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Math Help - find missing integer-algebraic division

  1. #1
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    find missing integer-algebraic division

    Hi:
    I have come across a question but have not been able to work out a method to solve this, yet
    When x^3+3x^2+ax+2 is divided by x 2 the remainder is 4. Find a.

    Any help. thanks
    err - most help. thanks
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  2. #2
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    use synthetic division ... you know the remainder is 4

    Code:
    2].......1........3........a.........2
    ..................2.......10.......20+2a
    ------------------------------------------
    .........1........5.....10+a.......22+2a
    22+2a = 4
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  3. #3
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    reply

    hmmm!
    don't know synthetic division. Is that high-school level?
    I think I am only supposed to be doing algebraic division
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  4. #4
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    got it

    hmmm!
    think i got it
    bring down the leading coefficient to the bottom row
    multiply c by the value just written on the bottom row
    add the column created in last step
    repeat until done
    write answer out and equate
    cool!
    thank you
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  5. #5
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    Quote Originally Posted by lemonz View Post
    hmmm!
    don't know synthetic division. Is that high-school level?
    I think I am only supposed to be doing algebraic division
    most definitely high school ...

    Synthetic Division
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  6. #6
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    re: why?

    why do you do this Skeeter?
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  7. #7
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    Quote Originally Posted by lemonz View Post
    Hi:
    I have come across a question but have not been able to work out a method to solve this, yet
    When x^3+3x^2+ax+2 is divided by x 2 the remainder is 4. Find a.

    Any help. thanks
    err - most help. thanks
    Basic theorem: if P(x) is a polynomial and dividing by x-b gives quotient Q(x) with remainder R (which is a constant, a number) then P(x)/(x-b)= Q(x)+ R/(x-b) so P(x)= Q(x)(x- b)+ R. Setting x= b, P(b)= Q(b)(b-b)+ R= R no matter what Q(b) is. That is, the remainder when P(x) is divided by x- b is the value of P(b).

    Here, P(2)= 2^4+ 3(2^2)+ a(2)+ 2= 16+ 3(4)+ 2a+ 2= 2a+ 30. What value of a makes that equal to 4?
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  8. #8
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    Quote Originally Posted by lemonz View Post
    why do you do this Skeeter?
    good question ... it's probably a form of insanity.
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  9. #9
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    re: nuttz!

    well, thanks crazy person
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  10. #10
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    So, is this right?

    x^3+bx^2-15x+3 divided by x+1 remainder 14: Find b

    1: 1 b -15 3
    1 1+b -14+b
    1 1+b -14+b 17+b

    17+b=14
    b=-3

    If this is so, it mean that in the case of the last column, where -14 is under the 3, I ignore the minus sign and just make addition with integer.
    But in the case of the 3rd column, I can't ignore the minus sign on the 15, cause it is the starting number?
    thanks
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  11. #11
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    re: synthetic division check

    So, is this right?

    x^3+bx^2-15x+3 divided by x+1 remainder 14: Find b

    1:::::1 b -15 3

    1 1+b -14+b

    1 1+b -14+b 17+b

    17+b=14
    b=-3

    If this is so, it mean that in the case of the last column, where -14 is under the 3, I ignore the minus sign and just make addition with integer.
    But in the case of the 3rd column, I can't ignore the minus sign on the 15, cause it is the starting number?
    thanks
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  12. #12
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    x^3+bx^2-15x+3 divided by x+1 remainder 14: Find b
    quick check ... see if f(-1) = 14 when b = -3

    f(x) = x^3-3x^2-15x+3

    f(-1) = \, ?
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