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today I found myself stumped on this problem:

A polynomial with integer coefficients is of the form $$9x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 15 = 0.$$Find the number of different possible rational roots of this polynomial.

What I have: The possible roots are the positive or negative factors of $15$ divided by those of $9$. Thus, we have $\pm\{1,3,5,15\}$, since this polynomial has integer coefficients. Then, it makes sense that the answer will be $8$. Is this correct, or did I make a mistake? This somehow doesn't seem right to me, I'm not sure though.

By the way, on my first go at this problem, I included fractions to get $48$, but that would include non-integer coefficients, I believe.

random guy 2000
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  • Polynomials with integer coefficients can have rational-number zeroes; for instance, $ \ 6x^2 -5x - 6 \ . $ That's why the Rational Zeroes Theorem has you divide all the factors of the constant terms by all the factors of the leading coefficient; in this example, $ \ \pm 1 \ $ are not the zeroes... –  Apr 23 '21 at 05:13

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Suppose $mdc(a, b) = 1$ $$9\left(\dfrac{a}{b}\right)^4 + a_3\left(\dfrac{a}{b}\right)^3 + a_2\left(\dfrac{a}{b}\right)^2 + a_1\left(\dfrac{a}{b}\right) + 15 = 0$$ $$9\left(\dfrac{a^4}{b^4}\right) + a_3\left(\dfrac{a^3}{b^3}\right) + a_2\left(\dfrac{a^2}{b^2}\right) + a_1\left(\dfrac{a}{b}\right) + 15 = 0$$ $$\times b^4$$ $$9a^4 +a_3a^3b + a_2a^2b^2 + a_1ab^3 + 15b^4 = 0$$ so $$9a^4 +a_3a^3b + a_2a^2b^2 + a_1ab^3 = - 15b^4$$ $$a(9a^3 +a_3a^2b + a_2ab^2 + a_1b^3) = - 15b^4$$ and $a | 15b^4 \longrightarrow a | 15$ ($mdc(a, b)=1$)

Similarly $$-9a^4 = a_3a^3b + a_2a^2b^2 + a_1ab^3 + 15b^4$$ $$-9a^4 = b(a_3a^3 + a_2a^2b+ a_1ab^2 + 15b^3)$$ and $b | 9$

if $a | 15 \longrightarrow a \in \pm\{1, 3, 5, 15 \}$

if $b | 9 \longrightarrow b \in \pm\{1, 3, 9\}$

so the possible rational roots of this polynomial is $\pm \{1, 3, 5, 15, \dfrac{1}{3}, \dfrac{5}{3}, \dfrac{1}{9}, \dfrac{5}{9}\}$, the answer is 16.

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    And indeed each of these is a root for suitable integer values of $a_1,a_2,a_3$. For example, $15$ is a root of $9 x^4 - 30376 x + 15$, and $1/9$ is a root of $9 x^4 - 10936 x^3 + 15$. – Robert Israel Apr 23 '21 at 04:19
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    Got it. I see why some fractional roots are allowed, now. Thank you so much! – random guy 2000 Apr 23 '21 at 16:44