Factoring a quadratic polynomial (absolute beginner level), are both answers correct?
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5
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I'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial:
$$x^2 + 3x - 10$$
And I'm given the task of finding the values of $a$ and $b$ in:
$$(x + a) (x + b)$$
Obviously the answer is:
$$(x + 5)(x - 2)$$
However the answer can be also:
$$(x - 2) (x + 5)$$
I just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b$' can be either $5$ or $-2$.
Therefore if a question asks what are the values of '$a$' and '$b$' both the following answers are correct:
Answer $1$
$a = -2$
$b = 5$
or
Answer $2$
$a = 5$
$b = -2$
I'm sure this is a completely obvious question, but I'm just a beginner in this.
polynomials quadratics factoring
 |Â
show 7 more comments
up vote
5
down vote
favorite
I'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial:
$$x^2 + 3x - 10$$
And I'm given the task of finding the values of $a$ and $b$ in:
$$(x + a) (x + b)$$
Obviously the answer is:
$$(x + 5)(x - 2)$$
However the answer can be also:
$$(x - 2) (x + 5)$$
I just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b$' can be either $5$ or $-2$.
Therefore if a question asks what are the values of '$a$' and '$b$' both the following answers are correct:
Answer $1$
$a = -2$
$b = 5$
or
Answer $2$
$a = 5$
$b = -2$
I'm sure this is a completely obvious question, but I'm just a beginner in this.
polynomials quadratics factoring
2
Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$.
â Matti P.
Aug 23 at 8:18
2
They are both valid answers, since the order of the factors doesn't matter.
â Ludvig Lindström
Aug 23 at 8:18
You teacher should have stated what context they want you to find $a$ and $b$, I have provided the answer below under the assumption that they meant the roots of the polynomial equation in $x$.
â Adam
Aug 23 at 8:54
Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case.
â Adam
Aug 23 at 8:58
@Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from Kahn Academy videos. I appreciate that there are many people here to give extra help.
â Zebrafish
Aug 23 at 9:03
 |Â
show 7 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial:
$$x^2 + 3x - 10$$
And I'm given the task of finding the values of $a$ and $b$ in:
$$(x + a) (x + b)$$
Obviously the answer is:
$$(x + 5)(x - 2)$$
However the answer can be also:
$$(x - 2) (x + 5)$$
I just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b$' can be either $5$ or $-2$.
Therefore if a question asks what are the values of '$a$' and '$b$' both the following answers are correct:
Answer $1$
$a = -2$
$b = 5$
or
Answer $2$
$a = 5$
$b = -2$
I'm sure this is a completely obvious question, but I'm just a beginner in this.
polynomials quadratics factoring
I'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial:
$$x^2 + 3x - 10$$
And I'm given the task of finding the values of $a$ and $b$ in:
$$(x + a) (x + b)$$
Obviously the answer is:
$$(x + 5)(x - 2)$$
However the answer can be also:
$$(x - 2) (x + 5)$$
I just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b$' can be either $5$ or $-2$.
Therefore if a question asks what are the values of '$a$' and '$b$' both the following answers are correct:
Answer $1$
$a = -2$
$b = 5$
or
Answer $2$
$a = 5$
$b = -2$
I'm sure this is a completely obvious question, but I'm just a beginner in this.
polynomials quadratics factoring
edited Aug 23 at 17:16
Daniel Buck
2,5151625
2,5151625
asked Aug 23 at 8:15
Zebrafish
2058
2058
2
Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$.
â Matti P.
Aug 23 at 8:18
2
They are both valid answers, since the order of the factors doesn't matter.
â Ludvig Lindström
Aug 23 at 8:18
You teacher should have stated what context they want you to find $a$ and $b$, I have provided the answer below under the assumption that they meant the roots of the polynomial equation in $x$.
â Adam
Aug 23 at 8:54
Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case.
â Adam
Aug 23 at 8:58
@Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from Kahn Academy videos. I appreciate that there are many people here to give extra help.
â Zebrafish
Aug 23 at 9:03
 |Â
show 7 more comments
2
Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$.
â Matti P.
Aug 23 at 8:18
2
They are both valid answers, since the order of the factors doesn't matter.
â Ludvig Lindström
Aug 23 at 8:18
You teacher should have stated what context they want you to find $a$ and $b$, I have provided the answer below under the assumption that they meant the roots of the polynomial equation in $x$.
â Adam
Aug 23 at 8:54
Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case.
â Adam
Aug 23 at 8:58
@Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from Kahn Academy videos. I appreciate that there are many people here to give extra help.
â Zebrafish
Aug 23 at 9:03
2
2
Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$.
â Matti P.
Aug 23 at 8:18
Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$.
â Matti P.
Aug 23 at 8:18
2
2
They are both valid answers, since the order of the factors doesn't matter.
â Ludvig Lindström
Aug 23 at 8:18
They are both valid answers, since the order of the factors doesn't matter.
â Ludvig Lindström
Aug 23 at 8:18
You teacher should have stated what context they want you to find $a$ and $b$, I have provided the answer below under the assumption that they meant the roots of the polynomial equation in $x$.
â Adam
Aug 23 at 8:54
You teacher should have stated what context they want you to find $a$ and $b$, I have provided the answer below under the assumption that they meant the roots of the polynomial equation in $x$.
â Adam
Aug 23 at 8:54
Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case.
â Adam
Aug 23 at 8:58
Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case.
â Adam
Aug 23 at 8:58
@Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from Kahn Academy videos. I appreciate that there are many people here to give extra help.
â Zebrafish
Aug 23 at 9:03
@Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from Kahn Academy videos. I appreciate that there are many people here to give extra help.
â Zebrafish
Aug 23 at 9:03
 |Â
show 7 more comments
3 Answers
3
active
oldest
votes
up vote
11
down vote
accepted
Yes, you are correct. Since $(x+5)(x-2) = (x-2)(x+5) = x^2 + 3x-10$, we note that $a$ and $b$ may either take the values $(5,-2)$ or $(-2,5)$.
I would consider providing just one of the two solutions to be insufficient, since the question itself ask for the values of $a$ and $b$, but nowhere mentions that they are unique. However, any question saying "find the values of $a$ and $b$" is wrong with the word "the" : they are assuming uniqueness of $a$ and $b$, which is not the case.The question as quoted by you includes the word "the" , and this is misleading.
Yes, actually the statement "find 'the' values of a and b" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you.
â Zebrafish
Aug 23 at 8:26
You are welcome!
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 9:56
I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently?
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 14:18
add a comment |Â
up vote
3
down vote
For commutative property of product we have that
$$(x + 5)(x - 2)=(x - 2)(x + 5)$$
note that also
$$(-x + 2)(-x - 5)$$
is a correct factorization.
add a comment |Â
up vote
1
down vote
You are right.
$$(x+a)(x+b)=x^2+(a+b)x+ab$$
and by identification with $x^2+3x-10$,
$$begincasesa+b=3,\ab=-10.endcases$$
This is a non-linear system of equations, and given commutativity of addition and multiplication, it is clear that if $(u,v)$ is a solution, so is $(v,u)$.
Now one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write
$$3a=(a+b)a=a^2+ab=a^2-10$$
which is the original equation (with a sign reversal)
$$a^2-3a-10=0.$$
To be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots.
So there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
Yes, you are correct. Since $(x+5)(x-2) = (x-2)(x+5) = x^2 + 3x-10$, we note that $a$ and $b$ may either take the values $(5,-2)$ or $(-2,5)$.
I would consider providing just one of the two solutions to be insufficient, since the question itself ask for the values of $a$ and $b$, but nowhere mentions that they are unique. However, any question saying "find the values of $a$ and $b$" is wrong with the word "the" : they are assuming uniqueness of $a$ and $b$, which is not the case.The question as quoted by you includes the word "the" , and this is misleading.
Yes, actually the statement "find 'the' values of a and b" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you.
â Zebrafish
Aug 23 at 8:26
You are welcome!
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 9:56
I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently?
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 14:18
add a comment |Â
up vote
11
down vote
accepted
Yes, you are correct. Since $(x+5)(x-2) = (x-2)(x+5) = x^2 + 3x-10$, we note that $a$ and $b$ may either take the values $(5,-2)$ or $(-2,5)$.
I would consider providing just one of the two solutions to be insufficient, since the question itself ask for the values of $a$ and $b$, but nowhere mentions that they are unique. However, any question saying "find the values of $a$ and $b$" is wrong with the word "the" : they are assuming uniqueness of $a$ and $b$, which is not the case.The question as quoted by you includes the word "the" , and this is misleading.
Yes, actually the statement "find 'the' values of a and b" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you.
â Zebrafish
Aug 23 at 8:26
You are welcome!
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 9:56
I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently?
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 14:18
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
Yes, you are correct. Since $(x+5)(x-2) = (x-2)(x+5) = x^2 + 3x-10$, we note that $a$ and $b$ may either take the values $(5,-2)$ or $(-2,5)$.
I would consider providing just one of the two solutions to be insufficient, since the question itself ask for the values of $a$ and $b$, but nowhere mentions that they are unique. However, any question saying "find the values of $a$ and $b$" is wrong with the word "the" : they are assuming uniqueness of $a$ and $b$, which is not the case.The question as quoted by you includes the word "the" , and this is misleading.
Yes, you are correct. Since $(x+5)(x-2) = (x-2)(x+5) = x^2 + 3x-10$, we note that $a$ and $b$ may either take the values $(5,-2)$ or $(-2,5)$.
I would consider providing just one of the two solutions to be insufficient, since the question itself ask for the values of $a$ and $b$, but nowhere mentions that they are unique. However, any question saying "find the values of $a$ and $b$" is wrong with the word "the" : they are assuming uniqueness of $a$ and $b$, which is not the case.The question as quoted by you includes the word "the" , and this is misleading.
edited Aug 23 at 16:15
answered Aug 23 at 8:22
ðÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
33.4k22870
33.4k22870
Yes, actually the statement "find 'the' values of a and b" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you.
â Zebrafish
Aug 23 at 8:26
You are welcome!
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 9:56
I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently?
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 14:18
add a comment |Â
Yes, actually the statement "find 'the' values of a and b" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you.
â Zebrafish
Aug 23 at 8:26
You are welcome!
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 9:56
I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently?
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 14:18
Yes, actually the statement "find 'the' values of a and b" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you.
â Zebrafish
Aug 23 at 8:26
Yes, actually the statement "find 'the' values of a and b" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you.
â Zebrafish
Aug 23 at 8:26
You are welcome!
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 9:56
You are welcome!
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 9:56
I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently?
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 14:18
I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently?
â Ã°ÃÂÃÂþý òÃÂûûð þûþàüÃÂûûñÃÂÃÂó
Aug 23 at 14:18
add a comment |Â
up vote
3
down vote
For commutative property of product we have that
$$(x + 5)(x - 2)=(x - 2)(x + 5)$$
note that also
$$(-x + 2)(-x - 5)$$
is a correct factorization.
add a comment |Â
up vote
3
down vote
For commutative property of product we have that
$$(x + 5)(x - 2)=(x - 2)(x + 5)$$
note that also
$$(-x + 2)(-x - 5)$$
is a correct factorization.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For commutative property of product we have that
$$(x + 5)(x - 2)=(x - 2)(x + 5)$$
note that also
$$(-x + 2)(-x - 5)$$
is a correct factorization.
For commutative property of product we have that
$$(x + 5)(x - 2)=(x - 2)(x + 5)$$
note that also
$$(-x + 2)(-x - 5)$$
is a correct factorization.
edited Aug 23 at 8:27
answered Aug 23 at 8:20
gimusi
70.1k73786
70.1k73786
add a comment |Â
add a comment |Â
up vote
1
down vote
You are right.
$$(x+a)(x+b)=x^2+(a+b)x+ab$$
and by identification with $x^2+3x-10$,
$$begincasesa+b=3,\ab=-10.endcases$$
This is a non-linear system of equations, and given commutativity of addition and multiplication, it is clear that if $(u,v)$ is a solution, so is $(v,u)$.
Now one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write
$$3a=(a+b)a=a^2+ab=a^2-10$$
which is the original equation (with a sign reversal)
$$a^2-3a-10=0.$$
To be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots.
So there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$.
add a comment |Â
up vote
1
down vote
You are right.
$$(x+a)(x+b)=x^2+(a+b)x+ab$$
and by identification with $x^2+3x-10$,
$$begincasesa+b=3,\ab=-10.endcases$$
This is a non-linear system of equations, and given commutativity of addition and multiplication, it is clear that if $(u,v)$ is a solution, so is $(v,u)$.
Now one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write
$$3a=(a+b)a=a^2+ab=a^2-10$$
which is the original equation (with a sign reversal)
$$a^2-3a-10=0.$$
To be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots.
So there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are right.
$$(x+a)(x+b)=x^2+(a+b)x+ab$$
and by identification with $x^2+3x-10$,
$$begincasesa+b=3,\ab=-10.endcases$$
This is a non-linear system of equations, and given commutativity of addition and multiplication, it is clear that if $(u,v)$ is a solution, so is $(v,u)$.
Now one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write
$$3a=(a+b)a=a^2+ab=a^2-10$$
which is the original equation (with a sign reversal)
$$a^2-3a-10=0.$$
To be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots.
So there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$.
You are right.
$$(x+a)(x+b)=x^2+(a+b)x+ab$$
and by identification with $x^2+3x-10$,
$$begincasesa+b=3,\ab=-10.endcases$$
This is a non-linear system of equations, and given commutativity of addition and multiplication, it is clear that if $(u,v)$ is a solution, so is $(v,u)$.
Now one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write
$$3a=(a+b)a=a^2+ab=a^2-10$$
which is the original equation (with a sign reversal)
$$a^2-3a-10=0.$$
To be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots.
So there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$.
answered Aug 23 at 8:37
Yves Daoust
113k665207
113k665207
add a comment |Â
add a comment |Â
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2
Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$.
â Matti P.
Aug 23 at 8:18
2
They are both valid answers, since the order of the factors doesn't matter.
â Ludvig Lindström
Aug 23 at 8:18
You teacher should have stated what context they want you to find $a$ and $b$, I have provided the answer below under the assumption that they meant the roots of the polynomial equation in $x$.
â Adam
Aug 23 at 8:54
Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case.
â Adam
Aug 23 at 8:58
@Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from Kahn Academy videos. I appreciate that there are many people here to give extra help.
â Zebrafish
Aug 23 at 9:03