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I was reading some book which had this question:

Q. The number of [quadratic] polynomials having zeros $-2$ and $5$ is:

(A) 1
(B) 2
(C) 3
(D) More than three?

Sol. (A) 1.

But according to me there should be an infinite amount of polynomials. All these polynomials have the zeroes $-2$ and $5$:

\begin{align*} (x+2)(x-5)&=0,\\ 2(x+2)(x-5)&=0,\\ 3(x+2)(x-5)&=0,\\ 4(x+2)(x-5)&=0,\\ 5(x+2)(x-5)&=0, \text{etc}. \end{align*}

According to me, if we are given the two zeroes of a quadratic polynomial, then we can find $\infty$ polynomials with those two zeroes. I do not know why the answer $1$ is given in the book. Maybe it is a misprint? So am I right or the book is right?

Clarification: I just wanted to ask if these polynomials are considered to be the same or different.

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  • 2
    $\begingroup$ You are right. Though check that the book may have made a note somewhere that only monic polynomials are being considered. $\endgroup$ – David Peterson Feb 14 '14 at 8:56
  • $\begingroup$ @DavidPeterson There is no note. There are only Question and answers in the book. $\endgroup$ – Kartik Feb 14 '14 at 8:58
  • $\begingroup$ All those are equivalent; constants in $\mathbb{Q}$ are units in $\mathbb{Q}[X]$ $\endgroup$ – FireGarden Feb 14 '14 at 9:08
  • $\begingroup$ @FireGarden What is $\mathbb{Q}[X]$? $\endgroup$ – Kartik Feb 14 '14 at 9:10
  • $\begingroup$ @FireGarden: They are equivalent (modulo the implied equivalence relation), but they are still distinct. $\endgroup$ – user14972 Feb 14 '14 at 9:14
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You are right, the book is mistaken (perhaps they are taking polynomials which are constant multiples of each other as equal).

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