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Let $g(x)$ be a polynomial of degree $d$ and $h(x)$ be a polynomial of degree $l$ over field $F$. If $\deg(\gcd(g,h)) > 0$ then there exist two polynomials $a(x)$, $b(x)$ of degrees $< l,d$ respectively such that $a(x)g(x) + b(x)h(x) = 0 $ and conversely.

By using extended Euclidean we can say that if $\deg(\gcd(g,h)) > 0$ then there exist two polynomials $a(x)$, $b(x)$ of degrees $< l,d$ respectively such that $a(x)g(x) + b(x)h(x) = \gcd(g,h)$. I am not getting how it is $0$. And the second thing is how to prove the second direction.

Reference : http://www.math.rutgers.edu/~sk1233/courses/ANT-F14/lec10.pdf (4th Page)

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    Set $d=\gcd(g,h)$ with $\deg d\ge 1$, and write $g=dg_1$, $h=dh_1$. Then take $a=h_1$ and $b=-g_1$. For the second suppose $\gcd(g,h)=1$ and from $ag+bh=0$ conclude $g\mid b$, so $\deg b\ge\deg g=d$. That's all. – 2017-02-19

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It's a dimension counting exercise. Write $F[x]_d$ for the space of polynomials of degree at most $d$. Then $(a,b)\mapsto ag+bh$ is a linear map $F[x]_{l-1}\times F[x]_{d-1}\to F[x]_{l+d-1}$. Note carefully that the two spaces have the same dimension, so the map is onto if and only if it is injective.

I'll let you try to finish it yourself.

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    @ Harald Hanche-Olsen I still have a doubt why $a(x).g(x) + b(x).h(x) = 0$ instead of $a(x).g(x) + b(x).h(x) = gcd(g(x),h(x))$ – 2017-02-18
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    $ag+bh$ always has the factor $\gcd(g,h)$. So if $\gcd(g,h)$ has degree at least $1$, then the map in my answer is not onto. And if it is not onto, it is not injective, meaning there is a nontrivial kernel. Take $(a,b)$ in that kernel. – 2017-02-18