Question

Given\[a=t^4-3t^3+3t^2-3t+2\]\[b=t^3-10t^2+23t-14\]Find d=gcd(a,b). I found the the gcd already and it is \[d=t^2-3t+2\]Now I am supposed to find \[d=ax+by\]How do I solve for x and y?

Answer 1

\[\frac{a}{t^{2}+1} = d\]
\[\frac{b}{t-7} = d\]
\[\rightarrow \frac{a}{2(t^{2}+1)} + \frac{b}{2(t-7)} = \frac{d}{2}+\frac{d}{2} = d\]
Therefore
\[x = \frac{1}{2(t^{2}+1)}\]
\[y = \frac{1}{2(t-7)}\]