Question

Let $f(x)$ be a quartic polynomial with integer coefficients and four integer roots. Suppose the constant term of $f(x)$ is $6$.(a) Is it possible for $x=3$ to be a root of $f(x)$?(b) Is it possible for $x=3$ to be a double root of $f(x)$?2

Answer 1

By the rational root theorem, the answer to (a) is yes, and the answer to (b) is no. This is because 3 divides 6, but 3^2 = 9 does not divide 6.

Answer 2

It is given that, f(x) is a quartic polynomial with integer coefficients and four integer roots.

Constant Term of f(x)=6

(a)

By rational root theorem , the roots of the polynomial can be

      $\pm 1,\pm 2,\pm 3,\pm 6$

So, Yes x=3, can be root of f(x).

(b)

Yes, a quartic Polynomial , can have double root as x=3, that is two roots equal to , x=3.

For example,Consider the Polynomial

  $\rightarrow (x^2-9)(x^2-\frac{6}{9})=0\\\\\rightarrow x^4-\frac{6x^2}{9}-9x^2+6=0\\\\\rightarrow x^4-\frac{87x^2}{9}+6=0$

has two zeroes equal to 3.