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)?

Prove your answers.

Guest Aug 9, 2019

#1**+1 **

(a) Is it possible for x=3 to be a root of f(x)?

Yes.....suppose we have

(x - 1)^2 ( x - 2) ( x - 3)......x = 3 is a root.....and expanding this we get......x^4 - 7 x^3 + 17 x^2 - 17 x + 6

(b) Is it possible for x = 3 to be a double root of f(x)?

Let's suppose that it is possible

Let the other roots be a, b where a, b are integers....and we have that

(x - a) ( x - b) ( x - 3) ( x - 3) =

(x^2 - (a + b) x + ab) ( x^2 - 6x + 9) =

x^4 -(a + b)x^3 + abx^2 - 6x^3 + 6(a+b)x^2 -6abx + 9x^2 - 9(a + b)x + 9ab

This implies that 9ab = 6 ⇒ ab = 6 / 9 ⇒ ab = 2 / 3

But a,b are integers.....and integers multiplied together must produce an integer......

So...it is not possible fro x = 3 to be a double-root in this case

CPhill Aug 9, 2019