Dear Uncle Colin,
How would you go about factorising $6x^2 - xy - y^2 + 7x - y + 2$?
- Argh! Getting Nowhere. Expression Simplification Impossible.
Hi, AGNESI, and thanks for your message!
I have, historically, not been a fan of these. However, I’ve recently come across a method that makes a bit of sense to me, and I figured it was worth sharing. (I have a second method in case you don’t like the first.)
The plan is:
- Consider what happens if you let each variable in turn be zero
- Combine the resulting factors as best you can.
A bit at a time
If $x=0$, we have $-y^2 - y + 2$, which I would rewrite as $-(y^2 + y - 2)$ so I don’t lose a minus sign, then as $(y+2)(1-y)$.
If $y=0$, we have $6x^2 + 7x + 2$, which is $(3x + 2)(2x+1)$.
If $x \gg 0$ and $y \gg 0$, we have $6x^2 - xy - y^2$, which is $(3x + y)(2x-y)$.
Now we can combine the brackets: $3x+y$, $3x + 2$ and $y+2$ look like they fit together, so we can make them $(3x+y+2)$.
What’s left over? $2x - y$, $2x+1$ and $-y+1$ also fit together, so the other factor is $(2x - y + 1)$.
We should probably check that, but I’m in a ‘Wolfram|Alpha says it’s right’ mood rather than a ‘multiply it all out’ one.
You could also start by factorising just the quadratic terms: again, $6x^2 - xy - y^2 = (3x+y)(2x-y)$. To get the linear and constant terms, we will need to add a constant to each of these brackets, making $(3x+y+a)(2x-y+b)$.
Then set up a multiplication grid (other expansion methods are available):
Then, comparing coefficients, we know that $ab=2$, $2a + 3b = 7$ and $b -a = -1$. These are satisfied by $a=2$ and $b=1$, so the factorisation is $(3x + y + 2)(2x - y + 1)$, as before.
Hope that helps!
- Uncle Colin
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