Find the GCD of ... Find the GCD of:
$y^2-10y+24+6x-9x^2$, $2y^4-18x^2y^2-48xy^2-32y^2$.
My Attempt:
$$
\begin{split}
1^{st} \text{expression} &= y^2-10y+24+6x-9x^2\\
&={(y)^2-2\cdot(y)\cdot5+(5)^2}-(5)^2+24+6x-9x^2\\
&=(y-5)^2-{25-24-6x+9x^2}\\
&=(y-5)^2-{(1)^2-2\cdot(1)\cdot(3x)+(3x)^2}\\
&=(y-5)^2-(1-3x)^2\\
&=(y-5+1-3x)(y-5-1+3x)\\
&=(y-3x-4)(y+3x-6).
\end{split}
$$
But, I could not factorize the $2^{nd}$ expression. Please help.
 A: The problem is that sometimes you can't find a factorization for the algebric expressions. If you can is just to use Mark's approach. 
If not, a more general way is use the same approach that we use to find the $\gcd$ between integer numbers. Which is based on Euclides's division algorithm. 
Step $1)$: Divide $$a(y)=2y^4-18x^2y^2-48xy^2-32y^2$$ by $$b(x)=y^2-10y+24+6x-9x^2$$ w.r.t. the variable $y$ and find the quotient $q_1(x)$ and the remainder $r_1(x)$. 
$$\text{So,}\quad  a(x)=b(x)q_1(x)+r_1(x)$$
Step $2)$: Divide $b(x)$ by $r_1(x)$ and get quotient $q_2(x)$ and remainder $r_2(x)$ 
$$\text{So,}\quad  b(x)=r_1(x)q_2(x)+r_2(x)$$
Step $3)$: Divide $r_1(x)$ by $r_2(x)$ and get quotient $q_3(x)$ and remainder $r_3(x)$ 
$$\text{So,}\quad  r_1(x)=r_2(x)q_3(x)+r_3(x)$$
and go on untill find $r_n(x)=0$. Then $\gcd(a(x),b(x))=r_{n-1}(x)$.
A: You don't need to factor that quartic expression from scratch; you just have to try out all factors of the quadratic expression.
Use synthetic division to see that 
$$
\frac{2y^4-18x^2y^2-48y^2x-32y^2}{y-3x-4} = 2y^2(y+3x+4)
$$
So the GCD is $(y-3x-4)$.
