Factoring cubic polynomial over R
$z^3-7z^2+14z-7=0$
I tried simplifying it to $z^3-7(z-1)^2=0$, but I don't think i can proceed from there. It sort of looked like geometric progression but it is not that either, and I don't see any other approach here. Apparently roots are $4cos^2(frac{pi}{14})$, $4cos^2(frac{3pi}{14})$,$4cos^2(frac{9pi}{14})$, but I don't know how to get them.
Thanks for help.
polynomials roots factoring
asked Nov 24 at 19:39
Dovla
849
|
show 5 more comments
$z^3-7z^2+14z-7=0$
I tried simplifying it to $z^3-7(z-1)^2=0$, but I don't think i can proceed from there. It sort of looked like geometric progression but it is not that either, and I don't see any other approach here. Apparently roots are $4cos^2(frac{pi}{14})$, $4cos^2(frac{3pi}{14})$,$4cos^2(frac{9pi}{14})$, but I don't know how to get them.
Thanks for help.
polynomials roots factoring
asked Nov 24 at 19:39
Dovla
849
2
You won't find the roots without Cardano's formulas or similar. Are you sure about the coefficients ?
– Yves Daoust
Nov 24 at 19:44
It seems like there is a typo somewhere since the roots of this equation are irrational according to WolframAplha.
– mrtaurho
Nov 24 at 19:45
1
Ye, its correct, I just checked the book and there are solutions, but not the method to get them, I will edit the post.
– Dovla
Nov 24 at 19:46
@YvesDaoust Not sure if you got a notification, but the OP updated the questions with the solutions, which are all in terms of $cos^2 dfrac {n pi}{14}$. Do you know of a way to solve polynomials and get this kind of answer? I remember a problem where you could find the values for stuff like $cos^2 dfrac {pi}{5}$ by multiplying different roots of unity somehow, but I don't remember how.
– Ovi
Nov 24 at 20:20
@ovi: this is a casus irreductibilis. The trick is to transform the polynomial by a linear transformation of the unknown, to the form $4x^3-3x=c$ (depress and rescale). This is as technical as Cardano.
– Yves Daoust
Nov 24 at 20:38
|
show 5 more comments
$z^3-7z^2+14z-7=0$
I tried simplifying it to $z^3-7(z-1)^2=0$, but I don't think i can proceed from there. It sort of looked like geometric progression but it is not that either, and I don't see any other approach here. Apparently roots are $4cos^2(frac{pi}{14})$, $4cos^2(frac{3pi}{14})$,$4cos^2(frac{9pi}{14})$, but I don't know how to get them.
Thanks for help.
polynomials roots factoring
asked Nov 24 at 19:39
Dovla
849
$z^3-7z^2+14z-7=0$
I tried simplifying it to $z^3-7(z-1)^2=0$, but I don't think i can proceed from there. It sort of looked like geometric progression but it is not that either, and I don't see any other approach here. Apparently roots are $4cos^2(frac{pi}{14})$, $4cos^2(frac{3pi}{14})$,$4cos^2(frac{9pi}{14})$, but I don't know how to get them.
Thanks for help.
polynomials roots factoring
polynomials roots factoring
asked Nov 24 at 19:39
Dovla
849
asked Nov 24 at 19:39
Dovla
849
edited Nov 24 at 20:15
asked Nov 24 at 19:39
Dovla
849
asked Nov 24 at 19:39
Dovla
849
asked Nov 24 at 19:39
Dovla
849
849
2
You won't find the roots without Cardano's formulas or similar. Are you sure about the coefficients ?
– Yves Daoust
Nov 24 at 19:44
It seems like there is a typo somewhere since the roots of this equation are irrational according to WolframAplha.
– mrtaurho
Nov 24 at 19:45
1
Ye, its correct, I just checked the book and there are solutions, but not the method to get them, I will edit the post.
– Dovla
Nov 24 at 19:46
@YvesDaoust Not sure if you got a notification, but the OP updated the questions with the solutions, which are all in terms of $cos^2 dfrac {n pi}{14}$. Do you know of a way to solve polynomials and get this kind of answer? I remember a problem where you could find the values for stuff like $cos^2 dfrac {pi}{5}$ by multiplying different roots of unity somehow, but I don't remember how.
– Ovi
Nov 24 at 20:20
@ovi: this is a casus irreductibilis. The trick is to transform the polynomial by a linear transformation of the unknown, to the form $4x^3-3x=c$ (depress and rescale). This is as technical as Cardano.
– Yves Daoust
Nov 24 at 20:38
|
show 5 more comments
2
You won't find the roots without Cardano's formulas or similar. Are you sure about the coefficients ?
– Yves Daoust
Nov 24 at 19:44
It seems like there is a typo somewhere since the roots of this equation are irrational according to WolframAplha.
– mrtaurho
Nov 24 at 19:45
1
Ye, its correct, I just checked the book and there are solutions, but not the method to get them, I will edit the post.
– Dovla
Nov 24 at 19:46
@YvesDaoust Not sure if you got a notification, but the OP updated the questions with the solutions, which are all in terms of $cos^2 dfrac {n pi}{14}$. Do you know of a way to solve polynomials and get this kind of answer? I remember a problem where you could find the values for stuff like $cos^2 dfrac {pi}{5}$ by multiplying different roots of unity somehow, but I don't remember how.
– Ovi
Nov 24 at 20:20
@ovi: this is a casus irreductibilis. The trick is to transform the polynomial by a linear transformation of the unknown, to the form $4x^3-3x=c$ (depress and rescale). This is as technical as Cardano.
– Yves Daoust
Nov 24 at 20:38
2
2
You won't find the roots without Cardano's formulas or similar. Are you sure about the coefficients ?
– Yves Daoust
Nov 24 at 19:44
You won't find the roots without Cardano's formulas or similar. Are you sure about the coefficients ?
– Yves Daoust
Nov 24 at 19:44
It seems like there is a typo somewhere since the roots of this equation are irrational according to WolframAplha.
– mrtaurho
Nov 24 at 19:45
It seems like there is a typo somewhere since the roots of this equation are irrational according to WolframAplha.
– mrtaurho
Nov 24 at 19:45
1
1
Ye, its correct, I just checked the book and there are solutions, but not the method to get them, I will edit the post.
– Dovla
Nov 24 at 19:46
Ye, its correct, I just checked the book and there are solutions, but not the method to get them, I will edit the post.
– Dovla
Nov 24 at 19:46
@YvesDaoust Not sure if you got a notification, but the OP updated the questions with the solutions, which are all in terms of $cos^2 dfrac {n pi}{14}$. Do you know of a way to solve polynomials and get this kind of answer? I remember a problem where you could find the values for stuff like $cos^2 dfrac {pi}{5}$ by multiplying different roots of unity somehow, but I don't remember how.
– Ovi
Nov 24 at 20:20
@YvesDaoust Not sure if you got a notification, but the OP updated the questions with the solutions, which are all in terms of $cos^2 dfrac {n pi}{14}$. Do you know of a way to solve polynomials and get this kind of answer? I remember a problem where you could find the values for stuff like $cos^2 dfrac {pi}{5}$ by multiplying different roots of unity somehow, but I don't remember how.
– Ovi
Nov 24 at 20:20
@ovi: this is a casus irreductibilis. The trick is to transform the polynomial by a linear transformation of the unknown, to the form $4x^3-3x=c$ (depress and rescale). This is as technical as Cardano.
– Yves Daoust
Nov 24 at 20:38
@ovi: this is a casus irreductibilis. The trick is to transform the polynomial by a linear transformation of the unknown, to the form $4x^3-3x=c$ (depress and rescale). This is as technical as Cardano.
– Yves Daoust
Nov 24 at 20:38
|
show 5 more comments
2 Answers
2
active
oldest
votes
Let $z=aw+b$. The equation turns to
$$a^3w^3+(3ba^2-7a^2)w^2+(3ab^2-14ab+14a)w+b^3-7b^2+14b-7=0.$$
Now, cancel the quadratic coefficient. This is achieved by $3b=7$, and
$$a^3w^3-frac73aw+frac7{27}=0.$$
Next, ensure that the ratio of the cubic coefficient over the linear one is $-dfrac43$. This is achieved by
$$a=frac{sqrt{28}}3,$$
and
$$frac{28}9frac{sqrt{28}}3w^3-frac73frac{sqrt{28}}3w+frac7{27}=0$$ or
$$4w^3-3w=-frac1{sqrt{28}}.$$
Finally, setting
$$w=costheta,$$
$$4cos^3theta-3costheta=cos 3theta=-frac1{sqrt{28}}$$ gives you the roots.
answered Nov 24 at 20:58
Yves Daoust
124k671221
add a comment |
The roots of $$ x^3 + x^2 - 2 x - 1 $$
are
$$ 2 cos frac{2 pi}{7} ; , ; ; 2 cos frac{4 pi}{7} ; , ; ;2 cos frac{8 pi}{7} ; . ; ; $$
This is pretty easy if we take $omega $ a 7th root of unity, then take $x = omega + frac{1}{omega}$ and calculate $ x^3 + x^2 - 2 x - 1 $ while demanding $omega^6 + omega^5 + omega^4 + omega^3 + omega^2 + omega + 1=0$
if we then take $x = 2 - z,$ we find
$$ -(x^3 + x^2 - 2 x - 1) = z^3 - 7 z^2 + 14 z - 7 $$
The image below comes from the book by REUSCHLE
answered Nov 24 at 20:50
Will Jagy
101k598198
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $z=aw+b$. The equation turns to
$$a^3w^3+(3ba^2-7a^2)w^2+(3ab^2-14ab+14a)w+b^3-7b^2+14b-7=0.$$
Now, cancel the quadratic coefficient. This is achieved by $3b=7$, and
$$a^3w^3-frac73aw+frac7{27}=0.$$
Next, ensure that the ratio of the cubic coefficient over the linear one is $-dfrac43$. This is achieved by
$$a=frac{sqrt{28}}3,$$
and
$$frac{28}9frac{sqrt{28}}3w^3-frac73frac{sqrt{28}}3w+frac7{27}=0$$ or
$$4w^3-3w=-frac1{sqrt{28}}.$$
Finally, setting
$$w=costheta,$$
$$4cos^3theta-3costheta=cos 3theta=-frac1{sqrt{28}}$$ gives you the roots.
answered Nov 24 at 20:58
Yves Daoust
124k671221
add a comment |
Let $z=aw+b$. The equation turns to
$$a^3w^3+(3ba^2-7a^2)w^2+(3ab^2-14ab+14a)w+b^3-7b^2+14b-7=0.$$
Now, cancel the quadratic coefficient. This is achieved by $3b=7$, and
$$a^3w^3-frac73aw+frac7{27}=0.$$
Next, ensure that the ratio of the cubic coefficient over the linear one is $-dfrac43$. This is achieved by
$$a=frac{sqrt{28}}3,$$
and
$$frac{28}9frac{sqrt{28}}3w^3-frac73frac{sqrt{28}}3w+frac7{27}=0$$ or
$$4w^3-3w=-frac1{sqrt{28}}.$$
Finally, setting
$$w=costheta,$$
$$4cos^3theta-3costheta=cos 3theta=-frac1{sqrt{28}}$$ gives you the roots.
answered Nov 24 at 20:58
Yves Daoust
124k671221
add a comment |
Let $z=aw+b$. The equation turns to
$$a^3w^3+(3ba^2-7a^2)w^2+(3ab^2-14ab+14a)w+b^3-7b^2+14b-7=0.$$
Now, cancel the quadratic coefficient. This is achieved by $3b=7$, and
$$a^3w^3-frac73aw+frac7{27}=0.$$
Next, ensure that the ratio of the cubic coefficient over the linear one is $-dfrac43$. This is achieved by
$$a=frac{sqrt{28}}3,$$
and
$$frac{28}9frac{sqrt{28}}3w^3-frac73frac{sqrt{28}}3w+frac7{27}=0$$ or
$$4w^3-3w=-frac1{sqrt{28}}.$$
Finally, setting
$$w=costheta,$$
$$4cos^3theta-3costheta=cos 3theta=-frac1{sqrt{28}}$$ gives you the roots.
answered Nov 24 at 20:58
Yves Daoust
124k671221
Let $z=aw+b$. The equation turns to
$$a^3w^3+(3ba^2-7a^2)w^2+(3ab^2-14ab+14a)w+b^3-7b^2+14b-7=0.$$
Now, cancel the quadratic coefficient. This is achieved by $3b=7$, and
$$a^3w^3-frac73aw+frac7{27}=0.$$
Next, ensure that the ratio of the cubic coefficient over the linear one is $-dfrac43$. This is achieved by
$$a=frac{sqrt{28}}3,$$
and
$$frac{28}9frac{sqrt{28}}3w^3-frac73frac{sqrt{28}}3w+frac7{27}=0$$ or
$$4w^3-3w=-frac1{sqrt{28}}.$$
Finally, setting
$$w=costheta,$$
$$4cos^3theta-3costheta=cos 3theta=-frac1{sqrt{28}}$$ gives you the roots.
answered Nov 24 at 20:58
Yves Daoust
124k671221
answered Nov 24 at 20:58
Yves Daoust
124k671221
answered Nov 24 at 20:58
Yves Daoust
124k671221
answered Nov 24 at 20:58
Yves Daoust
124k671221
124k671221
add a comment |
add a comment |
The roots of $$ x^3 + x^2 - 2 x - 1 $$
are
$$ 2 cos frac{2 pi}{7} ; , ; ; 2 cos frac{4 pi}{7} ; , ; ;2 cos frac{8 pi}{7} ; . ; ; $$
This is pretty easy if we take $omega $ a 7th root of unity, then take $x = omega + frac{1}{omega}$ and calculate $ x^3 + x^2 - 2 x - 1 $ while demanding $omega^6 + omega^5 + omega^4 + omega^3 + omega^2 + omega + 1=0$
if we then take $x = 2 - z,$ we find
$$ -(x^3 + x^2 - 2 x - 1) = z^3 - 7 z^2 + 14 z - 7 $$
The image below comes from the book by REUSCHLE
answered Nov 24 at 20:50
Will Jagy
101k598198
add a comment |
The roots of $$ x^3 + x^2 - 2 x - 1 $$
are
$$ 2 cos frac{2 pi}{7} ; , ; ; 2 cos frac{4 pi}{7} ; , ; ;2 cos frac{8 pi}{7} ; . ; ; $$
This is pretty easy if we take $omega $ a 7th root of unity, then take $x = omega + frac{1}{omega}$ and calculate $ x^3 + x^2 - 2 x - 1 $ while demanding $omega^6 + omega^5 + omega^4 + omega^3 + omega^2 + omega + 1=0$
if we then take $x = 2 - z,$ we find
$$ -(x^3 + x^2 - 2 x - 1) = z^3 - 7 z^2 + 14 z - 7 $$
The image below comes from the book by REUSCHLE
answered Nov 24 at 20:50
Will Jagy
101k598198
add a comment |
The roots of $$ x^3 + x^2 - 2 x - 1 $$
are
$$ 2 cos frac{2 pi}{7} ; , ; ; 2 cos frac{4 pi}{7} ; , ; ;2 cos frac{8 pi}{7} ; . ; ; $$
This is pretty easy if we take $omega $ a 7th root of unity, then take $x = omega + frac{1}{omega}$ and calculate $ x^3 + x^2 - 2 x - 1 $ while demanding $omega^6 + omega^5 + omega^4 + omega^3 + omega^2 + omega + 1=0$
if we then take $x = 2 - z,$ we find
$$ -(x^3 + x^2 - 2 x - 1) = z^3 - 7 z^2 + 14 z - 7 $$
The image below comes from the book by REUSCHLE
answered Nov 24 at 20:50
Will Jagy
101k598198
The roots of $$ x^3 + x^2 - 2 x - 1 $$
are
$$ 2 cos frac{2 pi}{7} ; , ; ; 2 cos frac{4 pi}{7} ; , ; ;2 cos frac{8 pi}{7} ; . ; ; $$
This is pretty easy if we take $omega $ a 7th root of unity, then take $x = omega + frac{1}{omega}$ and calculate $ x^3 + x^2 - 2 x - 1 $ while demanding $omega^6 + omega^5 + omega^4 + omega^3 + omega^2 + omega + 1=0$
if we then take $x = 2 - z,$ we find
$$ -(x^3 + x^2 - 2 x - 1) = z^3 - 7 z^2 + 14 z - 7 $$
The image below comes from the book by REUSCHLE
answered Nov 24 at 20:50
Will Jagy
101k598198
edited Nov 24 at 21:19
answered Nov 24 at 20:50
Will Jagy
101k598198
answered Nov 24 at 20:50
Will Jagy
101k598198
answered Nov 24 at 20:50
Will Jagy
101k598198
101k598198
add a comment |
add a comment |
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2
You won't find the roots without Cardano's formulas or similar. Are you sure about the coefficients ?
– Yves Daoust
Nov 24 at 19:44
It seems like there is a typo somewhere since the roots of this equation are irrational according to WolframAplha.
– mrtaurho
Nov 24 at 19:45
1
Ye, its correct, I just checked the book and there are solutions, but not the method to get them, I will edit the post.
– Dovla
Nov 24 at 19:46
@YvesDaoust Not sure if you got a notification, but the OP updated the questions with the solutions, which are all in terms of $cos^2 dfrac {n pi}{14}$. Do you know of a way to solve polynomials and get this kind of answer? I remember a problem where you could find the values for stuff like $cos^2 dfrac {pi}{5}$ by multiplying different roots of unity somehow, but I don't remember how.
– Ovi
Nov 24 at 20:20
@ovi: this is a casus irreductibilis. The trick is to transform the polynomial by a linear transformation of the unknown, to the form $4x^3-3x=c$ (depress and rescale). This is as technical as Cardano.
– Yves Daoust
Nov 24 at 20:38