1. ## numbers in brackets

Hi;
The problen is factor y^2-5y+6.

now I know how to factor this to (y-2)(y-3).

my issue is I could reverse the position of the 2 and the 3 without harmimg the outcome is that correct?

thanks.

2. Yes that will be fine.

Remember that $\displaystyle A\times B \equiv B \times A$.

Therefore $\displaystyle (y-2)(y-3)\equiv (y-3)(y-2)$.

3. You can only reverse it if the signs are the same and the number before the y is the same, and in this equation, yes you can.

If you had (y+2)(y-3) then you would not be able to switch the 2 & 3 because the signs are different (unless you moved the signs as well).
Also, if you had (2y-2)(y-3) then you would not be able to switch them because the number before the y is different.

But in your case, the signs and number before the y is the same, so you can reverse the positions of the 2 and 3.

4. So the same for this as well.
x^2+x-12 factors to (x+4)(x-3) or (x-3)(x+4) again same principle applies.

thanks for the help.

5. Correct.