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Show that f(x) = 0 has only one root

  1. Jul 6, 2016 #1
    1. The problem statement, all variables and given/known data

    f(x) = 2x3+ax2+bx+10
    When f(x)/(2x-1) the remainder is 12
    When f(x)/(x+1) there is no remainder
    a) Find the value of a and b
    b) Show that f(x) = 0 has only one root


    2. Relevant equations

    None

    3. The attempt at a solution


    a) (2x-1)=0
    x=1/2

    f(1/2) = 12 = 2(1/2)3+b(1/2)2+a(1/2)+10
    1/4+a/4+b/2+10=12
    1+a+2b+40=48
    a+2b=7

    (x+1)=0
    x=-1

    f(-1)= 0 = 2(-1)3+a(-1)2+b(-1)+10
    a-b+8=0
    b=a+8

    b=(7-2b)+8
    b=15-2b
    3b=15
    b=5

    a=7-2b
    a=7-10
    a=-3

    a=-3 and b=5

    b) 2x3-3x2+5x+10=0
    Now I need to factorise this, but I don't know how :frown:
    I tried using x as a common factor, but its not. 10 does not have a factor of x.
     
    Last edited: Jul 6, 2016
  2. jcsd
  3. Jul 6, 2016 #2
    You know x+1 is a factor since the f(-1)=0 or the division by x+1 doesn't leave any reminder.
     
  4. Jul 6, 2016 #3

    Ray Vickson

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    Homework Helper

    The formula you ##f(x) = 2x^3+ax^2 = bx + 10## makes no sense. Do you mean ##f(x) = 2x^3+ax^2 + bx + 10## or ##f(x) = 2x^3 + ax^2 - bx -10##, or what?
     
  5. Jul 6, 2016 #4
    So I use long division?

    (2x3-3x2 + 5x + 10)/(x+1)?

    I changed it :smile:
    I meant f(x) = 2x3+ax2 + bx + 10
     
  6. Jul 6, 2016 #5
    yes do the polynomial division and you 'll find the quotient polynomial of 2nd degree is such that it doesn't have real roots.
     
  7. Jul 6, 2016 #6
    I got (2x3-3x2 + 5x + 10)/(x+1) = 2x2-5x+10

    So 2x3-3x2 + 5x + 10 = (x+1)(2x2-5x+10)
     
  8. Jul 6, 2016 #7
    Very nice I guess you can see why ##2x^2-5x+10## does not have any real roots... So you original function written as the product of those two polynomials has how many real roots?
     
  9. Jul 6, 2016 #8
    One :smile:
    Thanks
     
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