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Suppose that $f(x) \in \mathbb Z[x] $ has an integer root. Does it mean $f(x)$ is reducible in $\mathbb Z[x]$?

Mohan
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3 Answers3

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No. $x-2$ is irreducible but has an integer root $2$.

If the degree of $f$ is greater than one, then yes. If $a$ is a root of $f(x)$, carry out synthetic division by $x-a$. You will get $f(x) = (x-a)g(x) + r$, and since $f(a) = 0$, $r=0$.

user7530
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    Why is $g(x)$ an integer polynomial? – Mohan Nov 19 '11 at 10:28
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    Look at the algorithm for performing polynomial division: at every step, to find the next term in the quotient, you are dividing an integer by the leading coefficient of $x-a$ (i.e. 1). Clearly this always gives an integer. – user7530 Nov 19 '11 at 10:36
  • I get it now. Thanks. – Mohan Nov 19 '11 at 10:39
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If the degree of the polynomial is 1, then it is irreductible by definition. But if the degree is greater than one and it has an integer root $r$, then the polynomial is reductible, because it can be written in the form $f(x)=(x-r)g(x)+r(x)$, where $r(x)$ is a constant (by remainder division theorem). Plugging $x=r$ yields that $r(x)=0$ and therefore $f(x)=(x-r)g(x)$.

Beni Bogosel
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No. Counter example: $f(x)=x$.

Phira
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