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Math Help - Polynomials

  1. #1
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    Polynomials

    Given that 3x^3 + 4x^2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C for all values of x, find the values of A, B and C.

    What I did:

    f(x) = 3x^3 + 4x^2 - 13x - 12

    A = 3 (by inspection)

    C = f(3) = 66

    My questions:
    1. Why is it not possible to find C by subsituting -(1/3) into x, since (3x + 1) would = 0?
    2. How do I solve for B? (Ans: 4)
    Last edited by fActor; September 19th 2012 at 11:46 PM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Polynomials

    You could expand the right side of the given equation and equate like coefficients to find B.
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  3. #3
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    Re: Polynomials

    Quote Originally Posted by MarkFL2 View Post
    You could expand the right side of the given equation and equate like coefficients to find B.
    I guess you mean this.

    3x^3 + 4x^2 - 13x - 12 = (3x^2 - 8x - 3)(x + B) + C

    4 = 3B - 8
    3B = 12
    B = 4

    Why can't I seem to get the answer when I compare coefficients of x, though?
    -13x = -8Bx - 3x
    -13 = -8B - 3
    8B = 16
    B = 2

    And someone please, please answer my first question... I'm really confused.....
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Polynomials

    Are you sure you aren't given:

    3x^3+4x^2-13x-12=(Ax+1)(x+B)(x-3)+Cx ?

    In this case, expanding the right side, we have:

    3x^3+4x^2-13x-12=Ax^3+(AB-3A+1)x^2-(3AB-B-C+3)x-(3B)

    This implies:

    A=3

    3B-9+1=4\:\therefore\:B=4

    36-4-C+3=13\:\therefore\:C=22
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  5. #5
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    Re: Polynomials

    Nope, there's no wrong with the question.
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  6. #6
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    Re: Polynomials

    3x^3 + 4x^2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C
    There are no constants A,B,C that make this equation true for all x. This can be proven by contradiction from your own reasoning. A must be 3, so that means C must equal both f(3) and f(-1/3), however f(3) is not equal to f(-1/3).
    Last edited by SworD; September 20th 2012 at 10:15 PM.
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  7. #7
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    Re: Polynomials

    hmm, i am intrigued.

    suppose: 3x3 + 4x2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C

    this means: 3x3 + 4x2 - 13x - 12 = Ax3 + (AB - 3A + 1)x2 + (-3AB + B - 3)x - 3B + C

    equating coefficients:

    A = 3
    AB - 3A + 1 = 4
    -3AB + B - 3 = -13
    -3B + C = -12

    the last equation gives us: B = (C+12)/3. so we have 2 equations in C:

    3((C+12)/3) - 9 + 1 = 4, which leads to: C = 0.

    (-9)((C+12)/3) + (C+12)/3 - 3 = -13, which leads to: C = -33/4.

    this makes no sense at all....i strongly suggest you check the problem again.
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  8. #8
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    Re: Polynomials

    I checked again... no problems. The answers I got/given might not be correct though. I think I may ask my teacher tomorrow...
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  9. #9
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    Re: Polynomials

    Quote Originally Posted by fActor View Post
    Given that 3x^3 + 4x^2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C for all values of x, find the values of A, B and C.

    What I did:

    f(x) = 3x^3 + 4x^2 - 13x - 12

    A = 3 (by inspection)

    C = f(3) = 66

    My questions:
    1. Why is it not possible to find C by subsituting -(1/3) into x, since (3x + 1) would = 0?
    2. How do I solve for B? (Ans: 4)
    Your reasoning to get A and C are perfect. Your question #1 is very good - it shows you're thinking.

    For reasons that will become apparent, let me do #2 first:
    Since you've found C, and you're looking for B, notice that B shows up in the constant term. You get the constant term by setting x=0:

    Plug in x=0 into: 3x^3 + 4x^2 - 13x - 12 = (3x + 1)(x + B)(x - 3) + 66, and you get

    -12 = (1)(B)(-3) + 66, so -12 = -3B +66, so -3B = -78, so B = 26.

    So you're done (um... not really): 3x^3 + 4x^2 - 13x - 12 = (3x + 1)(x + 26)(x -3) + 66.

    What about your concern over problem #1? That would worry me too. So let's check the answer:

    (3x + 1)(x + 26)(x -3) + 66 = (3x + 1)(x^2 + 23x - 78)

    = 3x^3 + (1 + 69)x^2 + (23 - 234)x - 78 = 3x^3 + 70x^2 - 211x - 78.

    Ooops. What did we do wrong?

    Nothing.

    The problem is itself wrong. There ARE no such numbers A, B, and C that make those polymonials equal. The setup declared otherwise, and we did everything right, GIVEN their false assertion that those two polynomials can be made equal.

    And that gives the answer to your #1: "Yes, you're absolutely right. Plugging in x = -1/3 should've caused the polynomial to evaluate to 66. But since there was a false assumption to begin with, everything after that error is without valid mathematical justification."

    If you let me begin by assuming 1 = 0, I can pretty much prove anything. (In fact, I can prove anything in one line. I think it's called "ex contradictione non sequitur quodlibet")

    Here's what's going on. A 3rd degree polynomial has 4 coefficients. Thus when you equate those polynomials, you'll have to equate them coefficient by coefficient. That gives you 4 equations. How many unknowns? Just 3: A, B, and C. Whenever you have more equations than unknowns, you run the risk of having no solutions. And that's what's happened here.

    Working it out:

    (Ax + 1)(x + B)(x -3) + C = (Ax + 1)(x^2 + (B-3)x -3B) + C

    = A x^3  + ( A(B-3) + 1) x^2 +  ( -3AB + (B-3) ) x + (-3B) + C

    = A x^3  + ( AB -3A + 1) x^2 + ( -3AB + B -3 ) x + ( C - 3B )

    So if that's going to equal 3x^3 + 4x^2 - 13x - 12, then A, B, and C must satisfy all 4 of these equations:

    A = 3, AB -3A + 1 = 4, -3AB + B -3 = -13, and C - 3B = -12.

    Plugging in A = 3 and other simplifying gives 3 equations in the 2 unknonws B and C:

    3B - 9 + 1 = 4, -9B + B - 3 = -13, C - 3B = -12.

    Just the first two become: 3B  = 12, and -8B = -10.

    Thus B = 4, and B = 5/4, and... well, that's enough to show that the problem has no solution.
    Last edited by johnsomeone; September 21st 2012 at 02:09 AM.
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  10. #10
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    Re: Polynomials

    This is what I did. I divided 3x^3+4x^2-13x-12 by x-3 to get  3x^2+13x+26+\frac{66}{x-3} the 66 is your C value because it was added on. So I solved for what the true equation is by solving for z.  -12 = 66 + z => z = -78 => 3x^2+13x+26+\frac{66}{x-3} = (3x^2+13x+26)(x-3)+66 By using the quadratic equation on  3x^2+13x+26 you can solve for your other unknown values.
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