Submitted by Mafs on Thu, 01/12/2017 - 21:19
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
If gotten isn't a word
Take the word got
No, your answer is perfectly
No, your answer is perfectly valid. It just differs by factors of -1 in two of the linear factors. You would not lose any marks for this.
factors of -1
Somebody mentioned this factors of -1 thing to me, but I don't fully understand what it means
Consider $(3y-x-5)=(-1)(-3y+x
Consider $(3y-x-5)=(-1)(-3y+x+5)$ and $(2y-x-1)=(-1)(-2y+x+1)$ so, $$(3y-x-5)(2y-x-1)=(-1)(-3y+x+5)(-1)(-2y+x+1)=(-3y+x+5)(-2y+x+1),$$ so your answer is the same as the one given in the mark scheme. Note that your linear factors can only be determined up to a sign when you have 2 or more of them.