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)?Prove your answers.

Answer 1

Answer:

  (a) yes

  (b) no; see below

Step-by-step explanation:

(a) Integer roots of the quartic will be integer divisors of 6. One of the divisors of 6 is 3, so 3 is a possible root.

(b) In order for 3 to be a double root, it would have to be a double factor of 6. The only integer factors of 6 are 1, 2, 3, 6. (3² = 9 is not one.)

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The quartic can be written as ...

  k(x -a)(x -b)(x -c)(x -d) . . . . . where a, b, c, d, k are integers

The constant term will be kabcd, of which each of the roots is a factor. If the constant is 6 and one root is d=3, then we must have

  kabcd = 3kabc = 6

  kabc = 6/3 = 2

Among these four integer factors, there must be an even number of minus signs, and one that has the value ±2. Another root whose value is 3 will not satisfy the requirements.