Factorise the following p(x) as the product of a linear term and a quadratic polynomial with no real roots.











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Factorise the following p(x) functions as the product of a linear term and a quadratic polynomial with no real roots (or if there are real roots, factorise to irreducible form and find them):
1. $$p(x)=x^3+x$$
2. $$p(x)=x^4+x^2$$
3. $$p(x)=x^3-x$$






linear-algebra polynomials roots factoring irreducible-polynomials







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  • What have you tried? Where are you confused?
    – Dzoooks
    Nov 22 at 4:45















up vote
-1
down vote

favorite
1












Factorise the following p(x) functions as the product of a linear term and a quadratic polynomial with no real roots (or if there are real roots, factorise to irreducible form and find them):
1. $$p(x)=x^3+x$$
2. $$p(x)=x^4+x^2$$
3. $$p(x)=x^3-x$$






linear-algebra polynomials roots factoring irreducible-polynomials







share|cite|improve this question






















  • What have you tried? Where are you confused?
    – Dzoooks
    Nov 22 at 4:45













up vote
-1
down vote

favorite
1









up vote
-1
down vote

favorite
1






1





Factorise the following p(x) functions as the product of a linear term and a quadratic polynomial with no real roots (or if there are real roots, factorise to irreducible form and find them):
1. $$p(x)=x^3+x$$
2. $$p(x)=x^4+x^2$$
3. $$p(x)=x^3-x$$






linear-algebra polynomials roots factoring irreducible-polynomials







share|cite|improve this question













Factorise the following p(x) functions as the product of a linear term and a quadratic polynomial with no real roots (or if there are real roots, factorise to irreducible form and find them):
1. $$p(x)=x^3+x$$
2. $$p(x)=x^4+x^2$$
3. $$p(x)=x^3-x$$






linear-algebra polynomials roots factoring irreducible-polynomials




linear-algebra polynomials roots factoring irreducible-polynomials






share|cite|improve this question













share|cite|improve this question











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asked Nov 22 at 4:36









Kayla Martin

636




636












  • What have you tried? Where are you confused?
    – Dzoooks
    Nov 22 at 4:45


















  • What have you tried? Where are you confused?
    – Dzoooks
    Nov 22 at 4:45
















What have you tried? Where are you confused?
– Dzoooks
Nov 22 at 4:45




What have you tried? Where are you confused?
– Dzoooks
Nov 22 at 4:45










2 Answers
2






active

oldest

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up vote
1
down vote



accepted










Really, these polynomials target a single skill: taking out a common factor. For example, with $x^2+x$, both terms have $x$ as a factor, so you can divide it out (inverse distributive property of sorts) and get $x(x+1)$.



$$$$





  1. $$x^3+x = x(x^2+1)$$
    Roots are $0$, $i$, and $-i$.


  2. $$x^4+x^2 = x^2(x^2+1)$$
    Roots are $0$, $i$, and $-i$.


  3. $$x^3-x = x(x^2-1) = x(x+1)(x-1)$$
    Roots are $0$, $1$, and $-1$.






share|cite|improve this answer





















  • Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
    – Kayla Martin
    Nov 22 at 4:54










  • @KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
    – idea
    Nov 22 at 4:59










  • Okay, thank you!
    – Kayla Martin
    Nov 22 at 5:01


















up vote
1
down vote













I think it's important you understand this idea for more difficult cases than you ask about, so my answer will touch on a more general method for solving these problems, and show you how it works for slightly more difficult cases.



We have that if for some polynomial $p(x)$, $p(a)=0$, then:



$$p(x)=(x-a)cdot q(x)$$



Where $q(x)$ is a polynomial one degree smaller than $p(x)$ (the degree of a polynomial just means the largest power of $x$ in it).



If you can find any point where a polynomial $p(x) $ has a zero, you can take out the factor in this way and find the smaller polynomial $q(x)$, which can either be solved or reduced again.



Let me demonstrate first on (b)



We have that $p(x)=x^4-x^2$
I like to test values $0,1,-1,2,-2$ etc... Here, it's plain that $p (0)=0$, so we take out $(x-0)$, or just $x$. We now have that:



$$x^4-x^2=xcdot q(x)$$



From which we get that $q(x)=x^3-x$. But $q(0)=0$, so we can take out another factor $x$, giving $$x^4-x^2=x^2cdot r (x)$$ It follows that $r(x)=x^2+1$, a quadratic we can solve by the Quadratic Formula.



This is a rather bland example, so I'll demonstrate the procedure on a better example.



Suppose we have $$f(x)=x^3+4x^2+7x+4$$



Plugging in values for $x$, we get that $f(-1)=0$



So we have that:



$$x^3+4x^2+7x+4=(x+1)cdot g(x)$$
Or rather,



$$x^3+4x^2+7x+4=(x+1)(x^2+px+q)$$ for some real $p,q$.



Expanding this out we get:



$$x^3+4x^2+7x+4=x^3+(p+1)x^2+(p+q)x+q$$



From which it follows that $p=3, q=4$.



So: $$f(x)=(x+1)(x^2+3x+4)$$, which is a quadratic we can solve (even if the solutions aren't real)






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    2 Answers
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    accepted










    Really, these polynomials target a single skill: taking out a common factor. For example, with $x^2+x$, both terms have $x$ as a factor, so you can divide it out (inverse distributive property of sorts) and get $x(x+1)$.



    $$$$





    1. $$x^3+x = x(x^2+1)$$
      Roots are $0$, $i$, and $-i$.


    2. $$x^4+x^2 = x^2(x^2+1)$$
      Roots are $0$, $i$, and $-i$.


    3. $$x^3-x = x(x^2-1) = x(x+1)(x-1)$$
      Roots are $0$, $1$, and $-1$.






    share|cite|improve this answer





















    • Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
      – Kayla Martin
      Nov 22 at 4:54










    • @KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
      – idea
      Nov 22 at 4:59










    • Okay, thank you!
      – Kayla Martin
      Nov 22 at 5:01















    up vote
    1
    down vote



    accepted










    Really, these polynomials target a single skill: taking out a common factor. For example, with $x^2+x$, both terms have $x$ as a factor, so you can divide it out (inverse distributive property of sorts) and get $x(x+1)$.



    $$$$





    1. $$x^3+x = x(x^2+1)$$
      Roots are $0$, $i$, and $-i$.


    2. $$x^4+x^2 = x^2(x^2+1)$$
      Roots are $0$, $i$, and $-i$.


    3. $$x^3-x = x(x^2-1) = x(x+1)(x-1)$$
      Roots are $0$, $1$, and $-1$.






    share|cite|improve this answer





















    • Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
      – Kayla Martin
      Nov 22 at 4:54










    • @KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
      – idea
      Nov 22 at 4:59










    • Okay, thank you!
      – Kayla Martin
      Nov 22 at 5:01













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    Really, these polynomials target a single skill: taking out a common factor. For example, with $x^2+x$, both terms have $x$ as a factor, so you can divide it out (inverse distributive property of sorts) and get $x(x+1)$.



    $$$$





    1. $$x^3+x = x(x^2+1)$$
      Roots are $0$, $i$, and $-i$.


    2. $$x^4+x^2 = x^2(x^2+1)$$
      Roots are $0$, $i$, and $-i$.


    3. $$x^3-x = x(x^2-1) = x(x+1)(x-1)$$
      Roots are $0$, $1$, and $-1$.






    share|cite|improve this answer












    Really, these polynomials target a single skill: taking out a common factor. For example, with $x^2+x$, both terms have $x$ as a factor, so you can divide it out (inverse distributive property of sorts) and get $x(x+1)$.



    $$$$





    1. $$x^3+x = x(x^2+1)$$
      Roots are $0$, $i$, and $-i$.


    2. $$x^4+x^2 = x^2(x^2+1)$$
      Roots are $0$, $i$, and $-i$.


    3. $$x^3-x = x(x^2-1) = x(x+1)(x-1)$$
      Roots are $0$, $1$, and $-1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 22 at 4:49









    Christopher Marley

    905115




    905115












    • Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
      – Kayla Martin
      Nov 22 at 4:54










    • @KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
      – idea
      Nov 22 at 4:59










    • Okay, thank you!
      – Kayla Martin
      Nov 22 at 5:01


















    • Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
      – Kayla Martin
      Nov 22 at 4:54










    • @KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
      – idea
      Nov 22 at 4:59










    • Okay, thank you!
      – Kayla Martin
      Nov 22 at 5:01
















    Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
    – Kayla Martin
    Nov 22 at 4:54




    Ah, see this is exactly what I thought to do at first, but then the wording of the question threw me. What do they mean by "product of a linear term"? Isn't 0 a real root? Is it possible to take 1 and 2 further so that the terms only give complex roots and no real roots? Or is that impossible?
    – Kayla Martin
    Nov 22 at 4:54












    @KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
    – idea
    Nov 22 at 4:59




    @KaylaMartin if p(x) has a real root, it will be there however further you take it. Linear term will obviously give you a real root. So, it is impossible.
    – idea
    Nov 22 at 4:59












    Okay, thank you!
    – Kayla Martin
    Nov 22 at 5:01




    Okay, thank you!
    – Kayla Martin
    Nov 22 at 5:01










    up vote
    1
    down vote













    I think it's important you understand this idea for more difficult cases than you ask about, so my answer will touch on a more general method for solving these problems, and show you how it works for slightly more difficult cases.



    We have that if for some polynomial $p(x)$, $p(a)=0$, then:



    $$p(x)=(x-a)cdot q(x)$$



    Where $q(x)$ is a polynomial one degree smaller than $p(x)$ (the degree of a polynomial just means the largest power of $x$ in it).



    If you can find any point where a polynomial $p(x) $ has a zero, you can take out the factor in this way and find the smaller polynomial $q(x)$, which can either be solved or reduced again.



    Let me demonstrate first on (b)



    We have that $p(x)=x^4-x^2$
    I like to test values $0,1,-1,2,-2$ etc... Here, it's plain that $p (0)=0$, so we take out $(x-0)$, or just $x$. We now have that:



    $$x^4-x^2=xcdot q(x)$$



    From which we get that $q(x)=x^3-x$. But $q(0)=0$, so we can take out another factor $x$, giving $$x^4-x^2=x^2cdot r (x)$$ It follows that $r(x)=x^2+1$, a quadratic we can solve by the Quadratic Formula.



    This is a rather bland example, so I'll demonstrate the procedure on a better example.



    Suppose we have $$f(x)=x^3+4x^2+7x+4$$



    Plugging in values for $x$, we get that $f(-1)=0$



    So we have that:



    $$x^3+4x^2+7x+4=(x+1)cdot g(x)$$
    Or rather,



    $$x^3+4x^2+7x+4=(x+1)(x^2+px+q)$$ for some real $p,q$.



    Expanding this out we get:



    $$x^3+4x^2+7x+4=x^3+(p+1)x^2+(p+q)x+q$$



    From which it follows that $p=3, q=4$.



    So: $$f(x)=(x+1)(x^2+3x+4)$$, which is a quadratic we can solve (even if the solutions aren't real)






    share|cite|improve this answer

























      up vote
      1
      down vote













      I think it's important you understand this idea for more difficult cases than you ask about, so my answer will touch on a more general method for solving these problems, and show you how it works for slightly more difficult cases.



      We have that if for some polynomial $p(x)$, $p(a)=0$, then:



      $$p(x)=(x-a)cdot q(x)$$



      Where $q(x)$ is a polynomial one degree smaller than $p(x)$ (the degree of a polynomial just means the largest power of $x$ in it).



      If you can find any point where a polynomial $p(x) $ has a zero, you can take out the factor in this way and find the smaller polynomial $q(x)$, which can either be solved or reduced again.



      Let me demonstrate first on (b)



      We have that $p(x)=x^4-x^2$
      I like to test values $0,1,-1,2,-2$ etc... Here, it's plain that $p (0)=0$, so we take out $(x-0)$, or just $x$. We now have that:



      $$x^4-x^2=xcdot q(x)$$



      From which we get that $q(x)=x^3-x$. But $q(0)=0$, so we can take out another factor $x$, giving $$x^4-x^2=x^2cdot r (x)$$ It follows that $r(x)=x^2+1$, a quadratic we can solve by the Quadratic Formula.



      This is a rather bland example, so I'll demonstrate the procedure on a better example.



      Suppose we have $$f(x)=x^3+4x^2+7x+4$$



      Plugging in values for $x$, we get that $f(-1)=0$



      So we have that:



      $$x^3+4x^2+7x+4=(x+1)cdot g(x)$$
      Or rather,



      $$x^3+4x^2+7x+4=(x+1)(x^2+px+q)$$ for some real $p,q$.



      Expanding this out we get:



      $$x^3+4x^2+7x+4=x^3+(p+1)x^2+(p+q)x+q$$



      From which it follows that $p=3, q=4$.



      So: $$f(x)=(x+1)(x^2+3x+4)$$, which is a quadratic we can solve (even if the solutions aren't real)






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        I think it's important you understand this idea for more difficult cases than you ask about, so my answer will touch on a more general method for solving these problems, and show you how it works for slightly more difficult cases.



        We have that if for some polynomial $p(x)$, $p(a)=0$, then:



        $$p(x)=(x-a)cdot q(x)$$



        Where $q(x)$ is a polynomial one degree smaller than $p(x)$ (the degree of a polynomial just means the largest power of $x$ in it).



        If you can find any point where a polynomial $p(x) $ has a zero, you can take out the factor in this way and find the smaller polynomial $q(x)$, which can either be solved or reduced again.



        Let me demonstrate first on (b)



        We have that $p(x)=x^4-x^2$
        I like to test values $0,1,-1,2,-2$ etc... Here, it's plain that $p (0)=0$, so we take out $(x-0)$, or just $x$. We now have that:



        $$x^4-x^2=xcdot q(x)$$



        From which we get that $q(x)=x^3-x$. But $q(0)=0$, so we can take out another factor $x$, giving $$x^4-x^2=x^2cdot r (x)$$ It follows that $r(x)=x^2+1$, a quadratic we can solve by the Quadratic Formula.



        This is a rather bland example, so I'll demonstrate the procedure on a better example.



        Suppose we have $$f(x)=x^3+4x^2+7x+4$$



        Plugging in values for $x$, we get that $f(-1)=0$



        So we have that:



        $$x^3+4x^2+7x+4=(x+1)cdot g(x)$$
        Or rather,



        $$x^3+4x^2+7x+4=(x+1)(x^2+px+q)$$ for some real $p,q$.



        Expanding this out we get:



        $$x^3+4x^2+7x+4=x^3+(p+1)x^2+(p+q)x+q$$



        From which it follows that $p=3, q=4$.



        So: $$f(x)=(x+1)(x^2+3x+4)$$, which is a quadratic we can solve (even if the solutions aren't real)






        share|cite|improve this answer












        I think it's important you understand this idea for more difficult cases than you ask about, so my answer will touch on a more general method for solving these problems, and show you how it works for slightly more difficult cases.



        We have that if for some polynomial $p(x)$, $p(a)=0$, then:



        $$p(x)=(x-a)cdot q(x)$$



        Where $q(x)$ is a polynomial one degree smaller than $p(x)$ (the degree of a polynomial just means the largest power of $x$ in it).



        If you can find any point where a polynomial $p(x) $ has a zero, you can take out the factor in this way and find the smaller polynomial $q(x)$, which can either be solved or reduced again.



        Let me demonstrate first on (b)



        We have that $p(x)=x^4-x^2$
        I like to test values $0,1,-1,2,-2$ etc... Here, it's plain that $p (0)=0$, so we take out $(x-0)$, or just $x$. We now have that:



        $$x^4-x^2=xcdot q(x)$$



        From which we get that $q(x)=x^3-x$. But $q(0)=0$, so we can take out another factor $x$, giving $$x^4-x^2=x^2cdot r (x)$$ It follows that $r(x)=x^2+1$, a quadratic we can solve by the Quadratic Formula.



        This is a rather bland example, so I'll demonstrate the procedure on a better example.



        Suppose we have $$f(x)=x^3+4x^2+7x+4$$



        Plugging in values for $x$, we get that $f(-1)=0$



        So we have that:



        $$x^3+4x^2+7x+4=(x+1)cdot g(x)$$
        Or rather,



        $$x^3+4x^2+7x+4=(x+1)(x^2+px+q)$$ for some real $p,q$.



        Expanding this out we get:



        $$x^3+4x^2+7x+4=x^3+(p+1)x^2+(p+q)x+q$$



        From which it follows that $p=3, q=4$.



        So: $$f(x)=(x+1)(x^2+3x+4)$$, which is a quadratic we can solve (even if the solutions aren't real)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 5:50









        Rhys Hughes

        4,6651327




        4,6651327






























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            Quarter-circle Tiles

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            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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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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