Show that $f(x)=x^{1000}-x^{500}+x^{100}+x+1=0$ has no rational roots. [closed]

Show that $f(x)=x^{1000}-x^{500}+x^{100}+x+1=0$ has no rational roots.

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If there a rational root, let it be $\frac pq$,where $(p,q)=1$ [$(p,q)$ means the H.C.F. of $p$ and $q$] and $q\ne 0$. then $q$ should divide coefficient of the leading tern and $p$ should divide the constant term.

Thus, $q|1 \implies q=\pm1$ and $p|1 \implies p=\pm1$

Thus, $$\frac{p}{q}=\pm1$$ Now, If the root $\frac{p}{q}=1$ $$f(1)=1-1+1+1+1=3\ne0$$ So, $1$ is not a root .

If $\frac pq=-1$, then $$f(-1)=1\ne0$$ Hence,$-1$ is not a root.

Thus, there exist no rational root for the given polynomial.

By the rational root theorem you need only consider $1$ and $-1$

Since the constant and the leading coefficient of the polynomial are both $1$, the rational root theorem tells us that $\pm1$ are the only possible rational roots. It is easy to check that these are in fact not roots of your polynomial.

• does the rational root theorem says that only integral factors must be checked – Blaise Thunderstorm Dec 24 '16 at 16:00
• @BlaiseThunderstorm I don't understand the question – Dominik Dec 24 '16 at 17:25
• I mean shouldn't you check$\frac{1}{2}$ or other ratios also ? – Blaise Thunderstorm Dec 25 '16 at 7:28
• @BlaiseThunderstorm The rational root theorem tells us that if $\frac{p}{q}$ is a root of this polynomial with coprime integers $p, q$, then $p | 1$ and $q | 1$. This only leaves $\pm1$ as possible roots. – Dominik Dec 25 '16 at 8:26

The Rational Roots theorem says that if there is a rational roots then it should be $1$ or $-1$. Testing both we see that $f(1)=3$ and $f(-1)=1$ and then there is no rational roots.