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!










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!










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!










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






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share|cite|improve this question











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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

    oldest

    votes








    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

    5,9771936




    5,9771936












    • $begingroup$
      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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