pairs.

Zeroes of monic polynomial can be imaginary also,also imaginary root occur in pairs.

In the Statement, it is given that x-a divides an integer Monic polynomial.

By looking at function 2, and function 3, we can conclude that both the roots of the integral monic polynomial will be either an integer , irrational or Imaginary.

→But, it is given that (x-a) divides the integer monic polynomial.

  x-a=0,gives , x=a,

But, it is given that, a is rational.

→Integers⊂ Rational number

x-a ,divides the integral monic polynomial,⇒Means, the monic polynomial has integral roots.

→Since , imaginary and Irrational root Occur in pairs,so both the roots of the monic polynomial will either be imaginary or irrational or Integral.

But here,one root is an integer,so another root will be also an integer. which is,explained in equation 2.

So, a will be an integer, if x-a divides an integer monic polynomial.