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For the question " If $f(x)$ is a polynomial with constant term $10$ having a factor $(x-k)$ where $k$ is an integer, then find the possible value of $k$", the options given are $-20, 20, 8$ and $5$.

I get that by remainder theorem, $k$ should divide $f(x)$, so, $f(k)$ would be zero.

Since $f(x)$ is not given, I assumed that to be $[x/2 + 10]$ to satisfy the first option. Same way, other polynomials can also be imagined and all options would satisfy. But the answer is given as only $5$.

If they had asked least positive value of $k$ then I may have understood but that is not the case. They have clearly asked 'possible value of $k$', so all the options should be correct.

But as is usually the case, I miss some fine detail, so, need your help yo figure that out. Thanks.

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2 Answers 2

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You're correct: for the problem as stated, $\,k\,$ can be any integer, e.g. $\ 10/k\ (x-k) = 10x/k - 10.\ $

Probably intended was that the polynomial has integer coefficients. Then the result follows by the Rational Root Test, or Gauss's Lemma, or the division algorithm.

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  • $\begingroup$ I see, thats very interesting! If the leading coefficient is a fraction, all the given options are a possibility xD (every integer is a possibile root). Question does need some revision! thanks for pointing it out :) $\endgroup$
    – AgentS
    Sep 5, 2014 at 14:52
  • $\begingroup$ @ganeshie8 Yes. Note, in particular, that the Rational Root Test applies only to polynomials with integer coefficients, so it cannot be applied to the problem as currently stated. $\endgroup$
    – Bill Dubuque
    Sep 5, 2014 at 15:01
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HINT

rational root test

What are the factors of constant term, $10$ ?

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  • $\begingroup$ 1, 2, 5, 10. So either of these coud have been the 'possible zero', not really the actual zero, because of this test. That's interesting. Thanks. $\endgroup$
    – aarbee
    Sep 5, 2014 at 12:04
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    $\begingroup$ Exactly ! and notice that we don't need to know the actual leading coefficient to find the possible integer roots; constant term is sufficient $\endgroup$
    – AgentS
    Sep 5, 2014 at 12:05
  • $\begingroup$ @ganeshie8 This is not true without further hypotheses - see my answer. $\endgroup$
    – Bill Dubuque
    Sep 5, 2014 at 14:45

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