Polynomial Function given roots and a point












0












$begingroup$


A 3rd degree polynomial has roots at x=-2i and x=5. The y-intercept is (0,25). Write an equation for this function in factored form with real coefficients.



My first hack is to say that in factored form this is (x-5)(X2+4) as that gets me the roots, but that has y-intercept (0,-20). What am I missing?



Thanks!






polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
    $endgroup$
    – jerH
    Dec 14 '18 at 2:11










  • $begingroup$
    The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 2:15
















0












$begingroup$


A 3rd degree polynomial has roots at x=-2i and x=5. The y-intercept is (0,25). Write an equation for this function in factored form with real coefficients.



My first hack is to say that in factored form this is (x-5)(X2+4) as that gets me the roots, but that has y-intercept (0,-20). What am I missing?



Thanks!






polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
    $endgroup$
    – jerH
    Dec 14 '18 at 2:11










  • $begingroup$
    The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 2:15














0












0








0





$begingroup$


A 3rd degree polynomial has roots at x=-2i and x=5. The y-intercept is (0,25). Write an equation for this function in factored form with real coefficients.



My first hack is to say that in factored form this is (x-5)(X2+4) as that gets me the roots, but that has y-intercept (0,-20). What am I missing?



Thanks!






polynomials







share|cite|improve this question









$endgroup$




A 3rd degree polynomial has roots at x=-2i and x=5. The y-intercept is (0,25). Write an equation for this function in factored form with real coefficients.



My first hack is to say that in factored form this is (x-5)(X2+4) as that gets me the roots, but that has y-intercept (0,-20). What am I missing?



Thanks!






polynomials




polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 2:02









jerHjerH

1104




1104












  • $begingroup$
    Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
    $endgroup$
    – jerH
    Dec 14 '18 at 2:11










  • $begingroup$
    The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 2:15


















  • $begingroup$
    Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
    $endgroup$
    – jerH
    Dec 14 '18 at 2:11










  • $begingroup$
    The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 2:15
















$begingroup$
Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
$endgroup$
– jerH
Dec 14 '18 at 2:11




$begingroup$
Duh! Thanks. -2i was how the problem was written, but yeah, they come in pairs don't they...if you want to stick that in as an answer I'll gladly accept it
$endgroup$
– jerH
Dec 14 '18 at 2:11












$begingroup$
The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
$endgroup$
– SmileyCraft
Dec 14 '18 at 2:15




$begingroup$
The way it states that the polynomial has real coefficients is a bit vague, but sure, then $-2i$ gives $+2i$ as well.
$endgroup$
– SmileyCraft
Dec 14 '18 at 2:15










2 Answers
2






active

oldest

votes


















2












$begingroup$

To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...




The Roots:



On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as



$$f(x) = a(x - 5)(x - 2i)(x + 2i)$$



up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.



Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).




Finding $a$:



So now you know that the polynomial has the form



$$f(x) = a(x-5)(x^2 + 4)$$



The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.



(Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)



We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.



Plugging in these values, then, we see



$$25 = a(-5)(4) ;;; Rightarrow ;;; 25 = a(-20) ;;; Rightarrow ;;; a = frac{25}{-20} = -frac{5}{4}$$



Thus, the polynomial in its factored form is



$$f(x) = left( -frac{5}{4} right)(x-5)(x^2 + 4)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 3:11






  • 1




    $begingroup$
    Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
    $endgroup$
    – Eevee Trainer
    Dec 14 '18 at 3:13



















1












$begingroup$

You solve the problem by simply multiplying the polynomial by $frac{25}{-20}$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...




    The Roots:



    On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as



    $$f(x) = a(x - 5)(x - 2i)(x + 2i)$$



    up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.



    Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).




    Finding $a$:



    So now you know that the polynomial has the form



    $$f(x) = a(x-5)(x^2 + 4)$$



    The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.



    (Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)



    We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.



    Plugging in these values, then, we see



    $$25 = a(-5)(4) ;;; Rightarrow ;;; 25 = a(-20) ;;; Rightarrow ;;; a = frac{25}{-20} = -frac{5}{4}$$



    Thus, the polynomial in its factored form is



    $$f(x) = left( -frac{5}{4} right)(x-5)(x^2 + 4)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
      $endgroup$
      – SmileyCraft
      Dec 14 '18 at 3:11






    • 1




      $begingroup$
      Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
      $endgroup$
      – Eevee Trainer
      Dec 14 '18 at 3:13
















    2












    $begingroup$

    To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...




    The Roots:



    On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as



    $$f(x) = a(x - 5)(x - 2i)(x + 2i)$$



    up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.



    Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).




    Finding $a$:



    So now you know that the polynomial has the form



    $$f(x) = a(x-5)(x^2 + 4)$$



    The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.



    (Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)



    We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.



    Plugging in these values, then, we see



    $$25 = a(-5)(4) ;;; Rightarrow ;;; 25 = a(-20) ;;; Rightarrow ;;; a = frac{25}{-20} = -frac{5}{4}$$



    Thus, the polynomial in its factored form is



    $$f(x) = left( -frac{5}{4} right)(x-5)(x^2 + 4)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
      $endgroup$
      – SmileyCraft
      Dec 14 '18 at 3:11






    • 1




      $begingroup$
      Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
      $endgroup$
      – Eevee Trainer
      Dec 14 '18 at 3:13














    2












    2








    2





    $begingroup$

    To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...




    The Roots:



    On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as



    $$f(x) = a(x - 5)(x - 2i)(x + 2i)$$



    up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.



    Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).




    Finding $a$:



    So now you know that the polynomial has the form



    $$f(x) = a(x-5)(x^2 + 4)$$



    The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.



    (Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)



    We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.



    Plugging in these values, then, we see



    $$25 = a(-5)(4) ;;; Rightarrow ;;; 25 = a(-20) ;;; Rightarrow ;;; a = frac{25}{-20} = -frac{5}{4}$$



    Thus, the polynomial in its factored form is



    $$f(x) = left( -frac{5}{4} right)(x-5)(x^2 + 4)$$






    share|cite|improve this answer









    $endgroup$



    To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...




    The Roots:



    On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as



    $$f(x) = a(x - 5)(x - 2i)(x + 2i)$$



    up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.



    Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).




    Finding $a$:



    So now you know that the polynomial has the form



    $$f(x) = a(x-5)(x^2 + 4)$$



    The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.



    (Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)



    We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.



    Plugging in these values, then, we see



    $$25 = a(-5)(4) ;;; Rightarrow ;;; 25 = a(-20) ;;; Rightarrow ;;; a = frac{25}{-20} = -frac{5}{4}$$



    Thus, the polynomial in its factored form is



    $$f(x) = left( -frac{5}{4} right)(x-5)(x^2 + 4)$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 14 '18 at 2:30









    Eevee TrainerEevee Trainer

    5,9471936




    5,9471936












    • $begingroup$
      In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
      $endgroup$
      – SmileyCraft
      Dec 14 '18 at 3:11






    • 1




      $begingroup$
      Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
      $endgroup$
      – Eevee Trainer
      Dec 14 '18 at 3:13


















    • $begingroup$
      In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
      $endgroup$
      – SmileyCraft
      Dec 14 '18 at 3:11






    • 1




      $begingroup$
      Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
      $endgroup$
      – Eevee Trainer
      Dec 14 '18 at 3:13
















    $begingroup$
    In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 3:11




    $begingroup$
    In my defense, from the question I thought it was quite clear that jerH has a good idea of what he's doing, so I thought they should be able to understand why my answer is correct.
    $endgroup$
    – SmileyCraft
    Dec 14 '18 at 3:11




    1




    1




    $begingroup$
    Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
    $endgroup$
    – Eevee Trainer
    Dec 14 '18 at 3:13




    $begingroup$
    Yeah I suppose I can sort of see that. I don't think he actually did, though, seeing as he described his method as a hack. (I feel like he came up with the $(x-5)(x^2 + 4)$ more by trial and error than a more thorough process.) That was my big hint that a detailed answer would be more appropriate.
    $endgroup$
    – Eevee Trainer
    Dec 14 '18 at 3:13











    1












    $begingroup$

    You solve the problem by simply multiplying the polynomial by $frac{25}{-20}$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You solve the problem by simply multiplying the polynomial by $frac{25}{-20}$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You solve the problem by simply multiplying the polynomial by $frac{25}{-20}$.






        share|cite|improve this answer









        $endgroup$



        You solve the problem by simply multiplying the polynomial by $frac{25}{-20}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 2:14









        SmileyCraftSmileyCraft

        3,591517




        3,591517






























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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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