Factors of a polynomial

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  • #1
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[tex]3x^2 + 4x + C \equiv A(x + 1)^2 + B(x + 1) + 7[/tex]
Find all values of A, B and C.
Could someone teach how to do this?
 

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  • #2
arildno
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Rewrite
[tex]x^{2}=((x+1)-1)^{2}[/tex]
and do the necessary operations.
 
  • #3
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Huh? Sorry I don't understand what u mean.
 
  • #4
arildno
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[tex]((x+1)-1)^{2}=(x+1)^{2}-2(x+1)+1[/tex]
 
  • #5
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How is [tex]((x+1)-1)^{2}=(x+1)^{2}-2(x+1)+1[/tex]
related to [tex]3x^2 + 4x + C \equiv A(x + 1)^2 + B(x + 1) + 7 ?[/tex]
Sorry but i don't get u.
 
  • #6
arildno
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It's equal to [tex]x^{2}[/tex]!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Make a similar rewriting of x:
[tex]x=((x+1)-1)[/tex]
Now, substitute these expressions for [tex]x,x^{2}[/tex] 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.
 
  • #7
Hurkyl
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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.
 
  • #8
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I probably wasn't clear. Firstly, how does [tex]x^{2}=((x+1)-1)^{2}.[/tex] Secondly, which ones do i substitute in for x? Is it [tex]((x+1)-1)^{2} = 3 or (x + 1)^2 ?[/tex]
 
  • #9
arildno
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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:
[tex]3x^{2}=3((x+1)-1)^{2}=3(x+1)^{2}-6(x+1)+3[/tex]
[tex]4x=4((x+1)-1)=4(x+1)-4[/tex]
[tex]C=C=C[/tex]
Now, add these equations together, downwards. The outermost terms then turn into:
[tex]3x^{2}+4x+C=3(x+1)^{2}-2(x+1)+(C-1)[/tex]
Do you understand this?
 
  • #10
Hurkyl
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What does (x + 1) - 1 equal? Then what does ((x+1) - 1)^2 equal?
 
  • #11
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I'm just curious, has this got something to do with modulus?
 
  • #12
arildno
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Sariaht said:
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"
 
  • #13
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<Sarcastic mode ON>
arildno you have to be a little more diplomatic, I think
<Sarcastic mode OFF>
 
  • #14
you guys are making the problem too complicated
since
[tex]3x^2 + 4x + C = A(x + 1)^2 + B(x + 1) + 7[/tex]
expand [tex]A(x + 1)^2 + B(x + 1) + 7[/tex]
then you have [tex]Ax^2 + 2Ax + A + Bx + B + 7[/tex]
and [tex]3x^2 + 4x + C = Ax^2 + 2Ax + A + Bx + B + 7[/tex]
equate the coeffients of the powers you have
[tex]3 = A[/tex] for [tex]x^2[/tex]
[tex]2A + B = 4[/tex] for [tex]x^1[/tex]
[tex]A + B + 7 = C[/tex] for [tex]x^0[/tex]
solve the system and you got the answer :smile:
 
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  • #15
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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!
 
  • #16
arildno
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I wasn't exactly pissed off; rather, I felt resignation sneak up on me..:wink:
 

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