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805+ (Hard)|   Algebra|      
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sherxon
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When x^2 + mx + n polynomial is divided by (x-m) 12 Jun 2021, 09:23
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95% (hard)

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34% (02:08) correct 66% (02:07) wrong based on 116 sessions

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When \(x^2 + mx + n\) polynomial is divided by (x-m) and (x-n), the remainders are m and n respectively. How many pairs of (m;n) are there?

A. 1
B. 2
C. 3
D. 4
E. 5
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C
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chetan2u
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When x^2 + mx + n polynomial is divided by (x-m) 13 Jun 2021, 06:39
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sherxon
When \(x^2 + mx + n\) polynomial is divided by (x-m) and (x-n), the remainders are m and n respectively. How many pairs of (m;n) are there?

A. 1
B. 2
C. 3
D. 4
E. 5
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1) When \(x^2 + mx + n\) polynomial is divided by (x-m), the remainder is m.
This means \(x^2 + mx + n-m\) is divisible by (x-m)
So, when we substitute x=m, the expression should give us a value of 0.

\(m^2 + m*m + n-m=0.......2m^2+n-m=0\).....(I)

2) When \(x^2 + mx + n\) polynomial is divided by (x-n), the remainder is n.
This means \(x^2 + mx + n-n\) is divisible by (x-n)
So, when we substitute x=n, the expression should give us a value of 0.

\(n^2+mn+n-n=0.........n(m+n)=0\)
So either n=0 or n=-m


Substitute these values of n in (I) to get pairs of m and n.

(a) n=0
\(2m^2+n-m=0......2m^2-m=0......m(2m-1)=0\)
Thus (m,n) can be (0,0) and (1/2,0)
(b) n=-m
\(2m^2+n-m=0......2m^2-2m=0......2m(m-1)=0\)
Thus (m,n) can be (0,0) and (1,1)

Thus 3 pairs possible
(0,0)
(1/2,0)
(1,1)


C
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samarpan.g28
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When x^2 + mx + n polynomial is divided by (x-m) 29 Dec 2023, 12:25
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chetan2u
sherxon
When \(x^2 + mx + n\) polynomial is divided by (x-m) and (x-n), the remainders are m and n respectively. How many pairs of (m;n) are there?

A. 1
B. 2
C. 3
D. 4
E. 5


1) When \(x^2 + mx + n\) polynomial is divided by (x-m), the remainder is m.
This means \(x^2 + mx + n-m\) is divisible by (x-m)
So, when we substitute x=m, the expression should give us a value of 0.

\(m^2 + m*m + n-m=0.......2m^2+n-m=0\).....(I)

2) When \(x^2 + mx + n\) polynomial is divided by (x-n), the remainder is n.
This means \(x^2 + mx + n-n\) is divisible by (x-n)
So, when we substitute x=n, the expression should give us a value of 0.

\(n^2+mn+n-n=0.........n(m+n)=0\)
So either n=0 or n=-m


Substitute these values of n in (I) to get pairs of m and n.

(a) n=0
\(2m^2+n-m=0......2m^2-m=0......m(2m-1)=0\)
Thus (m,n) can be (0,0) and (1/2,0)
(b) n=-m
\(2m^2+n-m=0......2m^2-2m=0......2m(m-1)=0\)
Thus (m,n) can be (0,0) and (1,1)

Thus 3 pairs possible
(0,0)
(1/2,0)
(1,1)


C
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Hi, please correct me if I am wrong. In case 2 where n=-m, once we get m=1, won't the corresponding value for n be -1 since n=-m? And the pair will be (1,-1)?
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When x^2 + mx + n polynomial is divided by (x-m) 01 Jan 2024, 00:45
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Why do we substitute x=m and x=n ?
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When x^2 + mx + n polynomial is divided by (x-m) 04 Feb 2024, 06:55
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NesnaSubedi
Why do we substitute x=m and x=n ?



 
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Hi, (x-m) and (x-n) are roots of the equation therefore if we substitute x=m and x=n, the equation will hold true i.e. it will become zero. Hope this helps you.
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When x^2 + mx + n polynomial is divided by (x-m) 04 Feb 2024, 07:53
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Given: When \(x^2 + mx + n\) polynomial is divided by (x-m) and (x-n), the remainders are m and n respectively.Asked: How many pairs of (m;n) are there?

When \(x^2 + mx + n\) polynomial is divided by (x-m), the remainders is m
m^2 + m^2 + n = m
2m^2 - m +n = 0

When \(x^2 + mx + n\) polynomial is divided by (x-n), the remainders is n
n^2 + mn + n = n
n^2 + mn = 0
Either n = 0 or
m + n = 0

Case 1: n = 0;
2m^2 - m +n = 0
2m^2 - m = 0 ;
m(2m -1)=0;
m=0 or m=1/2
(m,n) = {(0,0),(1/2,0)}

Case 2: n=-m
2m^2 - m +n = 0
2m^2 - m -m =0
m^2 - m =0
m(m-1)=0
m = 0 or m=1
(m,n) = [(0,0),(1,-1)}

(m,n) = {(0,0),(1/2,0),(1,-1)} :
3 ordered pairs are possible

IMO C
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When x^2 + mx + n polynomial is divided by (x-m) 07 Jul 2025, 05:54
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