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Calculus-Remainder Theorem

  1. Jun 19, 2005 #1
    Find the value of m so that when x^3+5x^2+6x+11 is divided by (x+m) the remainder is 3.

    Im having so much diffculty with this and it's so frustrating can anyoen help

    Thomas
     
  2. jcsd
  3. Jun 19, 2005 #2
    Do you know how to divide polynomials? For example if you divide ax^2+bx+c by (x+d) what would be the remainder? There's an easy way for doing this. find value of d which makes x+d=0. Then replace that value with x in the original polynomial equation, and what you got is the remainder of division!

    Try doing this and let me know if there's any problem
     
  4. Jun 19, 2005 #3
    Sorry but I'm still not clear on that. I know that to find a value thats divisible in an equation you try f(x) and if that equals 0 in the equation its divisible. In my case its x^3+5x^2+6x+11 so the only things i try are the last number 11. (+-1, +-11) and that doesn't equal 0..........Explain more please
     
  5. Jun 19, 2005 #4
    You divide the equation by (x+m) right? What I'm telling that find the root of this function (x+m, that is) and then place the root into the polynomial. Then you will have the remainder, which is 3. Solve for m and you got it!

    The root for x+m=0 is obviously -m. now place -m in the place of x in the equation x^3+5x^2+6x+11. What I tell you is that setting x=-m in this equation gives you the remainder when you divide x^3+5x^2+6x+11 by (x+m).
     
  6. Jun 19, 2005 #5
    omg dude no I get another answer..........can you tell me your answer if it's so easy lol
     
  7. Jun 19, 2005 #6
    Is it 4? If not there should be something wrong with the answer...
     
  8. Jun 19, 2005 #7
    yup,
    f(x)=g(x)(x+m) +R
    put x=-m
    u will get
    after rearranging
    [tex] m^3 - 5m^2+6m-8=0 [/tex]
    solving u will have
    [tex](m-4)*(m^2-m+2)=0[/tex]
     
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