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If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and

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If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and  [#permalink]

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New post 17 Dec 2015, 10:08
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A
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  15% (low)

Question Stats:

76% (01:28) correct 24% (01:43) wrong based on 214 sessions

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Re: If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and  [#permalink]

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New post 17 Dec 2015, 11:02
1
Bunuel wrote:
If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and s are constants, then which of the following is a possible value of r − s?

A. 8
B. 2
C. − 2
D. − 3
E. − 5



I got A (8)

Is it the correct answer?
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Re: If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and  [#permalink]

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New post 17 Dec 2015, 11:06
solving the above equation gives
(x+3)(x-5)=0
possible values of r-s =3+5=8 and -5-3=-8
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Re: If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and  [#permalink]

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New post 17 Dec 2015, 15:19
2
Bunuel wrote:
If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and s are constants, then which of the following is a possible value of r − s?

A. 8
B. 2
C. − 2
D. − 3
E. − 5


Given: x^2 − 2x − 15 = (x + r)( x + s)
Factor to get: (x - 5)(x + 3) = (x + r)( x + s)
Rewrite as: (x + -5)(x + 3) = (x + r)( x + s)
So, it's possible that r = -5 and s = 3
Here, r - s = (-5) - 3
= -8
Not an answer choice

Try REVERSING the factorization:
x^2 − 2x − 15 = (x + r)( x + s)
Factor to get: (x + 3)(x - 5) = (x + r)( x + s)
Rewrite as: (x + 3)(x + -5) = (x + r)( x + s)
So, it's possible that r = 3 and s = -5
Here, r - s = 3 - (-5)
= 8

Answer: A

Cheers,
Brent
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Re: If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and  [#permalink]

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New post 17 Dec 2015, 21:16
2
Bunuel wrote:
If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and s are constants, then which of the following is a possible value of r − s?

A. 8
B. 2
C. − 2
D. − 3
E. − 5



We know that given ax^2 + bx + c = 0, Sum of the roots = -b/a and product of the roots = c/a.

The roots here are -r and -s.

-r - s = -(-2)/1 = r + s = -2

(-r)*(-s) = -15/1 = rs

So one of r and s is -5 and the other is 3. So r - s could be 8 or -8.

Answer (A)
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Re: If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and  [#permalink]

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New post 16 Jul 2017, 14:01
If \(x^2 − 2x − 15 = (x + r)( x + s)\) for all values of x, and if r and s are constants, then which of the following is a possible value of r − s?

\(x^2 − 2x − 15 = (x + r)( x + s)\)

\((x - 5)(x + 3) = (x + r)( x + s)\)

\(r = -5\) and \(s = 3\)

\(r - s = -5 - 3 = -8\) ===> Not available in the answer choice

Let's reverse the values:

\(r = 3\) and \(s = -5\)

\(r - s = 3 - (-5) = 3 + 5 = 8\)

this matches the value in option A

Hence, Answer is A

Did you like the answer? 1 Kudos Please :good
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Re: If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and  [#permalink]

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New post 20 Jul 2017, 16:38
Bunuel wrote:
If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and s are constants, then which of the following is a possible value of r − s?

A. 8
B. 2
C. − 2
D. − 3
E. − 5


We can factor x^2 − 2x − 15:

x^2 − 2x − 15 = (x - 5)(x + 3)

Setting this equal to (x + r)(x + s), we see that EITHER r = -5 and s = 3 OR r = 3 and s = -5. If the former, then (r - s) = -8, and if the latter, then (r - s) = 8. We see that, of those two possibilities, only 8 is in the given choices, so choice A is the correct answer.

Answer: A
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Re: If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and   [#permalink] 20 Jul 2017, 16:38
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