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.







share|cite|improve this question


















  • 2




    Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$.
    – Matti P.
    Aug 23 at 8:18






  • 2




    They are both valid answers, since the order of the factors doesn't matter.
    – 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 $x$.
    – 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.
    – 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.
    – Zebrafish
    Aug 23 at 9:03














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.







share|cite|improve this question


















  • 2




    Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$.
    – Matti P.
    Aug 23 at 8:18






  • 2




    They are both valid answers, since the order of the factors doesn't matter.
    – 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 $x$.
    – 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.
    – 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.
    – Zebrafish
    Aug 23 at 9:03












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.







share|cite|improve this question














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.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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 $(a,b) = (5,-2)$.
    – Matti P.
    Aug 23 at 8:18






  • 2




    They are both valid answers, since the order of the factors doesn't matter.
    – 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 $x$.
    – 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.
    – 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.
    – Zebrafish
    Aug 23 at 9:03












  • 2




    Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$.
    – Matti P.
    Aug 23 at 8:18






  • 2




    They are both valid answers, since the order of the factors doesn't matter.
    – 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 $x$.
    – 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.
    – 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.
    – 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 $(a,b) = (5,-2)$.
– 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 $(a,b) = (5,-2)$.
– Matti P.
Aug 23 at 8:18




2




2




They are both valid answers, since the order of the factors doesn't matter.
– Ludvig Lindström
Aug 23 at 8:18




They are both valid answers, since the order of the factors doesn't matter.
– 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 $x$.
– 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 $x$.
– 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.
– 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.
– 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.
– 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.
– Zebrafish
Aug 23 at 9:03










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.






share|cite|improve this answer






















  • 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.
    – Zebrafish
    Aug 23 at 8:26










  • You are welcome!
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    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?
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 23 at 14:18


















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.






share|cite|improve this answer





























    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 $a=5,b=-2$.






    share|cite|improve this answer




















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      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.






      share|cite|improve this answer






















      • 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.
        – Zebrafish
        Aug 23 at 8:26










      • You are welcome!
        – Ð°ÑÑ‚он вілла олоф мэллбэрг
        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?
        – Ð°ÑÑ‚он вілла олоф мэллбэрг
        Aug 23 at 14:18















      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.






      share|cite|improve this answer






















      • 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.
        – Zebrafish
        Aug 23 at 8:26










      • You are welcome!
        – Ð°ÑÑ‚он вілла олоф мэллбэрг
        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?
        – Ð°ÑÑ‚он вілла олоф мэллбэрг
        Aug 23 at 14:18













      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.






      share|cite|improve this answer














      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.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Aug 23 at 16:15

























      answered Aug 23 at 8:22









      астон вілла олоф мэллбэрг

      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.
        – Zebrafish
        Aug 23 at 8:26










      • You are welcome!
        – Ð°ÑÑ‚он вілла олоф мэллбэрг
        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?
        – Ð°ÑÑ‚он вілла олоф мэллбэрг
        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.
        – Zebrafish
        Aug 23 at 8:26










      • You are welcome!
        – Ð°ÑÑ‚он вілла олоф мэллбэрг
        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?
        – Ð°ÑÑ‚он вілла олоф мэллбэрг
        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.
      – 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.
      – Zebrafish
      Aug 23 at 8:26












      You are welcome!
      – Ð°ÑÑ‚он вілла олоф мэллбэрг
      Aug 23 at 9:56




      You are welcome!
      – Ð°ÑÑ‚он вілла олоф мэллбэрг
      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?
      – Ð°ÑÑ‚он вілла олоф мэллбэрг
      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?
      – Ð°ÑÑ‚он вілла олоф мэллбэрг
      Aug 23 at 14:18











      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.






      share|cite|improve this answer


























        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.






        share|cite|improve this answer
























          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 23 at 8:27

























          answered Aug 23 at 8:20









          gimusi

          70.1k73786




          70.1k73786




















              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 $a=5,b=-2$.






              share|cite|improve this answer
























                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 $a=5,b=-2$.






                share|cite|improve this answer






















                  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 $a=5,b=-2$.






                  share|cite|improve this answer












                  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 $a=5,b=-2$.







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                  answered Aug 23 at 8:37









                  Yves Daoust

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