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Factorising a cubic polynomial

  1. Jan 1, 2008 #1
    1. A cubic polynomial is given by (f)x = x3 + x2 - 10x + 8

    1) i) Show that (x -1) is a factor of f(x)

    ii) Factorise f(x) fully

    iii) Sketch the graph



    2. Factor theorem?



    3. Attempt at solution

    i) I know that f(1) = 0, so (x - 1) is a factor of (f)x = x3 + x2 - 10x + 8

    ii) I think i am correct in saying that since you know (x - 1) is a factor you can write it like this; ( x-1 )( x2 - 10x + 8 ), i just dont know where to go from there. I dont think you can factorise x2 - 10x + 8 any further?

    I presume that by factorising fully, they mean in the form (x-a)(x-b)(x-c), i just cant seem to get there.
     
  2. jcsd
  3. Jan 1, 2008 #2
    i) is correct

    ii) is not because you cannot just jump to that answer. To get the factored form, you must divide f(x) by x-1 either by long division or synthetic division and then get the other factor.
     
  4. Jan 1, 2008 #3

    HallsofIvy

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    How did you get that? Just by dropping the x3? Now you know that's not right! As Plasmasphere said, you could divide [itex]x^ x^2- 10x+ 8[/itex] by x-1. Or you can "work backwards. You want to find [itex]ax^2+ bx+ c[/itex] so that [itex](x-1)(ax^2+ bx+ c)= x^3+ x^2- 10x+ 8[/itex]. Obviously, since [itex]x(x^2)= x^3[/itex] and [itex]-1(-8)= 8[/itex], a= 1, c= -8 (NOT +8!). Okay, multiply out [itex](x-1)(x^2+ bx- 8)= x^3+ x^2- 20x+ 8[/itex] to see what b must be.

     
  5. Jan 2, 2008 #4
    Take a look here.
     
  6. Jan 2, 2008 #5
    http://en.wikipedia.org/wiki/Polynomial_long_division

    Is the wiki page for the technique used in the link given by kbaumen.

    You should divide the original polynomial by the given factor; that would leave you with a quadratic polynomial, which you should be able to factor without any difficulty.
     
  7. Jan 2, 2008 #6
    Thanks for all the help, i used the working backwards method, and i got the right answer according to the answer booklet, the sketch for part iii) was easy too.

    I'm not sure why i thought that i could write it as; ( x-1 )( x2 - 10x + 8 ), if you expand that it clearly will never give the correct answer. I was jumping two steps ahead.

    Cheers, those pages on polynomial long division should be helpful in the future too.
     
    Last edited: Jan 2, 2008
  8. Jan 2, 2008 #7
    I don't understand that step (?) just curious how you got there
     
    Last edited: Jan 2, 2008
  9. Jan 3, 2008 #8

    HallsofIvy

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    If you are wondering about the "20x", that was a typo. The right hand side should be just the original polynomial, [itex]x^3+ x^2-10x+ 8[/itex].
     
  10. Jan 3, 2008 #9
    Actually I meant the Horner's scheme to be used for factorizing the equation. Sorry, if I made that unclear.
     
  11. Jan 3, 2008 #10
    What if we take division of the polynomial ?
    ( x3 + x2 - 10x + 8 ) / ( x-1)
     
  12. Jan 3, 2008 #11

    HallsofIvy

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    Yes, that's already been discussed in several responses.
     
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