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Is the set of polynomials with rational roots a vector space?

The polynomial $p_i(x)=(x-{{a_i}/{b_i}})(q_i(x))$ is from that given set.

If $p_1(x) + p_2(x) = (x-{{a_1}/{b_1}})(q_1(x)) + (x-{{a_2}/{b_2}})(q_2(x))$ is the polynomial from that set, it is ok. Where $b_i$ can't be 0.

If $rp_1(x) = r(x-{{a_1}/{b_1}})(q_1(x))$ is the polynomial from that set, it is ok. Where $r$ is a scalar.

We are restricted that if those polynomials form a subspace, all of them must have rational roots.

Any suggestions how to find a contradiction? Thank you and sorry for my English.

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    $p(x) = x^2$, $q(x) = x^2-1$ have rational roots. $(p+q)(x) = 2x^2 -1$ has irrational roots. – 2017-02-07
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    Or $p(x) = x^2$, $q(x) = 1$, then consider $p+rq$ with $r \in \Bbb Q$. – 2017-02-07

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No. Both $x^2$ and $(x-1)^2$ have rational roots, but their sum $x^2+(x-1)^2$ does not (it doesn't have any real roots).

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    Thank you, so it isn't a vector space. – 2017-02-07