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Prove that polynomial of degree $4$ with real roots cannot have $pm 1$ as coefficients (IITJEE)
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So I was going through my 11th class package on Quadratic equations and I saw a question to prove that a polynomial of $4$th degree with all real roots cannot have $pm 1$ as all its coefficients.
I tried proving it using calculus, by showing that at least one consecutive maxima and minima will lie either above or below the x axis, but couldn't solve it using that.
I also tried using Descartes Rule of Signs but couldn't solve it with that too.
Any help?
calculus algebra-precalculus polynomials quadratics
edited Jul 16 at 15:43
StubbornAtom
3,79311134
asked Jul 16 at 15:21
James Adams
535
 |Â
show 4 more comments
up vote
10
down vote
favorite
So I was going through my 11th class package on Quadratic equations and I saw a question to prove that a polynomial of $4$th degree with all real roots cannot have $pm 1$ as all its coefficients.
I tried proving it using calculus, by showing that at least one consecutive maxima and minima will lie either above or below the x axis, but couldn't solve it using that.
I also tried using Descartes Rule of Signs but couldn't solve it with that too.
Any help?
calculus algebra-precalculus polynomials quadratics
edited Jul 16 at 15:43
StubbornAtom
3,79311134
asked Jul 16 at 15:21
James Adams
535
3
Boring / unenlightening approach: There are, essentially, only $16$ such polynomials. One could just go through each of them, with the help of WolframAlpha, and check. Would only take a minute or two.
– Arthur
Jul 16 at 15:25
1
I like the approaches taken by OP. Note that taking a second derivative might close down the number of cases to consider substantially.
– hardmath
Jul 16 at 15:27
1
To be precise, there are $32$ such polynomials
– asdf
Jul 16 at 15:27
4
@asdf But $f$ and $-f$ have the same roots, so we may WLOG assume the polynomial is monic. That's what I meant by "essentially". In fact, $f(x)$ and $f(-x)$ are closely related too, so $8$ cases left to check.
– Arthur
Jul 16 at 15:28
True, I thought you aren't narrowing the cases in any way before bashing it out
– asdf
Jul 16 at 15:29
 |Â
show 4 more comments
up vote
10
down vote
favorite
up vote
10
down vote
favorite
So I was going through my 11th class package on Quadratic equations and I saw a question to prove that a polynomial of $4$th degree with all real roots cannot have $pm 1$ as all its coefficients.
I tried proving it using calculus, by showing that at least one consecutive maxima and minima will lie either above or below the x axis, but couldn't solve it using that.
I also tried using Descartes Rule of Signs but couldn't solve it with that too.
Any help?
calculus algebra-precalculus polynomials quadratics
edited Jul 16 at 15:43
StubbornAtom
3,79311134
asked Jul 16 at 15:21
James Adams
535
So I was going through my 11th class package on Quadratic equations and I saw a question to prove that a polynomial of $4$th degree with all real roots cannot have $pm 1$ as all its coefficients.
I tried proving it using calculus, by showing that at least one consecutive maxima and minima will lie either above or below the x axis, but couldn't solve it using that.
I also tried using Descartes Rule of Signs but couldn't solve it with that too.
Any help?
calculus algebra-precalculus polynomials quadratics
edited Jul 16 at 15:43
StubbornAtom
3,79311134
asked Jul 16 at 15:21
James Adams
535
edited Jul 16 at 15:43
StubbornAtom
3,79311134
edited Jul 16 at 15:43
StubbornAtom
3,79311134
edited Jul 16 at 15:43
StubbornAtom
3,79311134
3,79311134
asked Jul 16 at 15:21
James Adams
535
asked Jul 16 at 15:21
James Adams
535
asked Jul 16 at 15:21
James Adams
535
535
3
Boring / unenlightening approach: There are, essentially, only $16$ such polynomials. One could just go through each of them, with the help of WolframAlpha, and check. Would only take a minute or two.
– Arthur
Jul 16 at 15:25
1
I like the approaches taken by OP. Note that taking a second derivative might close down the number of cases to consider substantially.
– hardmath
Jul 16 at 15:27
1
To be precise, there are $32$ such polynomials
– asdf
Jul 16 at 15:27
4
@asdf But $f$ and $-f$ have the same roots, so we may WLOG assume the polynomial is monic. That's what I meant by "essentially". In fact, $f(x)$ and $f(-x)$ are closely related too, so $8$ cases left to check.
– Arthur
Jul 16 at 15:28
True, I thought you aren't narrowing the cases in any way before bashing it out
– asdf
Jul 16 at 15:29
 |Â
show 4 more comments
3
Boring / unenlightening approach: There are, essentially, only $16$ such polynomials. One could just go through each of them, with the help of WolframAlpha, and check. Would only take a minute or two.
– Arthur
Jul 16 at 15:25
1
I like the approaches taken by OP. Note that taking a second derivative might close down the number of cases to consider substantially.
– hardmath
Jul 16 at 15:27
1
To be precise, there are $32$ such polynomials
– asdf
Jul 16 at 15:27
4
@asdf But $f$ and $-f$ have the same roots, so we may WLOG assume the polynomial is monic. That's what I meant by "essentially". In fact, $f(x)$ and $f(-x)$ are closely related too, so $8$ cases left to check.
– Arthur
Jul 16 at 15:28
True, I thought you aren't narrowing the cases in any way before bashing it out
– asdf
Jul 16 at 15:29
3
3
Boring / unenlightening approach: There are, essentially, only $16$ such polynomials. One could just go through each of them, with the help of WolframAlpha, and check. Would only take a minute or two.
– Arthur
Jul 16 at 15:25
Boring / unenlightening approach: There are, essentially, only $16$ such polynomials. One could just go through each of them, with the help of WolframAlpha, and check. Would only take a minute or two.
– Arthur
Jul 16 at 15:25
1
1
I like the approaches taken by OP. Note that taking a second derivative might close down the number of cases to consider substantially.
– hardmath
Jul 16 at 15:27
I like the approaches taken by OP. Note that taking a second derivative might close down the number of cases to consider substantially.
– hardmath
Jul 16 at 15:27
1
1
To be precise, there are $32$ such polynomials
– asdf
Jul 16 at 15:27
To be precise, there are $32$ such polynomials
– asdf
Jul 16 at 15:27
4
4
@asdf But $f$ and $-f$ have the same roots, so we may WLOG assume the polynomial is monic. That's what I meant by "essentially". In fact, $f(x)$ and $f(-x)$ are closely related too, so $8$ cases left to check.
– Arthur
Jul 16 at 15:28
@asdf But $f$ and $-f$ have the same roots, so we may WLOG assume the polynomial is monic. That's what I meant by "essentially". In fact, $f(x)$ and $f(-x)$ are closely related too, so $8$ cases left to check.
– Arthur
Jul 16 at 15:28
True, I thought you aren't narrowing the cases in any way before bashing it out
– asdf
Jul 16 at 15:29
True, I thought you aren't narrowing the cases in any way before bashing it out
– asdf
Jul 16 at 15:29
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
up vote
9
down vote
accepted
Let $f(x)$ be any quartic polynomial with coefficients from $ -1, +1 $. Replacing $f(x)$ by $-f(x)$ if necessary, we can assume $f(x)$ is monic. i.e.
$$f(x) = x^4 + ax^3 + bx^2 + cx + dquadtext with quad a,b,c,d in -1, +1 $$
If $f(x)$ has $4$ real roots $lambda_1,lambda_2,lambda_3,lambda_4$, then by Vieta's formula, we have
$$sum_i=1^4 lambda_i = -a, sum_1le i < jle 4 lambda_ilambda_j = b
quadtext and quadprod_i=1^4 lambda_i = d$$
Notice
$$sum_i=1^4 lambda_i^2 = left(sum_i=1^4lambda_iright)^2 - 2sum_1le i < j le 4lambda_ilambda_j = a^2 - 2b = 1 -2b$$
Since $sum_i=1^4 lambda_i^2 ge 0$, we need $b = -1$. As a result, $$sum_i=1^4 lambda_i^2 = 3$$
By AM $ge$ GM, this leads to
$$frac34 = frac14sum_i=1^4 lambda_i^2 ge left(prod_i=1^4 lambda_i^2right)^1/4 = (d^2)^1/4 = 1$$
This is impossible and hence $f(x)$ cannot has 4 real roots.
answered Jul 16 at 15:48
achille hui
91k5127246
You are not using at any point that the sum of product of triples is $pm1$, so the claim can be generalised?
– asdf
Jul 16 at 15:52
@asdf yes, the proof doesn't need the value of $c$.
– achille hui
Jul 16 at 15:53
Nice, $(+1)$ from me
– asdf
Jul 16 at 15:53
Achille thanks for answering. However I'm an 11th grader and this looks out of my depth(this is in fact from my textbook). So can you explain to me what you have in a more basic manner? Thanks.Again if this cannot be explained to an 11th grader then I would still consider this question as still unanswered since it was designed to be solved by an 11th grader. Thanks a lot again mate.
– James Adams
Jul 16 at 15:56
1
@JamesAdams I don't remember how long I take, it is not that long. Since you are in 11 grade, I stop using anything that need calculus and looks for pure algebraic solution. I first look at tools like Newton inequalities or Maclaurin inequalities for polynomials for hints of an elementary solution (these tools are beyond 11 grade) but that didn't work. I switch to use Vieta's formula and attempt to bound the roots directly. That work and the rest is filling the details.
– achille hui
Jul 17 at 16:11
 |Â
show 5 more comments
up vote
5
down vote
It can be assumed WLOG that the leading coefficient is $,+1,$, so $,P(x)=x^4pm x^3pm x^2pm xpm 1,$.
Then $,P''(x)=12x^2 pm 6x pm 2,$, and for the quadratic to have real roots it is necessary that the constant term be negative, so $,P(x)=x^4pm x^3 - x^2pm xpm 1,$.
$P(x),$ has all the roots real iff $,x^4 Pleft(frac1xright),$ has all real roots. By the same argument as above, the constant term of $,P(x),$ must have opposite sign as the coefficient of $,x^2,$.
This leaves $4$ cases to check $,P(x)=x^4pm x^3 - x^2pm x+1,$.
[ EDIT ]
- $P(x),$ has all the roots real iff $,Pleft(-xright),$ has all real roots, so it is enough to consider the case where the coefficient of $,x^3,$ is $+1$.
This leaves $2$ cases to check $,P(x)=x^4+ x^3 - x^2pm x+1,$.
answered Jul 16 at 15:52
dxiv
54.3k64797
@RossMillikan On the other hand, it is enough to consider one of the cases $,pm x^3,$, as noted in the latest edit.
– dxiv
Jul 16 at 16:08
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Let $f(x)$ be any quartic polynomial with coefficients from $ -1, +1 $. Replacing $f(x)$ by $-f(x)$ if necessary, we can assume $f(x)$ is monic. i.e.
$$f(x) = x^4 + ax^3 + bx^2 + cx + dquadtext with quad a,b,c,d in -1, +1 $$
If $f(x)$ has $4$ real roots $lambda_1,lambda_2,lambda_3,lambda_4$, then by Vieta's formula, we have
$$sum_i=1^4 lambda_i = -a, sum_1le i < jle 4 lambda_ilambda_j = b
quadtext and quadprod_i=1^4 lambda_i = d$$
Notice
$$sum_i=1^4 lambda_i^2 = left(sum_i=1^4lambda_iright)^2 - 2sum_1le i < j le 4lambda_ilambda_j = a^2 - 2b = 1 -2b$$
Since $sum_i=1^4 lambda_i^2 ge 0$, we need $b = -1$. As a result, $$sum_i=1^4 lambda_i^2 = 3$$
By AM $ge$ GM, this leads to
$$frac34 = frac14sum_i=1^4 lambda_i^2 ge left(prod_i=1^4 lambda_i^2right)^1/4 = (d^2)^1/4 = 1$$
This is impossible and hence $f(x)$ cannot has 4 real roots.
answered Jul 16 at 15:48
achille hui
91k5127246
You are not using at any point that the sum of product of triples is $pm1$, so the claim can be generalised?
– asdf
Jul 16 at 15:52
@asdf yes, the proof doesn't need the value of $c$.
– achille hui
Jul 16 at 15:53
Nice, $(+1)$ from me
– asdf
Jul 16 at 15:53
Achille thanks for answering. However I'm an 11th grader and this looks out of my depth(this is in fact from my textbook). So can you explain to me what you have in a more basic manner? Thanks.Again if this cannot be explained to an 11th grader then I would still consider this question as still unanswered since it was designed to be solved by an 11th grader. Thanks a lot again mate.
– James Adams
Jul 16 at 15:56
1
@JamesAdams I don't remember how long I take, it is not that long. Since you are in 11 grade, I stop using anything that need calculus and looks for pure algebraic solution. I first look at tools like Newton inequalities or Maclaurin inequalities for polynomials for hints of an elementary solution (these tools are beyond 11 grade) but that didn't work. I switch to use Vieta's formula and attempt to bound the roots directly. That work and the rest is filling the details.
– achille hui
Jul 17 at 16:11
 |Â
show 5 more comments
up vote
9
down vote
accepted
Let $f(x)$ be any quartic polynomial with coefficients from $ -1, +1 $. Replacing $f(x)$ by $-f(x)$ if necessary, we can assume $f(x)$ is monic. i.e.
$$f(x) = x^4 + ax^3 + bx^2 + cx + dquadtext with quad a,b,c,d in -1, +1 $$
If $f(x)$ has $4$ real roots $lambda_1,lambda_2,lambda_3,lambda_4$, then by Vieta's formula, we have
$$sum_i=1^4 lambda_i = -a, sum_1le i < jle 4 lambda_ilambda_j = b
quadtext and quadprod_i=1^4 lambda_i = d$$
Notice
$$sum_i=1^4 lambda_i^2 = left(sum_i=1^4lambda_iright)^2 - 2sum_1le i < j le 4lambda_ilambda_j = a^2 - 2b = 1 -2b$$
Since $sum_i=1^4 lambda_i^2 ge 0$, we need $b = -1$. As a result, $$sum_i=1^4 lambda_i^2 = 3$$
By AM $ge$ GM, this leads to
$$frac34 = frac14sum_i=1^4 lambda_i^2 ge left(prod_i=1^4 lambda_i^2right)^1/4 = (d^2)^1/4 = 1$$
This is impossible and hence $f(x)$ cannot has 4 real roots.
answered Jul 16 at 15:48
achille hui
91k5127246
You are not using at any point that the sum of product of triples is $pm1$, so the claim can be generalised?
– asdf
Jul 16 at 15:52
@asdf yes, the proof doesn't need the value of $c$.
– achille hui
Jul 16 at 15:53
Nice, $(+1)$ from me
– asdf
Jul 16 at 15:53
Achille thanks for answering. However I'm an 11th grader and this looks out of my depth(this is in fact from my textbook). So can you explain to me what you have in a more basic manner? Thanks.Again if this cannot be explained to an 11th grader then I would still consider this question as still unanswered since it was designed to be solved by an 11th grader. Thanks a lot again mate.
– James Adams
Jul 16 at 15:56
1
@JamesAdams I don't remember how long I take, it is not that long. Since you are in 11 grade, I stop using anything that need calculus and looks for pure algebraic solution. I first look at tools like Newton inequalities or Maclaurin inequalities for polynomials for hints of an elementary solution (these tools are beyond 11 grade) but that didn't work. I switch to use Vieta's formula and attempt to bound the roots directly. That work and the rest is filling the details.
– achille hui
Jul 17 at 16:11
 |Â
show 5 more comments
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Let $f(x)$ be any quartic polynomial with coefficients from $ -1, +1 $. Replacing $f(x)$ by $-f(x)$ if necessary, we can assume $f(x)$ is monic. i.e.
$$f(x) = x^4 + ax^3 + bx^2 + cx + dquadtext with quad a,b,c,d in -1, +1 $$
If $f(x)$ has $4$ real roots $lambda_1,lambda_2,lambda_3,lambda_4$, then by Vieta's formula, we have
$$sum_i=1^4 lambda_i = -a, sum_1le i < jle 4 lambda_ilambda_j = b
quadtext and quadprod_i=1^4 lambda_i = d$$
Notice
$$sum_i=1^4 lambda_i^2 = left(sum_i=1^4lambda_iright)^2 - 2sum_1le i < j le 4lambda_ilambda_j = a^2 - 2b = 1 -2b$$
Since $sum_i=1^4 lambda_i^2 ge 0$, we need $b = -1$. As a result, $$sum_i=1^4 lambda_i^2 = 3$$
By AM $ge$ GM, this leads to
$$frac34 = frac14sum_i=1^4 lambda_i^2 ge left(prod_i=1^4 lambda_i^2right)^1/4 = (d^2)^1/4 = 1$$
This is impossible and hence $f(x)$ cannot has 4 real roots.
answered Jul 16 at 15:48
achille hui
91k5127246
Let $f(x)$ be any quartic polynomial with coefficients from $ -1, +1 $. Replacing $f(x)$ by $-f(x)$ if necessary, we can assume $f(x)$ is monic. i.e.
$$f(x) = x^4 + ax^3 + bx^2 + cx + dquadtext with quad a,b,c,d in -1, +1 $$
If $f(x)$ has $4$ real roots $lambda_1,lambda_2,lambda_3,lambda_4$, then by Vieta's formula, we have
$$sum_i=1^4 lambda_i = -a, sum_1le i < jle 4 lambda_ilambda_j = b
quadtext and quadprod_i=1^4 lambda_i = d$$
Notice
$$sum_i=1^4 lambda_i^2 = left(sum_i=1^4lambda_iright)^2 - 2sum_1le i < j le 4lambda_ilambda_j = a^2 - 2b = 1 -2b$$
Since $sum_i=1^4 lambda_i^2 ge 0$, we need $b = -1$. As a result, $$sum_i=1^4 lambda_i^2 = 3$$
By AM $ge$ GM, this leads to
$$frac34 = frac14sum_i=1^4 lambda_i^2 ge left(prod_i=1^4 lambda_i^2right)^1/4 = (d^2)^1/4 = 1$$
This is impossible and hence $f(x)$ cannot has 4 real roots.
answered Jul 16 at 15:48
achille hui
91k5127246
edited Jul 16 at 15:59
answered Jul 16 at 15:48
achille hui
91k5127246
answered Jul 16 at 15:48
achille hui
91k5127246
answered Jul 16 at 15:48
achille hui
91k5127246
91k5127246
You are not using at any point that the sum of product of triples is $pm1$, so the claim can be generalised?
– asdf
Jul 16 at 15:52
@asdf yes, the proof doesn't need the value of $c$.
– achille hui
Jul 16 at 15:53
Nice, $(+1)$ from me
– asdf
Jul 16 at 15:53
Achille thanks for answering. However I'm an 11th grader and this looks out of my depth(this is in fact from my textbook). So can you explain to me what you have in a more basic manner? Thanks.Again if this cannot be explained to an 11th grader then I would still consider this question as still unanswered since it was designed to be solved by an 11th grader. Thanks a lot again mate.
– James Adams
Jul 16 at 15:56
1
@JamesAdams I don't remember how long I take, it is not that long. Since you are in 11 grade, I stop using anything that need calculus and looks for pure algebraic solution. I first look at tools like Newton inequalities or Maclaurin inequalities for polynomials for hints of an elementary solution (these tools are beyond 11 grade) but that didn't work. I switch to use Vieta's formula and attempt to bound the roots directly. That work and the rest is filling the details.
– achille hui
Jul 17 at 16:11
 |Â
show 5 more comments
You are not using at any point that the sum of product of triples is $pm1$, so the claim can be generalised?
– asdf
Jul 16 at 15:52
@asdf yes, the proof doesn't need the value of $c$.
– achille hui
Jul 16 at 15:53
Nice, $(+1)$ from me
– asdf
Jul 16 at 15:53
Achille thanks for answering. However I'm an 11th grader and this looks out of my depth(this is in fact from my textbook). So can you explain to me what you have in a more basic manner? Thanks.Again if this cannot be explained to an 11th grader then I would still consider this question as still unanswered since it was designed to be solved by an 11th grader. Thanks a lot again mate.
– James Adams
Jul 16 at 15:56
1
@JamesAdams I don't remember how long I take, it is not that long. Since you are in 11 grade, I stop using anything that need calculus and looks for pure algebraic solution. I first look at tools like Newton inequalities or Maclaurin inequalities for polynomials for hints of an elementary solution (these tools are beyond 11 grade) but that didn't work. I switch to use Vieta's formula and attempt to bound the roots directly. That work and the rest is filling the details.
– achille hui
Jul 17 at 16:11
You are not using at any point that the sum of product of triples is $pm1$, so the claim can be generalised?
– asdf
Jul 16 at 15:52
You are not using at any point that the sum of product of triples is $pm1$, so the claim can be generalised?
– asdf
Jul 16 at 15:52
@asdf yes, the proof doesn't need the value of $c$.
– achille hui
Jul 16 at 15:53
@asdf yes, the proof doesn't need the value of $c$.
– achille hui
Jul 16 at 15:53
Nice, $(+1)$ from me
– asdf
Jul 16 at 15:53
Nice, $(+1)$ from me
– asdf
Jul 16 at 15:53
Achille thanks for answering. However I'm an 11th grader and this looks out of my depth(this is in fact from my textbook). So can you explain to me what you have in a more basic manner? Thanks.Again if this cannot be explained to an 11th grader then I would still consider this question as still unanswered since it was designed to be solved by an 11th grader. Thanks a lot again mate.
– James Adams
Jul 16 at 15:56
Achille thanks for answering. However I'm an 11th grader and this looks out of my depth(this is in fact from my textbook). So can you explain to me what you have in a more basic manner? Thanks.Again if this cannot be explained to an 11th grader then I would still consider this question as still unanswered since it was designed to be solved by an 11th grader. Thanks a lot again mate.
– James Adams
Jul 16 at 15:56
1
1
@JamesAdams I don't remember how long I take, it is not that long. Since you are in 11 grade, I stop using anything that need calculus and looks for pure algebraic solution. I first look at tools like Newton inequalities or Maclaurin inequalities for polynomials for hints of an elementary solution (these tools are beyond 11 grade) but that didn't work. I switch to use Vieta's formula and attempt to bound the roots directly. That work and the rest is filling the details.
– achille hui
Jul 17 at 16:11
@JamesAdams I don't remember how long I take, it is not that long. Since you are in 11 grade, I stop using anything that need calculus and looks for pure algebraic solution. I first look at tools like Newton inequalities or Maclaurin inequalities for polynomials for hints of an elementary solution (these tools are beyond 11 grade) but that didn't work. I switch to use Vieta's formula and attempt to bound the roots directly. That work and the rest is filling the details.
– achille hui
Jul 17 at 16:11
 |Â
show 5 more comments
up vote
5
down vote
It can be assumed WLOG that the leading coefficient is $,+1,$, so $,P(x)=x^4pm x^3pm x^2pm xpm 1,$.
Then $,P''(x)=12x^2 pm 6x pm 2,$, and for the quadratic to have real roots it is necessary that the constant term be negative, so $,P(x)=x^4pm x^3 - x^2pm xpm 1,$.
$P(x),$ has all the roots real iff $,x^4 Pleft(frac1xright),$ has all real roots. By the same argument as above, the constant term of $,P(x),$ must have opposite sign as the coefficient of $,x^2,$.
This leaves $4$ cases to check $,P(x)=x^4pm x^3 - x^2pm x+1,$.
[ EDIT ]
- $P(x),$ has all the roots real iff $,Pleft(-xright),$ has all real roots, so it is enough to consider the case where the coefficient of $,x^3,$ is $+1$.
This leaves $2$ cases to check $,P(x)=x^4+ x^3 - x^2pm x+1,$.
answered Jul 16 at 15:52
dxiv
54.3k64797
@RossMillikan On the other hand, it is enough to consider one of the cases $,pm x^3,$, as noted in the latest edit.
– dxiv
Jul 16 at 16:08
add a comment |Â
up vote
5
down vote
It can be assumed WLOG that the leading coefficient is $,+1,$, so $,P(x)=x^4pm x^3pm x^2pm xpm 1,$.
Then $,P''(x)=12x^2 pm 6x pm 2,$, and for the quadratic to have real roots it is necessary that the constant term be negative, so $,P(x)=x^4pm x^3 - x^2pm xpm 1,$.
$P(x),$ has all the roots real iff $,x^4 Pleft(frac1xright),$ has all real roots. By the same argument as above, the constant term of $,P(x),$ must have opposite sign as the coefficient of $,x^2,$.
This leaves $4$ cases to check $,P(x)=x^4pm x^3 - x^2pm x+1,$.
[ EDIT ]
- $P(x),$ has all the roots real iff $,Pleft(-xright),$ has all real roots, so it is enough to consider the case where the coefficient of $,x^3,$ is $+1$.
This leaves $2$ cases to check $,P(x)=x^4+ x^3 - x^2pm x+1,$.
answered Jul 16 at 15:52
dxiv
54.3k64797
@RossMillikan On the other hand, it is enough to consider one of the cases $,pm x^3,$, as noted in the latest edit.
– dxiv
Jul 16 at 16:08
add a comment |Â
up vote
5
down vote
up vote
5
down vote
It can be assumed WLOG that the leading coefficient is $,+1,$, so $,P(x)=x^4pm x^3pm x^2pm xpm 1,$.
Then $,P''(x)=12x^2 pm 6x pm 2,$, and for the quadratic to have real roots it is necessary that the constant term be negative, so $,P(x)=x^4pm x^3 - x^2pm xpm 1,$.
$P(x),$ has all the roots real iff $,x^4 Pleft(frac1xright),$ has all real roots. By the same argument as above, the constant term of $,P(x),$ must have opposite sign as the coefficient of $,x^2,$.
This leaves $4$ cases to check $,P(x)=x^4pm x^3 - x^2pm x+1,$.
[ EDIT ]
- $P(x),$ has all the roots real iff $,Pleft(-xright),$ has all real roots, so it is enough to consider the case where the coefficient of $,x^3,$ is $+1$.
This leaves $2$ cases to check $,P(x)=x^4+ x^3 - x^2pm x+1,$.
answered Jul 16 at 15:52
dxiv
54.3k64797
It can be assumed WLOG that the leading coefficient is $,+1,$, so $,P(x)=x^4pm x^3pm x^2pm xpm 1,$.
Then $,P''(x)=12x^2 pm 6x pm 2,$, and for the quadratic to have real roots it is necessary that the constant term be negative, so $,P(x)=x^4pm x^3 - x^2pm xpm 1,$.
$P(x),$ has all the roots real iff $,x^4 Pleft(frac1xright),$ has all real roots. By the same argument as above, the constant term of $,P(x),$ must have opposite sign as the coefficient of $,x^2,$.
This leaves $4$ cases to check $,P(x)=x^4pm x^3 - x^2pm x+1,$.
[ EDIT ]
- $P(x),$ has all the roots real iff $,Pleft(-xright),$ has all real roots, so it is enough to consider the case where the coefficient of $,x^3,$ is $+1$.
This leaves $2$ cases to check $,P(x)=x^4+ x^3 - x^2pm x+1,$.
answered Jul 16 at 15:52
dxiv
54.3k64797
edited Jul 16 at 16:06
answered Jul 16 at 15:52
dxiv
54.3k64797
answered Jul 16 at 15:52
dxiv
54.3k64797
answered Jul 16 at 15:52
dxiv
54.3k64797
54.3k64797
@RossMillikan On the other hand, it is enough to consider one of the cases $,pm x^3,$, as noted in the latest edit.
– dxiv
Jul 16 at 16:08
add a comment |Â
@RossMillikan On the other hand, it is enough to consider one of the cases $,pm x^3,$, as noted in the latest edit.
– dxiv
Jul 16 at 16:08
@RossMillikan On the other hand, it is enough to consider one of the cases $,pm x^3,$, as noted in the latest edit.
– dxiv
Jul 16 at 16:08
@RossMillikan On the other hand, it is enough to consider one of the cases $,pm x^3,$, as noted in the latest edit.
– dxiv
Jul 16 at 16:08
add a comment |Â
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3
Boring / unenlightening approach: There are, essentially, only $16$ such polynomials. One could just go through each of them, with the help of WolframAlpha, and check. Would only take a minute or two.
– Arthur
Jul 16 at 15:25
1
I like the approaches taken by OP. Note that taking a second derivative might close down the number of cases to consider substantially.
– hardmath
Jul 16 at 15:27
1
To be precise, there are $32$ such polynomials
– asdf
Jul 16 at 15:27
4
@asdf But $f$ and $-f$ have the same roots, so we may WLOG assume the polynomial is monic. That's what I meant by "essentially". In fact, $f(x)$ and $f(-x)$ are closely related too, so $8$ cases left to check.
– Arthur
Jul 16 at 15:28
True, I thought you aren't narrowing the cases in any way before bashing it out
– asdf
Jul 16 at 15:29