Help me to find Bezout's number for two polynomial .

Question

$f(x)=2x^4+5x^3+6$
$g(x)=x^3+2x^2+5x+4$
I can find $gsd(f(x),g(x))=1$ in $Z_7[X]$ but can't solve right $gsd(f(x),g(x))=f(x)*u+g(x)*v$
because i can't get 1 when try solve left part of equation, i have
$f(x)=(2x+1)*g(x)+(2x^2+x+2)$
$g(x)=(4x+6)(2x^2+x+2)+(5x+6)$
$2x^2+x+2=(6x)(5x+6)+2$
$5x+6=(6x+3)*2+0$
so my GCD is 2 but how i understand it must be monic polynomial. then i find: $2=(2x^2+x+2)-(6x)(5x+6)=(2x^2+x+2)-(6x)(g(x)-(4x+6)(2x^2+x+2))=f(x)-(2x+1)*g(x)-(6x)(g(x)-(4x+6)(f(x)-(2x+1)*g(x)))$
then i open brackets
$2=(4x)f(x)+(6x^2+4x)g(x) $
$2*1=2(2x)f(x)+2(3x^2+2x)g(x)$ but it's not equal,yes?
can you understand it for me.