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If x^2 + 3x + c = (x + a)(x + b) for all x, what is the

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Senior Manager
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If x^2 + 3x + c = (x + a)(x + b) for all x, what is the [#permalink] New post 30 Nov 2004, 12:12
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A
B
C
D
E

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If x^2 + 3x + c = (x + a)(x + b) for all x, what is the value of c ?

(1) a = 1

(2) b = 2
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 [#permalink] New post 30 Nov 2004, 12:26
D

x^2 + 3x + c = (x + a)(x + b)
= x^2+ax+bx+ab

=> (a+b)x =3x ; c =ab

s[1]: a=1; b=3-a=2 => c=2
S[2]: b=2; a=3-2 =1 => c=2
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 [#permalink] New post 30 Nov 2004, 13:03
Another great one Dookie !

Nothing to add, D it is.
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 [#permalink] New post 30 Nov 2004, 13:11
Yeah, i liked this question because at first, i wasn't sure that it's ok to go from:
(a+b)x+ab = 3x + c
To
=> (a+b)x =3x ; c =ab.

But then i realized that it must be correct, since both describe ONE specific line.

And there can only be one linear equation for this line.
:idea:
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 [#permalink] New post 30 Nov 2004, 14:47
Is the OA D ?
Since c = ab => we need stmt 1 and stmt 2 to get c
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 [#permalink] New post 02 Dec 2004, 05:41
we know that
x^2+(a+b)x+ab= (x+a)(x+b)
so it is D
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 [#permalink] New post 03 Dec 2004, 07:58
Dookie wrote:
Yeah, i liked this question because at first, i wasn't sure that it's ok to go from:
(a+b)x+ab = 3x + c
To
=> (a+b)x =3x ; c =ab.

But then i realized that it must be correct, since both describe ONE specific line.

And there can only be one linear equation for this line.
:idea:


Hey Dookie,

This one is easier then you made it. Here's what I always teach my students:

whenever you've got this polynomial situation, there are always four elements to figure out. It'll always look like:

x^2 + bx + c = (x + d)(x + e).

And we always know that d + e = b and de = c

And of course we know how to factor the first to make the second if we have b and c, and we know how to mulitply the second to make the first if we've got d + e.

So the point is, we have enough informaiton as long as we have any two of b, c, d, and e to always get the other two. It's an absolute. Give me any two of those numbers, I'll give you the other two.

So in any problem, just look for two of them, and you'll get the other ones. You don't have to think about it - just do it.

Check out these example from the OG to back that up:

MC195
MC333
DS124
  [#permalink] 03 Dec 2004, 07:58
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