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Factoring A Cubic Function [closed]
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Can someone help me factor $;2x^3-1=0;$ ? I've never known how to factor cubic functions. There should be cubic roots involved.
factoring cubic-equations mathematica
edited Jul 19 at 11:17
achille hui
91k5127246
asked Jul 19 at 11:15
Alan Zhou
124
closed as off-topic by amWhy, user223391, Mostafa Ayaz, José Carlos Santos, Adrian Keister Jul 22 at 0:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Community, Mostafa Ayaz, Jos̩ Carlos Santos, Adrian Keister
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up vote
0
down vote
favorite
Can someone help me factor $;2x^3-1=0;$ ? I've never known how to factor cubic functions. There should be cubic roots involved.
factoring cubic-equations mathematica
edited Jul 19 at 11:17
achille hui
91k5127246
asked Jul 19 at 11:15
Alan Zhou
124
closed as off-topic by amWhy, user223391, Mostafa Ayaz, José Carlos Santos, Adrian Keister Jul 22 at 0:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Community, Mostafa Ayaz, Jos̩ Carlos Santos, Adrian Keister
2
Use $A^3-B^3=(A-B)(A^2+AB+B^2)
– user362325
Jul 19 at 11:19
Thanks. I've never learned this formula before.
– Alan Zhou
Jul 19 at 11:47
In case you wanted to know, here are the general formulas for the roots of a cubic: wolframalpha.com/input/?i=ax%5E3%2Bbx%5E2%2Bcx%2Bd%3D0 [Just a little kidding, it is possible to express them in a simpler way.]
– Yves Daoust
Jul 19 at 12:53
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can someone help me factor $;2x^3-1=0;$ ? I've never known how to factor cubic functions. There should be cubic roots involved.
factoring cubic-equations mathematica
edited Jul 19 at 11:17
achille hui
91k5127246
asked Jul 19 at 11:15
Alan Zhou
124
Can someone help me factor $;2x^3-1=0;$ ? I've never known how to factor cubic functions. There should be cubic roots involved.
factoring cubic-equations mathematica
edited Jul 19 at 11:17
achille hui
91k5127246
asked Jul 19 at 11:15
Alan Zhou
124
edited Jul 19 at 11:17
achille hui
91k5127246
edited Jul 19 at 11:17
achille hui
91k5127246
edited Jul 19 at 11:17
achille hui
91k5127246
91k5127246
asked Jul 19 at 11:15
Alan Zhou
124
asked Jul 19 at 11:15
Alan Zhou
124
asked Jul 19 at 11:15
Alan Zhou
124
124
closed as off-topic by amWhy, user223391, Mostafa Ayaz, José Carlos Santos, Adrian Keister Jul 22 at 0:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Community, Mostafa Ayaz, Jos̩ Carlos Santos, Adrian Keister
closed as off-topic by amWhy, user223391, Mostafa Ayaz, José Carlos Santos, Adrian Keister Jul 22 at 0:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Community, Mostafa Ayaz, Jos̩ Carlos Santos, Adrian Keister
2
Use $A^3-B^3=(A-B)(A^2+AB+B^2)
– user362325
Jul 19 at 11:19
Thanks. I've never learned this formula before.
– Alan Zhou
Jul 19 at 11:47
In case you wanted to know, here are the general formulas for the roots of a cubic: wolframalpha.com/input/?i=ax%5E3%2Bbx%5E2%2Bcx%2Bd%3D0 [Just a little kidding, it is possible to express them in a simpler way.]
– Yves Daoust
Jul 19 at 12:53
add a comment |Â
2
Use $A^3-B^3=(A-B)(A^2+AB+B^2)
– user362325
Jul 19 at 11:19
Thanks. I've never learned this formula before.
– Alan Zhou
Jul 19 at 11:47
In case you wanted to know, here are the general formulas for the roots of a cubic: wolframalpha.com/input/?i=ax%5E3%2Bbx%5E2%2Bcx%2Bd%3D0 [Just a little kidding, it is possible to express them in a simpler way.]
– Yves Daoust
Jul 19 at 12:53
2
2
Use $A^3-B^3=(A-B)(A^2+AB+B^2)
– user362325
Jul 19 at 11:19
Use $A^3-B^3=(A-B)(A^2+AB+B^2)
– user362325
Jul 19 at 11:19
Thanks. I've never learned this formula before.
– Alan Zhou
Jul 19 at 11:47
Thanks. I've never learned this formula before.
– Alan Zhou
Jul 19 at 11:47
In case you wanted to know, here are the general formulas for the roots of a cubic: wolframalpha.com/input/?i=ax%5E3%2Bbx%5E2%2Bcx%2Bd%3D0 [Just a little kidding, it is possible to express them in a simpler way.]
– Yves Daoust
Jul 19 at 12:53
In case you wanted to know, here are the general formulas for the roots of a cubic: wolframalpha.com/input/?i=ax%5E3%2Bbx%5E2%2Bcx%2Bd%3D0 [Just a little kidding, it is possible to express them in a simpler way.]
– Yves Daoust
Jul 19 at 12:53
add a comment |Â
3 Answers
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Hint: Set $u=sqrt[3]2x$. Then $2x^3-1=u^3-1=(u-1)(u^2+u+1)$.
answered Jul 19 at 12:40
lhf
156k9160366
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0
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This equation is quite simple to factor completely as it involves the cube roots (slightly modified) of unity. The three cube roots of unity are (by inspection): 1, $-frac1 2+isqrt3/2$, $-frac1 2-isqrt3/2$, sometimes denoted: 1, $omega_1, omega_2$. (You should do some simple (complex) algebra to verify that when each of these is cubed it produces the result of 1. The roots of your equation are the same but multiplied by the cube-root of 2.
answered Jul 19 at 12:03
Dr Peter McGowan
4637
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You can calculate one root immediatly, since $2x^3-1=0Leftrightarrow x=sqrt[3]frac23$
Now you can calculate: $(2x^3-1)div left(x-sqrt[3]frac23right)$
To reduce it to a quadratic term and then go ahead with the pq-formula.
answered Jul 19 at 12:45
Cornman
2,47921128
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: Set $u=sqrt[3]2x$. Then $2x^3-1=u^3-1=(u-1)(u^2+u+1)$.
answered Jul 19 at 12:40
lhf
156k9160366
add a comment |Â
up vote
1
down vote
Hint: Set $u=sqrt[3]2x$. Then $2x^3-1=u^3-1=(u-1)(u^2+u+1)$.
answered Jul 19 at 12:40
lhf
156k9160366
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Set $u=sqrt[3]2x$. Then $2x^3-1=u^3-1=(u-1)(u^2+u+1)$.
answered Jul 19 at 12:40
lhf
156k9160366
Hint: Set $u=sqrt[3]2x$. Then $2x^3-1=u^3-1=(u-1)(u^2+u+1)$.
answered Jul 19 at 12:40
lhf
156k9160366
answered Jul 19 at 12:40
lhf
156k9160366
answered Jul 19 at 12:40
lhf
156k9160366
answered Jul 19 at 12:40
lhf
156k9160366
156k9160366
add a comment |Â
add a comment |Â
up vote
0
down vote
This equation is quite simple to factor completely as it involves the cube roots (slightly modified) of unity. The three cube roots of unity are (by inspection): 1, $-frac1 2+isqrt3/2$, $-frac1 2-isqrt3/2$, sometimes denoted: 1, $omega_1, omega_2$. (You should do some simple (complex) algebra to verify that when each of these is cubed it produces the result of 1. The roots of your equation are the same but multiplied by the cube-root of 2.
answered Jul 19 at 12:03
Dr Peter McGowan
4637
add a comment |Â
up vote
0
down vote
This equation is quite simple to factor completely as it involves the cube roots (slightly modified) of unity. The three cube roots of unity are (by inspection): 1, $-frac1 2+isqrt3/2$, $-frac1 2-isqrt3/2$, sometimes denoted: 1, $omega_1, omega_2$. (You should do some simple (complex) algebra to verify that when each of these is cubed it produces the result of 1. The roots of your equation are the same but multiplied by the cube-root of 2.
answered Jul 19 at 12:03
Dr Peter McGowan
4637
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This equation is quite simple to factor completely as it involves the cube roots (slightly modified) of unity. The three cube roots of unity are (by inspection): 1, $-frac1 2+isqrt3/2$, $-frac1 2-isqrt3/2$, sometimes denoted: 1, $omega_1, omega_2$. (You should do some simple (complex) algebra to verify that when each of these is cubed it produces the result of 1. The roots of your equation are the same but multiplied by the cube-root of 2.
answered Jul 19 at 12:03
Dr Peter McGowan
4637
This equation is quite simple to factor completely as it involves the cube roots (slightly modified) of unity. The three cube roots of unity are (by inspection): 1, $-frac1 2+isqrt3/2$, $-frac1 2-isqrt3/2$, sometimes denoted: 1, $omega_1, omega_2$. (You should do some simple (complex) algebra to verify that when each of these is cubed it produces the result of 1. The roots of your equation are the same but multiplied by the cube-root of 2.
answered Jul 19 at 12:03
Dr Peter McGowan
4637
answered Jul 19 at 12:03
Dr Peter McGowan
4637
answered Jul 19 at 12:03
Dr Peter McGowan
4637
answered Jul 19 at 12:03
Dr Peter McGowan
4637
4637
add a comment |Â
add a comment |Â
up vote
0
down vote
You can calculate one root immediatly, since $2x^3-1=0Leftrightarrow x=sqrt[3]frac23$
Now you can calculate: $(2x^3-1)div left(x-sqrt[3]frac23right)$
To reduce it to a quadratic term and then go ahead with the pq-formula.
answered Jul 19 at 12:45
Cornman
2,47921128
add a comment |Â
up vote
0
down vote
You can calculate one root immediatly, since $2x^3-1=0Leftrightarrow x=sqrt[3]frac23$
Now you can calculate: $(2x^3-1)div left(x-sqrt[3]frac23right)$
To reduce it to a quadratic term and then go ahead with the pq-formula.
answered Jul 19 at 12:45
Cornman
2,47921128
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can calculate one root immediatly, since $2x^3-1=0Leftrightarrow x=sqrt[3]frac23$
Now you can calculate: $(2x^3-1)div left(x-sqrt[3]frac23right)$
To reduce it to a quadratic term and then go ahead with the pq-formula.
answered Jul 19 at 12:45
Cornman
2,47921128
You can calculate one root immediatly, since $2x^3-1=0Leftrightarrow x=sqrt[3]frac23$
Now you can calculate: $(2x^3-1)div left(x-sqrt[3]frac23right)$
To reduce it to a quadratic term and then go ahead with the pq-formula.
answered Jul 19 at 12:45
Cornman
2,47921128
answered Jul 19 at 12:45
Cornman
2,47921128
answered Jul 19 at 12:45
Cornman
2,47921128
answered Jul 19 at 12:45
Cornman
2,47921128
2,47921128
add a comment |Â
add a comment |Â
2
Use $A^3-B^3=(A-B)(A^2+AB+B^2)
– user362325
Jul 19 at 11:19
Thanks. I've never learned this formula before.
– Alan Zhou
Jul 19 at 11:47
In case you wanted to know, here are the general formulas for the roots of a cubic: wolframalpha.com/input/?i=ax%5E3%2Bbx%5E2%2Bcx%2Bd%3D0 [Just a little kidding, it is possible to express them in a simpler way.]
– Yves Daoust
Jul 19 at 12:53