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705-805 (Hard)|   Algebra|      
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MathRevolution
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We have a polynomial x^2 + 4x - n for an integer n between 1 and 100, 18 Feb 2020, 02:56
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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75% (hard)

Question Stats:

61% (03:02) correct 39% (02:54) wrong based on 258 sessions

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We have a polynomial \(x^2 + 4x - n\) for an integer \(n\) between \(1\) and \(100\), inclusive. The polynomial \(x^2 + 4x - n\) is factored into \((x + a)(x + b)\) where \(a\) and \(b\) are integers. How many different values of \(n\) satisfy the polynomial \(x^2 + 4x – n\)?

A. \(5\)

B. \(6\)

C. \(7\)

D. \(8\)

E. \(9\)
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D
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We have a polynomial x^2 + 4x - n for an integer n between 1 and 100, 18 Feb 2020, 05:03
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MathRevolution
[GMAT math practice question]

We have a polynomial \(x^2 + 4x - n\) for an integer \(n\) between \(1\) and \(100\), inclusive. The polynomial \(x^2 + 4x - n\) is factored into \((x + a)(x + b)\) where \(a\) and \(b\) are integers. How many different values of \(n\) satisfy the polynomial \(x^2 + 4x – n\)?

A. \(5\)

B. \(6\)

C. \(7\)

D. \(8\)

E. \(9\)
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Solution:
    • \(n\) is an integer between \(1\) and \(100\).
    • \(x^2+4x – n\) is factored into \((x+a)(x+b)\)
      o Thus, \(x^2+4x-n = x^2+(a+b) + ab\)
         \(a + b = 4\), and \(ab = -n\)
         Either \(a\) is negative or \(b\) is negative.
    • Let’s consider \(b\) to be negative, because that won’t change the answer because we have to deal the product of \(a\) and \(b\).
    • Let’s see how many distinct values of \(n\) are possible.

For any other value of \(a\) or \(b\), \(n\) will be either less than \(1\) or greater than \(100\), so these are the only possible values of \(n\).
Hence, the distinct possible values of \(n\) are \(8\), the correct answer is Option D.
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We have a polynomial x^2 + 4x - n for an integer n between 1 and 100, 20 Feb 2020, 01:56
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=>

We can assume \(a < b\) without loss of generality.

Since we have \(a + b = 4\) and \(ab = -n\), we have the following pairs of \((a, b).\)

\((-1, 5), (-2, 6), (-3, 7), (-4, 8), (-5, 9), (-6, 10), (-7, 11), (-8, 12)\)

Then, the possible values of \(n\) are \(5, 12, 21, 32, 45, 60, 77\) and \(96.\)

We have \(8\) possible values of \(n\).

Therefore, the answer is D.
Answer: D
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We have a polynomial x^2 + 4x - n for an integer n between 1 and 100, 29 Nov 2025, 08:01
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We have a polynomial \(x^2 + 4x - n\) for an integer \(n\) between \(1\) and \(100\), inclusive. The polynomial \(x^2 + 4x - n\) is factored into \((x + a)(x + b)\) where \(a\) and \(b\) are integers.

How many different values of \(n\) satisfy the polynomial \(x^2 + 4x – n\)?

\(x^2 + 4x - n = (x+a)(x+b) = x^2 + (a+b)x + ab\)

a + b = 4
ab = - n

a, b, n are integers and 1 <=n <=100

b = 4 - a
a(4-a) = -n

n = a(a-4) >= 1
(a-2)^2 >= 5
a <= -1 or a>=5

n = a(a-4) <=100
(a-2)^2 <= 104

-8 <= a <= 12

-8 <= a <=-1 & 5<=a<=12

a = {-8,-7,-6,-5,-4,-3,-2,-1,5,6,7,8,9,10,11,12}
n = a(a-4) = {96, 77, 60, 45, 32, 21, 12, 5} : 8 different values of n

IMO D
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