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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)?

Aug 9, 2019

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(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   Aug 9, 2019