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More polynomial division

  1. Mar 23, 2007 #1


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    Some of you may have read my troubles that I had in this thread

    Now I am having more troubles with more advanced problems.

    I started with this problem:

    [4y^2 - 7y - 12] / [(y)(y+2)(y-3)]

    The problem is that I set it up the same way as I did in that thread and im able to solve for A, B, and C:
    A= 3
    B= 9/5
    C= 1/5

    so I integrate 3/y + (9/5)/(y+2) + (1/5)/(y-3)

    and i get

    3ln(y) + 9/5ln(y+2) + 1/5ln(y-3)

    Beautiful right? Well im suppost to find the integral over the area from 1 to 2.

    When I plug in 1 and 2 into the 1/5 ln (y-3) it yields a negetive number and you cant take the ln of a negetive number!!!!

    Im sure im making the mistake in the polynomial division somewhere but i dont know. Im thinking there is a way to simplify the original question first before i solve for A B and C.... right? any insite would be wonderful!
  2. jcsd
  3. Mar 23, 2007 #2


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    Ok so I forgot to mention that I also tried to take the absolute values of the value inside the ln and it still did not work.
  4. Mar 23, 2007 #3
    Oh ok. You didn't set up your equations right. With a denominator with three factors, it becomes (you can see this by yourself):

    [tex] A(y)(y+2) + B(y)(y-3) + C(y+2)(y-3) = 4y^2 - 7y - 12 [/tex]

    [tex] A(y^{2} + 2y) + B(y^{2} - 3y) + C(y^{2} - y - 6) = 4y^2 - 7y - 12 [/tex]

    [tex] (A + B + C)y^{2} + (2A - 3B - C)y - 6C = 4y^2 - 7y - 12 [/tex]

    If we want A, B and C to be constants, we compare each term with one another (equality of polynomials theorem). We find C = 2, and then we get [tex]2A - 3B - 2 = -7 [/tex] and [tex] A + B + 2 = 4 [/tex]. Now we have two unknowns with two equations and we solve.
    Last edited: Mar 23, 2007
  5. Mar 24, 2007 #4


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    What Werg22 is talking about is setting the coefficients of the same powers equal.

    Another very nice method is, since the equation must be true for all x, to take specific values for x. Since the left hand side of the equation Werg22 gave you involve factors y, y+2, and y- 3, let y= 0, -2, and 3 in succesion. Your equations become:
    x= 0 0+ 0+ C(2)(-3)= -12 so C= 2
    x=-2 0+ B(-2)(-5)+ 0= 4(-2)2- 7(-2)- 12= 16+14-12= 18 so B= 9/5
    x= 3 A(3)(5)+ 0+ 0= 4(3)2-7(3)-12= 36- 21-12= 3 so A= 1/5.
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