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Remainder theorem question - combine divisor

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    when f(x) is divided by (x+1), remainder is -9; when f(x) is divided by (x-3), remainder is -1; what is the remainder if f(x) is divided by (x+1)(x-3)?

    2. Relevant equations

    f(x) = divider * q(x) + remainder

    3. The attempt at a solution

    f(x) = (x+1) * a(x) -9
    f(x) = (x-3) * b(x) -1

    unable to perform further useful calculation
     
  2. jcsd
  3. Feb 13, 2012 #2

    eumyang

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    When you divide f(x) by (x+1)(x-3), a quadratic, the highest possible degree of the remainder polynomial is 1.

    So you have f(x) = (x+1)(x-3)*q(x) + (ax + b).

    By the Remainder Theorem...
    "when f(x) is divided by (x+1), remainder is -9; when f(x) is divided by (x-3), remainder is -1
    ... what does the above tell us? You'll then need to plug some things into x and f(x) above to come up with two equations involving a and b. Solve for a and b.
     
  4. Feb 13, 2012 #3
    Let's assume that a(x)=ax+b and b(x)=cx+d. This is a valid assumption, but it's not the only one that works (for example, you could have a higher order polynomial for a(x) and b(x)).

    Let a(x)=ax+b
    Let b(x)=cx+d

    Then
    f(x)=(x+1)(ax+b)-9 = ax2+(a+b)x+b-9
    f(x)=(x-3)(cx+d)-1 = cx2+(d-3c)x-3d-1

    We want to determine the coefficients a, b, c, and d that will make this work. Let's subtract the two equations.

    0=[ax2+(a+b)x+b-9] - [cx2+(d-3c)x-3d-1]
    0=(a-c)x2+(a+b+3c-d)x+(b+3d-8)

    Hence,

    (a-c)=0
    (a+b+3c-d)=0
    (b+3d-8)=0

    Three equations, four unknowns. Observe that there is not one unique solution, but a set of valid solutions.
     
  5. Feb 13, 2012 #4
    thanks eumyang

    ok so i give it another go

    i have

    f(-1) = -9
    f( 3) = -1

    thus

    f(x) = (x+1)(x-3)*q(x) + (ax + b)

    f(-1) = [(-1)+1][(-1)-3]*q(x) + [a(-1) + b]
    f(3) = [(3)+1][(3)-3]*q(x) + [a(3) + b]

    -9 = -a + b
    -1 = 3a + b

    a=2
    b=-7

    solution = 2x - 7

    would this be correct?
     
  6. Feb 13, 2012 #5
    thanks for help, but i cannot deduce a unique equation from this set of equations nor a set of rules

    this question came from an exam paper so there should be some sort of reasonable answer, i'm not sure if my solution of 2x - 7 is correct, now i'm trying to plug in something to verify
     
  7. Feb 13, 2012 #6
    the solution (2x-7) seems to be wrong

    for f(x) = (x+1)(x-3)*q(x) + (2x -7)

    i tried to plug in q(x) = (x+2)

    thus f(x) = x^3 + x + 1

    when f(x) is divided by (x+1) and (x-3), i don't get -9 and -1


    please help
     
  8. Feb 13, 2012 #7

    Dick

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    I get (x+1)(x-3)(x+2)+(2x -7)=x^3-5x-13. I think your example does work.
     
  9. Feb 13, 2012 #8
    thanks dick

    yes i made a mistake, when x-2 is plug in i got

    (x+1)(x-3)(x+2)+(2x -7)=x^3-5x-13

    but still it does not work?

    can someone tell me if 2x-7 is the correct answer

    and

    is there a unique answer to this question?
     
  10. Feb 13, 2012 #9

    Dick

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    Why doesn't it work? If I divide that polynomial by (x+1) I get remainder -9, if I divide by (x-3) I get remainder -1. And I don't see anything wrong with eumyang's original hint. Reread it and see if you disagree. 2x-7 is the unique correct answer. The x+2 you plugged in for q(x) to check isn't unique. You could have plugged anything in for that and it will still work.
     
  11. Feb 13, 2012 #10
    thanks Dick! i made some more arithmetic errors, i guess i'm too tired. now i got it.

    but in the 3rd post, Harrisonized mentioned there is a set a solution. i couldn't see a flaw in his work out. is it invalid to assume

    Let a(x)=ax+b
    Let b(x)=cx+d
     
  12. Feb 13, 2012 #11

    eumyang

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    Well, Harrisonized did say that
    But note that Harrisonized is trying to find the coefficients for the two quotient polynomials a(x) and b(x). (To be honest, I'm not sure why he did that, since you're being asked for the remainder polynomial.)
     
  13. Feb 13, 2012 #12
    Above, I found that

    (a-c)=0
    (a+b+3c-d)=0
    (b+3d-8)=0

    From the first equation, we have a=c. Since we have three equations and four unknowns, we have one degree of freedom, so we can rewrite all four coefficients a, b, c, and d in terms of a single variable. Let's choose that single variable to be a. (This is an arbitrary choice.) Solving the remaining two equations:

    (4a+b-d)=0 → b=d-4a
    (b+3d-8)=0 → d-4a+3d=8 → d=a+2
    b=d-4a → b=-3a+2

    Therefore,

    f(x)=(x+1)(ax+b)-9=(x+1)(ax-3a+2)-9=ax2+(-2a+2)x-3a-7
    f(x)=(x-3)(cx+d)-1=(x-3)(ax+a+2)-1=ax2+(-2a+2)x-3a-7

    I didn't have to plug into both to find f(x), but since both forms result in the same f(x), I confirm that the solutions found are correct. Since we have determined it up to a constant, a, we have found all quadratic solutions of what you described in your problem.

    Now you want the remainder, right? Divide through by (x+1)(x-3)=x2-2x-3.

    [ax2+(-2a+2)x-3a-7]/[x2-2x-3]
    = [ax2-2ax-3a]/[x2-2x-3] + [2x-7]/[x2-2x-3]
    = a + [2x-7]/[x2-2x-3]

    2x-7 is your unique remainder.
     
    Last edited: Feb 13, 2012
  14. Feb 13, 2012 #13

    Dick

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    Yes, but as eumyang already pointed out, you don't have to do this at all. In what way is this simpler than the original suggestion? It's certainly less general if you need to assume that f(x) is quadratic. Which you don't.
     
  15. Feb 13, 2012 #14
    It's just another way to solve the same problem. I ended up writing the rest of it, because the OP posted this

    and I wanted to show him that this method works as well.
     
  16. Feb 13, 2012 #15

    Dick

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    Yes, it is and it works. Because you know you'll get the same answer no matter what form you assume for the two polynomials. It's just not simpler than the original approach and confused the OP. That's my only problem with it.
     
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