1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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
    Last edited by a moderator: Apr 23, 2017
  6. Jan 2, 2008 #5

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    Yes, that's already been discussed in several responses.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook