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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.

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    you have found the GCD already – 2017-02-08
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    @Dr. Sonnhard Graubner, Where? How? please elaborate.. – 2017-02-08
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    What do $\,x,y,\,$ denote, indeterminates (variables), or integers, or ....? – 2017-02-08
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    @ Bill Dubuque, They are just variables.. – 2017-02-08

2 Answers 2

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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 $\gcd(a(x),b(x))=r_{n-1}(x)$.

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    @Euler: is it clear? – 2017-02-09
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    @ Arnaldo, Yes sir. It is.. – 2017-02-09
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    @ Arnaldo, Hello sir. Could you please help me with this question :: math.stackexchange.com/questions/2137609/prove-that-sec2-20-circ-sec2-40-circ-sec2-80-circ-textrm-36 – 2017-02-10
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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 $(y-3x-4)$.

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    @ Mark Fischler, why did you choose $y-3x-4$ as divisior? – 2017-02-08
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    @Euler note that for $x=0$ we get $2y^2(y^2-16)$ and $y-4$ vs $y-6$ for the divisors. If one factor has a chance to work then it is the one in $y-4$. – 2017-02-08
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    @ S, Ramanujan I actually tried the other factor $(y+3x-6) first. – 2017-02-10