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footprints
Oct16-04, 11:06 AM
3x^2 + 4x + C \equiv A(x + 1)^2 + B(x + 1) + 7
Find all values of A, B and C.
Could someone teach how to do this?

arildno
Oct16-04, 11:09 AM
Rewrite
x^{2}=((x+1)-1)^{2}
and do the necessary operations.

footprints
Oct16-04, 11:21 AM
Huh? Sorry I don't understand what u mean.

arildno
Oct16-04, 11:23 AM
((x+1)-1)^{2}=(x+1)^{2}-2(x+1)+1

footprints
Oct16-04, 11:32 AM
How is ((x+1)-1)^{2}=(x+1)^{2}-2(x+1)+1
related to 3x^2 + 4x + C \equiv A(x + 1)^2 + B(x + 1) + 7 ?
Sorry but i don't get u.

arildno
Oct16-04, 11:36 AM
It's equal to x^{2}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!
Make a similar rewriting of x:
x=((x+1)-1)
Now, substitute these expressions for x,x^{2} into your LEFT-HAND SIDE.
Reorganize the terms you get, and derive conditions so that your new expression equals your ORIGINAL RIGHT-HAND SIDE.
This will determine A,B,C.

Hurkyl
Oct16-04, 11:44 AM
Alternatively, you could expand the right hand side, and equate coefficients to solve for A, B, and C.
Or you could plug in a few values for x to generate equations.

footprints
Oct16-04, 11:47 AM
I probably wasn't clear. Firstly, how does x^{2}=((x+1)-1)^{2}. Secondly, which ones do i substitute in for x? Is it ((x+1)-1)^{2} = 3 or (x + 1)^2 ?

arildno
Oct16-04, 12:02 PM
You know that 1-1=0, right?
So (x+1)-1=x+1-1=x.
To be nice, I'll do this for once:
We have:
3x^{2}=3((x+1)-1)^{2}=3(x+1)^{2}-6(x+1)+3
4x=4((x+1)-1)=4(x+1)-4
C=C=C
Now, add these equations together, downwards. The outermost terms then turn into:
3x^{2}+4x+C=3(x+1)^{2}-2(x+1)+(C-1)
Do you understand this?

Hurkyl
Oct16-04, 12:05 PM
What does (x + 1) - 1 equal? Then what does ((x+1) - 1)^2 equal?

Sariaht
Oct16-04, 12:28 PM
I'm just curious, has this got something to do with modulus?

arildno
Oct16-04, 12:30 PM
I'm just curious, has this got something to do with modulus?
Not that I know of..
It is simply to substitute "equal for equal"

MiGUi
Oct16-04, 02:06 PM
<Sarcastic mode ON>
arildno you have to be a little more diplomatic, I think
<Sarcastic mode OFF>

Derivative86
Oct16-04, 04:19 PM
you guys are making the problem too complicated
since
3x^2 + 4x + C = A(x + 1)^2 + B(x + 1) + 7
expand A(x + 1)^2 + B(x + 1) + 7
then you have Ax^2 + 2Ax + A + Bx + B + 7
and 3x^2 + 4x + C = Ax^2 + 2Ax + A + Bx + B + 7
equate the coeffients of the powers you have
3 = A for x^2
2A + B = 4 for x^1
A + B + 7 = C for x^0
solve the system and you got the answer :smile:

footprints
Oct17-04, 12:53 AM
Ahh... Sorry but i was a bit slow on "(x+1)-1" part (sorry if i pissed u arildno :frown: ). Thanks for the help guys!

arildno
Oct17-04, 10:48 AM
I wasn't exactly pissed off; rather, I felt resignation sneak up on me..:wink: