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Two Questions On The Rational Root Theorem
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I have 2 questions on the rational root theorem:
Can the rational root theorem be used to find all the real roots of any polynomial, assuming it has whole number coefficients? Or are there situations where there are real roots of a polynomial with whole number coefficient where some of it's real roots cannot be found through the rational root theorem? I'm asking this because my textbook says that the rational root theorem can be used to find all the real roots of any polynomial.
In the same textbook however, there was a question requiring you to find the real roots of $x^4+3x^2+2=0$. One of the real roots was $pmsqrt2$. How does one find this using the rational root theorem?
Please try to keep the answers at the level of a high school Pre-calc student. Thanks.
algebra-precalculus polynomials
edited Aug 2 at 13:28
zzuussee
1,142419
asked Aug 2 at 13:25
Ethan Chan
591322
 |Â
show 1 more comment
up vote
1
down vote
favorite
I have 2 questions on the rational root theorem:
Can the rational root theorem be used to find all the real roots of any polynomial, assuming it has whole number coefficients? Or are there situations where there are real roots of a polynomial with whole number coefficient where some of it's real roots cannot be found through the rational root theorem? I'm asking this because my textbook says that the rational root theorem can be used to find all the real roots of any polynomial.
In the same textbook however, there was a question requiring you to find the real roots of $x^4+3x^2+2=0$. One of the real roots was $pmsqrt2$. How does one find this using the rational root theorem?
Please try to keep the answers at the level of a high school Pre-calc student. Thanks.
algebra-precalculus polynomials
edited Aug 2 at 13:28
zzuussee
1,142419
asked Aug 2 at 13:25
Ethan Chan
591322
You have 500+ rep. There's no reason why you shouldn't use MathJax: math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Aug 2 at 13:29
2
The rational root theorem is of no use for the irrational roots.
– lulu
Aug 2 at 13:30
@lulu So it can find all real roots, except roots that are irrational?
– Ethan Chan
Aug 2 at 13:32
1
@EthanChan that’s why it’s called the rational root theorem :)
– Oiler
Aug 2 at 13:32
1
Yes, it can find all the rational roots.
– lulu
Aug 2 at 13:39
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have 2 questions on the rational root theorem:
Can the rational root theorem be used to find all the real roots of any polynomial, assuming it has whole number coefficients? Or are there situations where there are real roots of a polynomial with whole number coefficient where some of it's real roots cannot be found through the rational root theorem? I'm asking this because my textbook says that the rational root theorem can be used to find all the real roots of any polynomial.
In the same textbook however, there was a question requiring you to find the real roots of $x^4+3x^2+2=0$. One of the real roots was $pmsqrt2$. How does one find this using the rational root theorem?
Please try to keep the answers at the level of a high school Pre-calc student. Thanks.
algebra-precalculus polynomials
edited Aug 2 at 13:28
zzuussee
1,142419
asked Aug 2 at 13:25
Ethan Chan
591322
I have 2 questions on the rational root theorem:
Can the rational root theorem be used to find all the real roots of any polynomial, assuming it has whole number coefficients? Or are there situations where there are real roots of a polynomial with whole number coefficient where some of it's real roots cannot be found through the rational root theorem? I'm asking this because my textbook says that the rational root theorem can be used to find all the real roots of any polynomial.
In the same textbook however, there was a question requiring you to find the real roots of $x^4+3x^2+2=0$. One of the real roots was $pmsqrt2$. How does one find this using the rational root theorem?
Please try to keep the answers at the level of a high school Pre-calc student. Thanks.
algebra-precalculus polynomials
edited Aug 2 at 13:28
zzuussee
1,142419
asked Aug 2 at 13:25
Ethan Chan
591322
edited Aug 2 at 13:28
zzuussee
1,142419
edited Aug 2 at 13:28
zzuussee
1,142419
edited Aug 2 at 13:28
zzuussee
1,142419
1,142419
asked Aug 2 at 13:25
Ethan Chan
591322
asked Aug 2 at 13:25
Ethan Chan
591322
asked Aug 2 at 13:25
Ethan Chan
591322
591322
You have 500+ rep. There's no reason why you shouldn't use MathJax: math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Aug 2 at 13:29
2
The rational root theorem is of no use for the irrational roots.
– lulu
Aug 2 at 13:30
@lulu So it can find all real roots, except roots that are irrational?
– Ethan Chan
Aug 2 at 13:32
1
@EthanChan that’s why it’s called the rational root theorem :)
– Oiler
Aug 2 at 13:32
1
Yes, it can find all the rational roots.
– lulu
Aug 2 at 13:39
 |Â
show 1 more comment
You have 500+ rep. There's no reason why you shouldn't use MathJax: math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Aug 2 at 13:29
2
The rational root theorem is of no use for the irrational roots.
– lulu
Aug 2 at 13:30
@lulu So it can find all real roots, except roots that are irrational?
– Ethan Chan
Aug 2 at 13:32
1
@EthanChan that’s why it’s called the rational root theorem :)
– Oiler
Aug 2 at 13:32
1
Yes, it can find all the rational roots.
– lulu
Aug 2 at 13:39
You have 500+ rep. There's no reason why you shouldn't use MathJax: math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Aug 2 at 13:29
You have 500+ rep. There's no reason why you shouldn't use MathJax: math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Aug 2 at 13:29
2
2
The rational root theorem is of no use for the irrational roots.
– lulu
Aug 2 at 13:30
The rational root theorem is of no use for the irrational roots.
– lulu
Aug 2 at 13:30
@lulu So it can find all real roots, except roots that are irrational?
– Ethan Chan
Aug 2 at 13:32
@lulu So it can find all real roots, except roots that are irrational?
– Ethan Chan
Aug 2 at 13:32
1
1
@EthanChan that’s why it’s called the rational root theorem :)
– Oiler
Aug 2 at 13:32
@EthanChan that’s why it’s called the rational root theorem :)
– Oiler
Aug 2 at 13:32
1
1
Yes, it can find all the rational roots.
– lulu
Aug 2 at 13:39
Yes, it can find all the rational roots.
– lulu
Aug 2 at 13:39
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Decided to throw an answer in because I thought that it was too long for a comment.
Can the Rational Root Theorem be used to find all the real roots of a polynomial (with integer coefficients)?
No (at least not directly). The Rational Root Theorem asserts that if $fracab$ is a root of $f$, where $a$ and $b$ are integers with $b neq 0$ and $gcd(a,b) = 1$, then $a$ divides the constant term of $f(x)$ and $b$ divides the lead coefficient of $f(x)$. Note that the Rational Root Theorem in no way guarantees the existence of rational roots. It does, however, give information about the nature of rational roots should they exist.
There are some instances where the Rational Root Theorem is sufficient to find all the real roots of a polynomial. For example, consider the polynomial $f(x) = x^4 - x^3 - 7x^2 + x + 6$. The Rational Root Theorem tells us that if $fracab$ is a root of $f(x)$, then $a$ divides 6 and $b$ divides $1$. Since the divisors of 6 are $pm 1 , pm 2 , pm 3 , pm 6$ and the divisors of $1$ are $pm 1$, the potential roots of $f(x)$ are $pm 1 , pm 2 , pm 3 , pm 6$. One checks that $f(pm1) = f(-2) = f(3) = 0$, which implies that we have indeed found all of the roots because the number of roots is less than or equal to the degree.
In the previous example, all of the roots of $f(x)$ were rational, which is why the Rational Root Theorem was able to show us the way in finding all of them. There are some instances where the Rational Root Theorem will not elucidate all of the roots, but it will give you enough to determine all of the real roots. Consider the polynomial $f(x) = x^5 - 2x^4 - 8x^3 + 16x^2 + 7x -14$. The Rational Root Theorem tells us that the possible roots of $f(x)$ are $pm 1 , pm 2, pm 7 , pm 14$. One checks that the only rational roots of $f(x)$ are $x = pm 1, 2$. However this of course implies that $(x - 2)(x - 1)(x + 1)$ is a factor of $f(x)$. You can then perform polynomial long division and show that $$f(x) = (x - 1)(x + 1)(x - 2)(x^2 - 7),$$
at which point it becomes apparent that the two remaining roots of $f(x)$ are $x = pm sqrt7$.
answered Aug 3 at 2:24
Oiler
1,8551720
add a comment |Â
up vote
2
down vote
Recall that Rational root theorem guarantees that each rational solution $x$ must be in the form $x = fracpq$ with
- p integer factor of the constant term $a_0$
- q integer factor of the leading coefficient $a_n$
and nothing more than this.
With reference to your example by $t=x^2$
$$x^4+3x^2+2=0 implies t^2+3t+2=0$$
by rational root theorem we can find roots $t=-1$ and $t=-2$ and then
$$x^4+3x^2+2=(x^2+1)(x^2+2)=(x+i)(x-i)(x+isqrt 2)(x-isqrt 2)$$
and of course $sqrt 2$ (as any real number) can't be a solution since $x^4+3x^2+2ge 2$.
answered Aug 2 at 13:31
gimusi
63.8k73480
add a comment |Â
up vote
0
down vote
I would guess that this is a typo of your textbook. The rational root test suffices, for a polynomial with integer coefficients, to find all rational roots, as you may list
- all the integer divisors $q$ of the leading coefficient $a_n$
- all the integer divisors $p$ of the trailing coefficient $a_0$
You can then form all combinations $fracpq$ and the rational root theorem guarantees you that, if there are rational roots, then they are of this form. Of course, you still have to check which combinations really apply to be a root.
Note that rational numbers are most certainly real numbers, but of course, this procedure can not speak about the irrational roots at all, as they can not be expressed as a fraction $fracpq$.
I have not checked you polynomial, but another good example could be $x^2-x-1$ with roots $frac1+sqrt52$ and $frac1-sqrt52$, i.e. the golden ration and its conjugate which are irrational as well.
However, if you find a rational root(as the rational root test does not guarantee that there are any), you may use it to reduce your problem/polynomial via polynomial division, which may help in finding the remaining roots.
answered Aug 2 at 13:33
zzuussee
1,142419
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Decided to throw an answer in because I thought that it was too long for a comment.
Can the Rational Root Theorem be used to find all the real roots of a polynomial (with integer coefficients)?
No (at least not directly). The Rational Root Theorem asserts that if $fracab$ is a root of $f$, where $a$ and $b$ are integers with $b neq 0$ and $gcd(a,b) = 1$, then $a$ divides the constant term of $f(x)$ and $b$ divides the lead coefficient of $f(x)$. Note that the Rational Root Theorem in no way guarantees the existence of rational roots. It does, however, give information about the nature of rational roots should they exist.
There are some instances where the Rational Root Theorem is sufficient to find all the real roots of a polynomial. For example, consider the polynomial $f(x) = x^4 - x^3 - 7x^2 + x + 6$. The Rational Root Theorem tells us that if $fracab$ is a root of $f(x)$, then $a$ divides 6 and $b$ divides $1$. Since the divisors of 6 are $pm 1 , pm 2 , pm 3 , pm 6$ and the divisors of $1$ are $pm 1$, the potential roots of $f(x)$ are $pm 1 , pm 2 , pm 3 , pm 6$. One checks that $f(pm1) = f(-2) = f(3) = 0$, which implies that we have indeed found all of the roots because the number of roots is less than or equal to the degree.
In the previous example, all of the roots of $f(x)$ were rational, which is why the Rational Root Theorem was able to show us the way in finding all of them. There are some instances where the Rational Root Theorem will not elucidate all of the roots, but it will give you enough to determine all of the real roots. Consider the polynomial $f(x) = x^5 - 2x^4 - 8x^3 + 16x^2 + 7x -14$. The Rational Root Theorem tells us that the possible roots of $f(x)$ are $pm 1 , pm 2, pm 7 , pm 14$. One checks that the only rational roots of $f(x)$ are $x = pm 1, 2$. However this of course implies that $(x - 2)(x - 1)(x + 1)$ is a factor of $f(x)$. You can then perform polynomial long division and show that $$f(x) = (x - 1)(x + 1)(x - 2)(x^2 - 7),$$
at which point it becomes apparent that the two remaining roots of $f(x)$ are $x = pm sqrt7$.
answered Aug 3 at 2:24
Oiler
1,8551720
add a comment |Â
up vote
1
down vote
accepted
Decided to throw an answer in because I thought that it was too long for a comment.
Can the Rational Root Theorem be used to find all the real roots of a polynomial (with integer coefficients)?
No (at least not directly). The Rational Root Theorem asserts that if $fracab$ is a root of $f$, where $a$ and $b$ are integers with $b neq 0$ and $gcd(a,b) = 1$, then $a$ divides the constant term of $f(x)$ and $b$ divides the lead coefficient of $f(x)$. Note that the Rational Root Theorem in no way guarantees the existence of rational roots. It does, however, give information about the nature of rational roots should they exist.
There are some instances where the Rational Root Theorem is sufficient to find all the real roots of a polynomial. For example, consider the polynomial $f(x) = x^4 - x^3 - 7x^2 + x + 6$. The Rational Root Theorem tells us that if $fracab$ is a root of $f(x)$, then $a$ divides 6 and $b$ divides $1$. Since the divisors of 6 are $pm 1 , pm 2 , pm 3 , pm 6$ and the divisors of $1$ are $pm 1$, the potential roots of $f(x)$ are $pm 1 , pm 2 , pm 3 , pm 6$. One checks that $f(pm1) = f(-2) = f(3) = 0$, which implies that we have indeed found all of the roots because the number of roots is less than or equal to the degree.
In the previous example, all of the roots of $f(x)$ were rational, which is why the Rational Root Theorem was able to show us the way in finding all of them. There are some instances where the Rational Root Theorem will not elucidate all of the roots, but it will give you enough to determine all of the real roots. Consider the polynomial $f(x) = x^5 - 2x^4 - 8x^3 + 16x^2 + 7x -14$. The Rational Root Theorem tells us that the possible roots of $f(x)$ are $pm 1 , pm 2, pm 7 , pm 14$. One checks that the only rational roots of $f(x)$ are $x = pm 1, 2$. However this of course implies that $(x - 2)(x - 1)(x + 1)$ is a factor of $f(x)$. You can then perform polynomial long division and show that $$f(x) = (x - 1)(x + 1)(x - 2)(x^2 - 7),$$
at which point it becomes apparent that the two remaining roots of $f(x)$ are $x = pm sqrt7$.
answered Aug 3 at 2:24
Oiler
1,8551720
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Decided to throw an answer in because I thought that it was too long for a comment.
Can the Rational Root Theorem be used to find all the real roots of a polynomial (with integer coefficients)?
No (at least not directly). The Rational Root Theorem asserts that if $fracab$ is a root of $f$, where $a$ and $b$ are integers with $b neq 0$ and $gcd(a,b) = 1$, then $a$ divides the constant term of $f(x)$ and $b$ divides the lead coefficient of $f(x)$. Note that the Rational Root Theorem in no way guarantees the existence of rational roots. It does, however, give information about the nature of rational roots should they exist.
There are some instances where the Rational Root Theorem is sufficient to find all the real roots of a polynomial. For example, consider the polynomial $f(x) = x^4 - x^3 - 7x^2 + x + 6$. The Rational Root Theorem tells us that if $fracab$ is a root of $f(x)$, then $a$ divides 6 and $b$ divides $1$. Since the divisors of 6 are $pm 1 , pm 2 , pm 3 , pm 6$ and the divisors of $1$ are $pm 1$, the potential roots of $f(x)$ are $pm 1 , pm 2 , pm 3 , pm 6$. One checks that $f(pm1) = f(-2) = f(3) = 0$, which implies that we have indeed found all of the roots because the number of roots is less than or equal to the degree.
In the previous example, all of the roots of $f(x)$ were rational, which is why the Rational Root Theorem was able to show us the way in finding all of them. There are some instances where the Rational Root Theorem will not elucidate all of the roots, but it will give you enough to determine all of the real roots. Consider the polynomial $f(x) = x^5 - 2x^4 - 8x^3 + 16x^2 + 7x -14$. The Rational Root Theorem tells us that the possible roots of $f(x)$ are $pm 1 , pm 2, pm 7 , pm 14$. One checks that the only rational roots of $f(x)$ are $x = pm 1, 2$. However this of course implies that $(x - 2)(x - 1)(x + 1)$ is a factor of $f(x)$. You can then perform polynomial long division and show that $$f(x) = (x - 1)(x + 1)(x - 2)(x^2 - 7),$$
at which point it becomes apparent that the two remaining roots of $f(x)$ are $x = pm sqrt7$.
answered Aug 3 at 2:24
Oiler
1,8551720
Decided to throw an answer in because I thought that it was too long for a comment.
Can the Rational Root Theorem be used to find all the real roots of a polynomial (with integer coefficients)?
No (at least not directly). The Rational Root Theorem asserts that if $fracab$ is a root of $f$, where $a$ and $b$ are integers with $b neq 0$ and $gcd(a,b) = 1$, then $a$ divides the constant term of $f(x)$ and $b$ divides the lead coefficient of $f(x)$. Note that the Rational Root Theorem in no way guarantees the existence of rational roots. It does, however, give information about the nature of rational roots should they exist.
There are some instances where the Rational Root Theorem is sufficient to find all the real roots of a polynomial. For example, consider the polynomial $f(x) = x^4 - x^3 - 7x^2 + x + 6$. The Rational Root Theorem tells us that if $fracab$ is a root of $f(x)$, then $a$ divides 6 and $b$ divides $1$. Since the divisors of 6 are $pm 1 , pm 2 , pm 3 , pm 6$ and the divisors of $1$ are $pm 1$, the potential roots of $f(x)$ are $pm 1 , pm 2 , pm 3 , pm 6$. One checks that $f(pm1) = f(-2) = f(3) = 0$, which implies that we have indeed found all of the roots because the number of roots is less than or equal to the degree.
In the previous example, all of the roots of $f(x)$ were rational, which is why the Rational Root Theorem was able to show us the way in finding all of them. There are some instances where the Rational Root Theorem will not elucidate all of the roots, but it will give you enough to determine all of the real roots. Consider the polynomial $f(x) = x^5 - 2x^4 - 8x^3 + 16x^2 + 7x -14$. The Rational Root Theorem tells us that the possible roots of $f(x)$ are $pm 1 , pm 2, pm 7 , pm 14$. One checks that the only rational roots of $f(x)$ are $x = pm 1, 2$. However this of course implies that $(x - 2)(x - 1)(x + 1)$ is a factor of $f(x)$. You can then perform polynomial long division and show that $$f(x) = (x - 1)(x + 1)(x - 2)(x^2 - 7),$$
at which point it becomes apparent that the two remaining roots of $f(x)$ are $x = pm sqrt7$.
answered Aug 3 at 2:24
Oiler
1,8551720
edited Aug 3 at 2:30
answered Aug 3 at 2:24
Oiler
1,8551720
answered Aug 3 at 2:24
Oiler
1,8551720
answered Aug 3 at 2:24
Oiler
1,8551720
1,8551720
add a comment |Â
add a comment |Â
up vote
2
down vote
Recall that Rational root theorem guarantees that each rational solution $x$ must be in the form $x = fracpq$ with
- p integer factor of the constant term $a_0$
- q integer factor of the leading coefficient $a_n$
and nothing more than this.
With reference to your example by $t=x^2$
$$x^4+3x^2+2=0 implies t^2+3t+2=0$$
by rational root theorem we can find roots $t=-1$ and $t=-2$ and then
$$x^4+3x^2+2=(x^2+1)(x^2+2)=(x+i)(x-i)(x+isqrt 2)(x-isqrt 2)$$
and of course $sqrt 2$ (as any real number) can't be a solution since $x^4+3x^2+2ge 2$.
answered Aug 2 at 13:31
gimusi
63.8k73480
add a comment |Â
up vote
2
down vote
Recall that Rational root theorem guarantees that each rational solution $x$ must be in the form $x = fracpq$ with
- p integer factor of the constant term $a_0$
- q integer factor of the leading coefficient $a_n$
and nothing more than this.
With reference to your example by $t=x^2$
$$x^4+3x^2+2=0 implies t^2+3t+2=0$$
by rational root theorem we can find roots $t=-1$ and $t=-2$ and then
$$x^4+3x^2+2=(x^2+1)(x^2+2)=(x+i)(x-i)(x+isqrt 2)(x-isqrt 2)$$
and of course $sqrt 2$ (as any real number) can't be a solution since $x^4+3x^2+2ge 2$.
answered Aug 2 at 13:31
gimusi
63.8k73480
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Recall that Rational root theorem guarantees that each rational solution $x$ must be in the form $x = fracpq$ with
- p integer factor of the constant term $a_0$
- q integer factor of the leading coefficient $a_n$
and nothing more than this.
With reference to your example by $t=x^2$
$$x^4+3x^2+2=0 implies t^2+3t+2=0$$
by rational root theorem we can find roots $t=-1$ and $t=-2$ and then
$$x^4+3x^2+2=(x^2+1)(x^2+2)=(x+i)(x-i)(x+isqrt 2)(x-isqrt 2)$$
and of course $sqrt 2$ (as any real number) can't be a solution since $x^4+3x^2+2ge 2$.
answered Aug 2 at 13:31
gimusi
63.8k73480
Recall that Rational root theorem guarantees that each rational solution $x$ must be in the form $x = fracpq$ with
- p integer factor of the constant term $a_0$
- q integer factor of the leading coefficient $a_n$
and nothing more than this.
With reference to your example by $t=x^2$
$$x^4+3x^2+2=0 implies t^2+3t+2=0$$
by rational root theorem we can find roots $t=-1$ and $t=-2$ and then
$$x^4+3x^2+2=(x^2+1)(x^2+2)=(x+i)(x-i)(x+isqrt 2)(x-isqrt 2)$$
and of course $sqrt 2$ (as any real number) can't be a solution since $x^4+3x^2+2ge 2$.
answered Aug 2 at 13:31
gimusi
63.8k73480
edited Aug 2 at 13:36
answered Aug 2 at 13:31
gimusi
63.8k73480
answered Aug 2 at 13:31
gimusi
63.8k73480
answered Aug 2 at 13:31
gimusi
63.8k73480
63.8k73480
add a comment |Â
add a comment |Â
up vote
0
down vote
I would guess that this is a typo of your textbook. The rational root test suffices, for a polynomial with integer coefficients, to find all rational roots, as you may list
- all the integer divisors $q$ of the leading coefficient $a_n$
- all the integer divisors $p$ of the trailing coefficient $a_0$
You can then form all combinations $fracpq$ and the rational root theorem guarantees you that, if there are rational roots, then they are of this form. Of course, you still have to check which combinations really apply to be a root.
Note that rational numbers are most certainly real numbers, but of course, this procedure can not speak about the irrational roots at all, as they can not be expressed as a fraction $fracpq$.
I have not checked you polynomial, but another good example could be $x^2-x-1$ with roots $frac1+sqrt52$ and $frac1-sqrt52$, i.e. the golden ration and its conjugate which are irrational as well.
However, if you find a rational root(as the rational root test does not guarantee that there are any), you may use it to reduce your problem/polynomial via polynomial division, which may help in finding the remaining roots.
answered Aug 2 at 13:33
zzuussee
1,142419
add a comment |Â
up vote
0
down vote
I would guess that this is a typo of your textbook. The rational root test suffices, for a polynomial with integer coefficients, to find all rational roots, as you may list
- all the integer divisors $q$ of the leading coefficient $a_n$
- all the integer divisors $p$ of the trailing coefficient $a_0$
You can then form all combinations $fracpq$ and the rational root theorem guarantees you that, if there are rational roots, then they are of this form. Of course, you still have to check which combinations really apply to be a root.
Note that rational numbers are most certainly real numbers, but of course, this procedure can not speak about the irrational roots at all, as they can not be expressed as a fraction $fracpq$.
I have not checked you polynomial, but another good example could be $x^2-x-1$ with roots $frac1+sqrt52$ and $frac1-sqrt52$, i.e. the golden ration and its conjugate which are irrational as well.
However, if you find a rational root(as the rational root test does not guarantee that there are any), you may use it to reduce your problem/polynomial via polynomial division, which may help in finding the remaining roots.
answered Aug 2 at 13:33
zzuussee
1,142419
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I would guess that this is a typo of your textbook. The rational root test suffices, for a polynomial with integer coefficients, to find all rational roots, as you may list
- all the integer divisors $q$ of the leading coefficient $a_n$
- all the integer divisors $p$ of the trailing coefficient $a_0$
You can then form all combinations $fracpq$ and the rational root theorem guarantees you that, if there are rational roots, then they are of this form. Of course, you still have to check which combinations really apply to be a root.
Note that rational numbers are most certainly real numbers, but of course, this procedure can not speak about the irrational roots at all, as they can not be expressed as a fraction $fracpq$.
I have not checked you polynomial, but another good example could be $x^2-x-1$ with roots $frac1+sqrt52$ and $frac1-sqrt52$, i.e. the golden ration and its conjugate which are irrational as well.
However, if you find a rational root(as the rational root test does not guarantee that there are any), you may use it to reduce your problem/polynomial via polynomial division, which may help in finding the remaining roots.
answered Aug 2 at 13:33
zzuussee
1,142419
I would guess that this is a typo of your textbook. The rational root test suffices, for a polynomial with integer coefficients, to find all rational roots, as you may list
- all the integer divisors $q$ of the leading coefficient $a_n$
- all the integer divisors $p$ of the trailing coefficient $a_0$
You can then form all combinations $fracpq$ and the rational root theorem guarantees you that, if there are rational roots, then they are of this form. Of course, you still have to check which combinations really apply to be a root.
Note that rational numbers are most certainly real numbers, but of course, this procedure can not speak about the irrational roots at all, as they can not be expressed as a fraction $fracpq$.
I have not checked you polynomial, but another good example could be $x^2-x-1$ with roots $frac1+sqrt52$ and $frac1-sqrt52$, i.e. the golden ration and its conjugate which are irrational as well.
However, if you find a rational root(as the rational root test does not guarantee that there are any), you may use it to reduce your problem/polynomial via polynomial division, which may help in finding the remaining roots.
answered Aug 2 at 13:33
zzuussee
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edited Aug 2 at 13:41
answered Aug 2 at 13:33
zzuussee
1,142419
answered Aug 2 at 13:33
zzuussee
1,142419
answered Aug 2 at 13:33
zzuussee
1,142419
1,142419
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You have 500+ rep. There's no reason why you shouldn't use MathJax: math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Aug 2 at 13:29
2
The rational root theorem is of no use for the irrational roots.
– lulu
Aug 2 at 13:30
@lulu So it can find all real roots, except roots that are irrational?
– Ethan Chan
Aug 2 at 13:32
1
@EthanChan that’s why it’s called the rational root theorem :)
– Oiler
Aug 2 at 13:32
1
Yes, it can find all the rational roots.
– lulu
Aug 2 at 13:39