Why does this Polynomial have two valid factored forms?
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When factoring $3 + 4a - 7a^2$ why are there two possible factors? $(-7a -3)(a - 1)$ and $(3 + 7a)(1 - a)$. What is the principle? I assume that the two factors are inverse in some way.
polynomials factoring
asked Dec 22 '18 at 16:24
patawa91patawa91
125
$endgroup$
add a comment |
$begingroup$
When factoring $3 + 4a - 7a^2$ why are there two possible factors? $(-7a -3)(a - 1)$ and $(3 + 7a)(1 - a)$. What is the principle? I assume that the two factors are inverse in some way.
polynomials factoring
asked Dec 22 '18 at 16:24
patawa91patawa91
125
$endgroup$
$begingroup$
You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
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– lulu
Dec 22 '18 at 16:25
1
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If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
$endgroup$
– lulu
Dec 22 '18 at 16:26
1
$begingroup$
Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
$endgroup$
– badjohn
Dec 22 '18 at 16:33
add a comment |
$begingroup$
When factoring $3 + 4a - 7a^2$ why are there two possible factors? $(-7a -3)(a - 1)$ and $(3 + 7a)(1 - a)$. What is the principle? I assume that the two factors are inverse in some way.
polynomials factoring
asked Dec 22 '18 at 16:24
patawa91patawa91
125
$endgroup$
When factoring $3 + 4a - 7a^2$ why are there two possible factors? $(-7a -3)(a - 1)$ and $(3 + 7a)(1 - a)$. What is the principle? I assume that the two factors are inverse in some way.
polynomials factoring
polynomials factoring
asked Dec 22 '18 at 16:24
patawa91patawa91
125
asked Dec 22 '18 at 16:24
patawa91patawa91
125
edited Dec 22 '18 at 16:29
patawa91
asked Dec 22 '18 at 16:24
patawa91patawa91
125
asked Dec 22 '18 at 16:24
patawa91patawa91
125
asked Dec 22 '18 at 16:24
patawa91patawa91
125
125
$begingroup$
You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
$endgroup$
– lulu
Dec 22 '18 at 16:25
1
$begingroup$
If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
$endgroup$
– lulu
Dec 22 '18 at 16:26
1
$begingroup$
Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
$endgroup$
– badjohn
Dec 22 '18 at 16:33
add a comment |
$begingroup$
You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
$endgroup$
– lulu
Dec 22 '18 at 16:25
1
$begingroup$
If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
$endgroup$
– lulu
Dec 22 '18 at 16:26
1
$begingroup$
Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
$endgroup$
– badjohn
Dec 22 '18 at 16:33
$begingroup$
You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
$endgroup$
– lulu
Dec 22 '18 at 16:25
$begingroup$
You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
$endgroup$
– lulu
Dec 22 '18 at 16:25
1
1
$begingroup$
If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
$endgroup$
– lulu
Dec 22 '18 at 16:26
$begingroup$
If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
$endgroup$
– lulu
Dec 22 '18 at 16:26
1
1
$begingroup$
Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
$endgroup$
– badjohn
Dec 22 '18 at 16:33
$begingroup$
Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
$endgroup$
– badjohn
Dec 22 '18 at 16:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
(-7a -3) and (3 + 7a) are "equivalent" in the sense that they differ by a multiplicative constant (which is -1 in this case). The same is true for (a - 1) and (1 - a).
Other instance is: Ax + By + C = 0 and - Ax - By - C = 0 are two different equations but they represent the same straight line.
answered Dec 22 '18 at 16:28
MickMick
11.8k21641
$endgroup$
add a comment |
$begingroup$
The only difference between the two is one is multiplied by -1. If you were to find the solution for a. You get a= -3/7 for both.
Ex 1.
-7a-3 = 0
a = -3/7
Ex 2.
3+7a=0
a = -3/7
answered Dec 22 '18 at 16:35
muhe31muhe31
85
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
(-7a -3) and (3 + 7a) are "equivalent" in the sense that they differ by a multiplicative constant (which is -1 in this case). The same is true for (a - 1) and (1 - a).
Other instance is: Ax + By + C = 0 and - Ax - By - C = 0 are two different equations but they represent the same straight line.
answered Dec 22 '18 at 16:28
MickMick
11.8k21641
$endgroup$
add a comment |
$begingroup$
(-7a -3) and (3 + 7a) are "equivalent" in the sense that they differ by a multiplicative constant (which is -1 in this case). The same is true for (a - 1) and (1 - a).
Other instance is: Ax + By + C = 0 and - Ax - By - C = 0 are two different equations but they represent the same straight line.
answered Dec 22 '18 at 16:28
MickMick
11.8k21641
$endgroup$
add a comment |
$begingroup$
(-7a -3) and (3 + 7a) are "equivalent" in the sense that they differ by a multiplicative constant (which is -1 in this case). The same is true for (a - 1) and (1 - a).
Other instance is: Ax + By + C = 0 and - Ax - By - C = 0 are two different equations but they represent the same straight line.
answered Dec 22 '18 at 16:28
MickMick
11.8k21641
$endgroup$
(-7a -3) and (3 + 7a) are "equivalent" in the sense that they differ by a multiplicative constant (which is -1 in this case). The same is true for (a - 1) and (1 - a).
Other instance is: Ax + By + C = 0 and - Ax - By - C = 0 are two different equations but they represent the same straight line.
answered Dec 22 '18 at 16:28
MickMick
11.8k21641
edited Dec 22 '18 at 16:35
answered Dec 22 '18 at 16:28
MickMick
11.8k21641
answered Dec 22 '18 at 16:28
MickMick
11.8k21641
answered Dec 22 '18 at 16:28
MickMick
11.8k21641
11.8k21641
add a comment |
add a comment |
$begingroup$
The only difference between the two is one is multiplied by -1. If you were to find the solution for a. You get a= -3/7 for both.
Ex 1.
-7a-3 = 0
a = -3/7
Ex 2.
3+7a=0
a = -3/7
answered Dec 22 '18 at 16:35
muhe31muhe31
85
$endgroup$
add a comment |
$begingroup$
The only difference between the two is one is multiplied by -1. If you were to find the solution for a. You get a= -3/7 for both.
Ex 1.
-7a-3 = 0
a = -3/7
Ex 2.
3+7a=0
a = -3/7
answered Dec 22 '18 at 16:35
muhe31muhe31
85
$endgroup$
add a comment |
$begingroup$
The only difference between the two is one is multiplied by -1. If you were to find the solution for a. You get a= -3/7 for both.
Ex 1.
-7a-3 = 0
a = -3/7
Ex 2.
3+7a=0
a = -3/7
answered Dec 22 '18 at 16:35
muhe31muhe31
85
$endgroup$
The only difference between the two is one is multiplied by -1. If you were to find the solution for a. You get a= -3/7 for both.
Ex 1.
-7a-3 = 0
a = -3/7
Ex 2.
3+7a=0
a = -3/7
answered Dec 22 '18 at 16:35
muhe31muhe31
85
answered Dec 22 '18 at 16:35
muhe31muhe31
85
answered Dec 22 '18 at 16:35
muhe31muhe31
85
answered Dec 22 '18 at 16:35
muhe31muhe31
85
85
add a comment |
add a comment |
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$begingroup$
You can always multiply and divide by a unit (an invertible element). Here you just multiplied each term by $-1$ (noting that $frac 1{-1}=-1$).
$endgroup$
– lulu
Dec 22 '18 at 16:25
1
$begingroup$
If you don't require integer coefficients then you could also get, e.g., $(6+14a)(frac 12-frac a2)$ and so on.
$endgroup$
– lulu
Dec 22 '18 at 16:26
1
$begingroup$
Similarly $6 = 2 times 3$ but also $6 = -2 times -3$. This is not regarded as a violation of unique factorisation.
$endgroup$
– badjohn
Dec 22 '18 at 16:33