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Find the remainder of a polynomial

  1. Jan 19, 2006 #1
    7a) f(x0 = 2x3+32-6x+1
    Find the remainder when F(x) is divided by (2x-1)
    b) (i) Find the remainder when f(x) is divided by (x+2)
    (ii) Hence, or otherwise solve the equation: 2x3+32-6x-8=0
    giving your answer to two decimal places.

    This is what i've done:
    a) r= -1 (i did f(1/2) to get that)
    b) (1) r = 9 ( i did f(-2) for that)
    could someone please help me with part (ii)?
     
  2. jcsd
  3. Jan 19, 2006 #2

    NateTG

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    I don't think you understand what the problem is asking for.

    For example
    [tex]x^2-4[/tex]
    divided by
    [tex]x+2[/tex]
    is
    [tex]x-2[/tex]
    because
    [tex](x-2)(x+2)=x^2-4[/itex]

    You should be doing the division with x's.
     
  4. Jan 20, 2006 #3

    HallsofIvy

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    No, NateTG, what he did was perfectly feasible. To find the remainder, r, when polynomial P(x) is divided by (x-a) you CAN do the division, but it is simpler to use the fact that P(x)= (x-a)Q(x)+ r so that P(a)= r. Of course, dividing by x-1/2 will give the same remainder as dividing by 2x-1.

    discombobulated, you twice wrote "2x3+32-6x+1".
    Didn't you really mean "2x3+ 3x2- 6x+ 1"?

    For the third part, I'm not sure about the "Hence" but look at the coefficients. Those tell you that there is one very simple answer!
     
  5. Jan 21, 2006 #4
    yes i know part a) and bi) are correct because i checked it by dividing too, but i find the other way easier. HallsofIvy, yes i did mean 2x3+ 3x2- 6x+ 1. (i should also point out that i'm not a, 'he'!) I don't understand what you're trying to say about the coefficients sorry.
    I noticed that the remainder (9) has been taken away so the factors of 2x3+ 3x2- 6x-8 are (x+2) and (2x2 - x-4)
    so i tried this:
    (x+2)((2x2 - x-4)=0
    (=-2
    (2x2 - x-4=0
    then by completing the square i got:
    x= 1.69 or x= -1.19
    does this make sense?
     
  6. Jan 21, 2006 #5

    HallsofIvy

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    Boy, did I mess up! I didn't notice that the equation you wanted to solve in (ii) was not the orginal equation! You are completly correct. Because you already new that 9 was the remainder when
    2x3+ 3x2- 6x+1 was divided by (x+ 2), you knew that 2x3+ 3x2- 6x+1= (x+2)(something)+ 9.
    Subtracting 9 from both sides tells you that
    2x3+ 3x2- 6x- 8= (x+2)(something).
    And that means that x= -2 is solution. Yes, that "something" is 2x2 - x+ 4. Solving that by completing the square tells you that
    x= 1.69 and x= -1.19.

    My "hint" was for solving the wrong equation! Notice that the coefficients in the original equation are 2, 3, -6, 1 and they add up to 0. That tells you that x= 1 is a solution. Of course, you could factor out x-1 to reduce to a quadratic and solve that, either by completing the square or using the quadratic formula.
     
  7. Jan 23, 2006 #6
    thanks for your help anyway!
     
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