06 Jun 06
Originally posted by XanthosNZWhat, you think this was my homework, now. XanthousNZ that is just for fun. And I am only 4th Form and this is 6th Form Maths, Level 2 XanthousNZ Level2.
I do get homework. But I do it myself and even if I were to ask for help here I doubt most of you could manage to say anything useful.
Also, AThousandYoung is right. Hope you fail.
08 Jun 06
Originally posted by Knight SquareOne interesting pattern to note that may be helpful for quick calculations in the future:
Work this out:
(y+4)(y-7)(y+6)= (?)
and if you are really smart, factorise:
y3 - 54= (?)
(x+A)(x+B) = x^2 + (A+B)x + AB
(x+A)(x+B)(x+C) = x^3 + (A+B+C)x^2 + (AB+AC+AB)x + ABC
(x+A)(x+B)(x+C)(x+D) = x^4 + (A+B+C+D)x^3 + (AB+AC+AD+BC+BD+CD)x^2 + (ABC+ABD+ACD+BCD)x + ABCD
etc...
This pattern arises from the combinatoric process of selecting "n" x's and "m-n" constants from the factorization of a polynomial of degree "m". It's handy to remember, because it saves you making mechanical errors keeping track of all the x^2 and such.
08 Jun 06
1 edit
Originally posted by Knight SquareIf you want a quick estimate of Y3-54=?
Work this out:
(y+4)(y-7)(y+6)= (?)
and if you are really smart, factorise:
y3 - 54= (?)
just take the square root of 54 and divide by 2. The answer is usually accurate to 1 or 2 decimal places with small numbers like 54....3.67 in this case
EDIT: welcome back PB6
08 Jun 06
Originally posted by uzlessGood to be back! But I'm still going to lurk for a while. I just hate people in general, so I'm ramping up to full-blown posting.
If you want a quick estimate of Y3-54=?
just take the square root of 54 and divide by 2. The answer is usually accurate to 1 or 2 decimal places with small numbers like 54....3.67 in this case
EDIT: welcome back PB6