How many quadratic polynomials exist given the two zeroes? ($1$ or $infty$) The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What is the correspondence between combinatorial problems and the location of the zeroes of polynomials called?How to prove two polynomials have no zeroes in common?Finding a polynomial with product and sum of its zeroesExtensions of the Hermite Bielher and Hermite-Kakeya TheoremWhat is the lowest-degree function that passes through these points?Finding a cubic polynomial whose zeroes are the same as collectively of two other quadratic polynomials.Given some zeroes of a real polynomial of a given degree, how can one find the remaining zeroes?Only 12 polynomials exist with given propertiesMethods of approximating polynomial rootsGCD of two polynomials without Euclidean algorithm
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How many quadratic polynomials exist given the two zeroes? ($1$ or $infty$)
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What is the correspondence between combinatorial problems and the location of the zeroes of polynomials called?How to prove two polynomials have no zeroes in common?Finding a polynomial with product and sum of its zeroesExtensions of the Hermite Bielher and Hermite-Kakeya TheoremWhat is the lowest-degree function that passes through these points?Finding a cubic polynomial whose zeroes are the same as collectively of two other quadratic polynomials.Given some zeroes of a real polynomial of a given degree, how can one find the remaining zeroes?Only 12 polynomials exist with given propertiesMethods of approximating polynomial rootsGCD of two polynomials without Euclidean algorithm
$begingroup$
I was reading some book which had this question:
Q. The number of [quadratic] polynomials having zeros $-2$ and $5$ is:
(A) 1
(B) 2
(C) 3
(D) More than three?
Sol. (A) 1.
But according to me there should be an infinite amount of polynomials. All these polynomials have the zeroes $-2$ and $5$:
beginalign*
(x+2)(x-5)&=0,\
2(x+2)(x-5)&=0,\
3(x+2)(x-5)&=0,\
4(x+2)(x-5)&=0,\
5(x+2)(x-5)&=0, textetc.
endalign*
According to me, if we are given the two zeroes of a quadratic polynomial, then we can find $infty$ polynomials with those two zeroes. I do not know why the answer $1$ is given in the book. Maybe it is a misprint? So am I right or the book is right?
Clarification: I just wanted to ask if these polynomials are considered to be the same or different.
polynomials roots
edited Feb 14 '14 at 14:16
neofoxmulder
994616
asked Feb 14 '14 at 8:49
KartikKartik
888922
$endgroup$
|
show 3 more comments
$begingroup$
I was reading some book which had this question:
Q. The number of [quadratic] polynomials having zeros $-2$ and $5$ is:
(A) 1
(B) 2
(C) 3
(D) More than three?
Sol. (A) 1.
But according to me there should be an infinite amount of polynomials. All these polynomials have the zeroes $-2$ and $5$:
beginalign*
(x+2)(x-5)&=0,\
2(x+2)(x-5)&=0,\
3(x+2)(x-5)&=0,\
4(x+2)(x-5)&=0,\
5(x+2)(x-5)&=0, textetc.
endalign*
According to me, if we are given the two zeroes of a quadratic polynomial, then we can find $infty$ polynomials with those two zeroes. I do not know why the answer $1$ is given in the book. Maybe it is a misprint? So am I right or the book is right?
Clarification: I just wanted to ask if these polynomials are considered to be the same or different.
polynomials roots
edited Feb 14 '14 at 14:16
neofoxmulder
994616
asked Feb 14 '14 at 8:49
KartikKartik
888922
$endgroup$
2
$begingroup$
You are right. Though check that the book may have made a note somewhere that only monic polynomials are being considered.
$endgroup$
– David Peterson
Feb 14 '14 at 8:56
$begingroup$
@DavidPeterson There is no note. There are only Question and answers in the book.
$endgroup$
– Kartik
Feb 14 '14 at 8:58
$begingroup$
All those are equivalent; constants in $mathbbQ$ are units in $mathbbQ[X]$
$endgroup$
– FireGarden
Feb 14 '14 at 9:08
$begingroup$
@FireGarden What is $mathbbQ[X]$?
$endgroup$
– Kartik
Feb 14 '14 at 9:10
$begingroup$
@FireGarden: They are equivalent (modulo the implied equivalence relation), but they are still distinct.
$endgroup$
– Hurkyl
Feb 14 '14 at 9:14
|
show 3 more comments
$begingroup$
I was reading some book which had this question:
Q. The number of [quadratic] polynomials having zeros $-2$ and $5$ is:
(A) 1
(B) 2
(C) 3
(D) More than three?
Sol. (A) 1.
But according to me there should be an infinite amount of polynomials. All these polynomials have the zeroes $-2$ and $5$:
beginalign*
(x+2)(x-5)&=0,\
2(x+2)(x-5)&=0,\
3(x+2)(x-5)&=0,\
4(x+2)(x-5)&=0,\
5(x+2)(x-5)&=0, textetc.
endalign*
According to me, if we are given the two zeroes of a quadratic polynomial, then we can find $infty$ polynomials with those two zeroes. I do not know why the answer $1$ is given in the book. Maybe it is a misprint? So am I right or the book is right?
Clarification: I just wanted to ask if these polynomials are considered to be the same or different.
polynomials roots
edited Feb 14 '14 at 14:16
neofoxmulder
994616
asked Feb 14 '14 at 8:49
KartikKartik
888922
$endgroup$
I was reading some book which had this question:
Q. The number of [quadratic] polynomials having zeros $-2$ and $5$ is:
(A) 1
(B) 2
(C) 3
(D) More than three?
Sol. (A) 1.
But according to me there should be an infinite amount of polynomials. All these polynomials have the zeroes $-2$ and $5$:
beginalign*
(x+2)(x-5)&=0,\
2(x+2)(x-5)&=0,\
3(x+2)(x-5)&=0,\
4(x+2)(x-5)&=0,\
5(x+2)(x-5)&=0, textetc.
endalign*
According to me, if we are given the two zeroes of a quadratic polynomial, then we can find $infty$ polynomials with those two zeroes. I do not know why the answer $1$ is given in the book. Maybe it is a misprint? So am I right or the book is right?
Clarification: I just wanted to ask if these polynomials are considered to be the same or different.
polynomials roots
polynomials roots
edited Feb 14 '14 at 14:16
neofoxmulder
994616
asked Feb 14 '14 at 8:49
KartikKartik
888922
edited Feb 14 '14 at 14:16
neofoxmulder
994616
asked Feb 14 '14 at 8:49
KartikKartik
888922
edited Feb 14 '14 at 14:16
neofoxmulder
994616
edited Feb 14 '14 at 14:16
neofoxmulder
994616
edited Feb 14 '14 at 14:16
neofoxmulder
994616
994616
asked Feb 14 '14 at 8:49
KartikKartik
888922
asked Feb 14 '14 at 8:49
KartikKartik
888922
asked Feb 14 '14 at 8:49
KartikKartik
888922
888922
2
$begingroup$
You are right. Though check that the book may have made a note somewhere that only monic polynomials are being considered.
$endgroup$
– David Peterson
Feb 14 '14 at 8:56
$begingroup$
@DavidPeterson There is no note. There are only Question and answers in the book.
$endgroup$
– Kartik
Feb 14 '14 at 8:58
$begingroup$
All those are equivalent; constants in $mathbbQ$ are units in $mathbbQ[X]$
$endgroup$
– FireGarden
Feb 14 '14 at 9:08
$begingroup$
@FireGarden What is $mathbbQ[X]$?
$endgroup$
– Kartik
Feb 14 '14 at 9:10
$begingroup$
@FireGarden: They are equivalent (modulo the implied equivalence relation), but they are still distinct.
$endgroup$
– Hurkyl
Feb 14 '14 at 9:14
|
show 3 more comments
2
$begingroup$
You are right. Though check that the book may have made a note somewhere that only monic polynomials are being considered.
$endgroup$
– David Peterson
Feb 14 '14 at 8:56
$begingroup$
@DavidPeterson There is no note. There are only Question and answers in the book.
$endgroup$
– Kartik
Feb 14 '14 at 8:58
$begingroup$
All those are equivalent; constants in $mathbbQ$ are units in $mathbbQ[X]$
$endgroup$
– FireGarden
Feb 14 '14 at 9:08
$begingroup$
@FireGarden What is $mathbbQ[X]$?
$endgroup$
– Kartik
Feb 14 '14 at 9:10
$begingroup$
@FireGarden: They are equivalent (modulo the implied equivalence relation), but they are still distinct.
$endgroup$
– Hurkyl
Feb 14 '14 at 9:14
2
2
$begingroup$
You are right. Though check that the book may have made a note somewhere that only monic polynomials are being considered.
$endgroup$
– David Peterson
Feb 14 '14 at 8:56
$begingroup$
You are right. Though check that the book may have made a note somewhere that only monic polynomials are being considered.
$endgroup$
– David Peterson
Feb 14 '14 at 8:56
$begingroup$
@DavidPeterson There is no note. There are only Question and answers in the book.
$endgroup$
– Kartik
Feb 14 '14 at 8:58
$begingroup$
@DavidPeterson There is no note. There are only Question and answers in the book.
$endgroup$
– Kartik
Feb 14 '14 at 8:58
$begingroup$
All those are equivalent; constants in $mathbbQ$ are units in $mathbbQ[X]$
$endgroup$
– FireGarden
Feb 14 '14 at 9:08
$begingroup$
All those are equivalent; constants in $mathbbQ$ are units in $mathbbQ[X]$
$endgroup$
– FireGarden
Feb 14 '14 at 9:08
$begingroup$
@FireGarden What is $mathbbQ[X]$?
$endgroup$
– Kartik
Feb 14 '14 at 9:10
$begingroup$
@FireGarden What is $mathbbQ[X]$?
$endgroup$
– Kartik
Feb 14 '14 at 9:10
$begingroup$
@FireGarden: They are equivalent (modulo the implied equivalence relation), but they are still distinct.
$endgroup$
– Hurkyl
Feb 14 '14 at 9:14
$begingroup$
@FireGarden: They are equivalent (modulo the implied equivalence relation), but they are still distinct.
$endgroup$
– Hurkyl
Feb 14 '14 at 9:14
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
You are right, the book is mistaken (perhaps they are taking polynomials which are constant multiples of each other as equal).
answered Feb 14 '14 at 13:44
vonbrandvonbrand
20k63260
$endgroup$
add a comment |
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$begingroup$
You are right, the book is mistaken (perhaps they are taking polynomials which are constant multiples of each other as equal).
answered Feb 14 '14 at 13:44
vonbrandvonbrand
20k63260
$endgroup$
add a comment |
$begingroup$
You are right, the book is mistaken (perhaps they are taking polynomials which are constant multiples of each other as equal).
answered Feb 14 '14 at 13:44
vonbrandvonbrand
20k63260
$endgroup$
add a comment |
$begingroup$
You are right, the book is mistaken (perhaps they are taking polynomials which are constant multiples of each other as equal).
answered Feb 14 '14 at 13:44
vonbrandvonbrand
20k63260
$endgroup$
You are right, the book is mistaken (perhaps they are taking polynomials which are constant multiples of each other as equal).
answered Feb 14 '14 at 13:44
vonbrandvonbrand
20k63260
answered Feb 14 '14 at 13:44
vonbrandvonbrand
20k63260
answered Feb 14 '14 at 13:44
vonbrandvonbrand
20k63260
answered Feb 14 '14 at 13:44
vonbrandvonbrand
20k63260
20k63260
add a comment |
add a comment |
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$begingroup$
You are right. Though check that the book may have made a note somewhere that only monic polynomials are being considered.
$endgroup$
– David Peterson
Feb 14 '14 at 8:56
$begingroup$
@DavidPeterson There is no note. There are only Question and answers in the book.
$endgroup$
– Kartik
Feb 14 '14 at 8:58
$begingroup$
All those are equivalent; constants in $mathbbQ$ are units in $mathbbQ[X]$
$endgroup$
– FireGarden
Feb 14 '14 at 9:08
$begingroup$
@FireGarden What is $mathbbQ[X]$?
$endgroup$
– Kartik
Feb 14 '14 at 9:10
$begingroup$
@FireGarden: They are equivalent (modulo the implied equivalence relation), but they are still distinct.
$endgroup$
– Hurkyl
Feb 14 '14 at 9:14