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Task: Factorise $6y^3 -y^2 -21y + 2x^2 + 12x -4xy +x^2y -5xy^2 + 10 $ into 3 linear factors

Workings: let $$ f(x,y) = 6y^3 -y^2 -21y + 2x^2 + 12x -4xy +x^2y -5xy^2 + 10 $$ $$ f(x,-2) = 0 \Rightarrow f(x,y)= (y+2)(ay+cx+d)(by+ex+f) $$

At this stage, I noted the possible combinations a and b could take in the equation $ab = 6$. $ a = 3, b = 2 ; a = -3, b=-2$. I had two values a and b could take and didn't know which one to take, so I just took $a = 3$ and $b=2$. After going through a few simultaneous equations, I ended up with $$f(x,y) = (y+2)(3y-x-5)(2y-x-1)$$

The mark scheme ended up with $$f(x,y) = (y+2)(-2y+x+1)(-3y+x+5)$$ since they took $a=-3$ and $b=-2$.

Question: Have I gotten this question wrong? If so, explain why please

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    Since two of your factors are simply the negatives of two of the mark scheme factors, and $-1$ is a unit, and factorisation is only ever unique up to multiplication by units, you have a correct answer. You could even say (ahead of time), taking values for $a,b$, that the only difference will be the signs of the two factors, because you can always multiply both factors by $-1$. – 2017-01-12
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    no - it's giving the same roots and although answers might not appear exactly the same as in markschemes, they can be equivalent – 2017-01-12
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    @user405274 does this answer your question? – 2017-01-12
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    @MarkBennet Thanks, think I get the gist of what you're saying. Need to read it a couple more times... – 2017-01-12

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I do not know what a mark scheme is.

$6y^3 -y^2 -21y + 2x^2 + 12x -4xy +x^2y -5xy^2 + 10$

The main thing I see is that there is no $x^3$ term. This means that one of the terms must be of the form $y - A$ for some constant $A.$ In turn, this means that the collections of terms with no $x$ at all must be a multiple; that is, $6 y^3 - y^2 - 21 y+ 10 $ must have root $A.$ Indeed, you indicate $A = -2.$

What happens when $y = -2?$ We do get $6 y^3 - y^2 - 21 y+ 10 = -48 - 4 + 42 + 10 = -52 +52 = 0. $ Good so far. The rest is $$ 2x^2 + 12x -4xy +x^2y -5xy^2.$$ When $y = -2,$ this is $$ 2 x^2 + 12 x + 8 x - 2 x^2 - 20 x = 0. $$ This confirms that $y+2$ really does divide the original polynomial. I get $$ (y+2)\left( x^2 -5xy + 6 y^2 + 6x - 13 y + 5 \right) $$

The quadratic part of the remaining factor, $$ x^2 - 5 xy + 6 y^2 = (x-2y)(x-3y) $$ since it has a square discriminant $25 - 4 \cdot 6 = 1.$ A little more fiddling gives $$ x^2 -5xy + 6 y^2 + 6x - 13 y + 5 = (x-2y+1)(x-3y+5) $$

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    What does a square discriminant tell us – 2017-01-12
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    @user405274, it tells us that a binary quadratic form with integer coefficients does, indeed, factor with integer coefficients. Exactly the same information as asking about roots and factoring after setting $y$ to one, that is $x^2 - 5 x + 6 = (x-2)(x-3)$ When we put back the letter $y$ we get $ x^2 - 5 xy + 6 y^2 = (x-2y)(x-3y) $ – 2017-01-12
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    Thanks for this, I have one last question. I'm preparing for STEP which is an examination targeted at applicants for Cambridge, we are expected to solve questions in unfamiliar topics with elementary knowledge. In relation to this topic, I have learned only long division, basic factor and remainder theorem and factoring single variable polynomials. I would like to know what topic what polynomials with 2 variables such as these come under?? – 2017-01-12