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March 26, 2017

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Is this right?


Decompose into partial fractions.

(2x^2-24x+35)/(x^2+2x-7)(x-2)=
(Ax+B)/(x^2+2x-7)+C/(x-2)

=Ax^2+Bx-2Ax-2B+Cx^2+2xC-7C
=(A+C)x^2+Bx+2(Ax+Cx-B)-7C
A+C=2
2Ax+Bx+2Cx=24
-2B-7C=35
A=2-C
B=(35-7C)/(-2)

  • Algebra - ,

    My answers are
    A=2-C
    B=(35-7C)/(-2)

  • Algebra - ,

    I agree to here:
    Ax^2+Bx-2Ax-2B+Cx^2+2xC-7C
    but then
    (A+C)x^2 + (B-2A+2C) x +(-2B-7C)
    so
    A+C=2 so C=(2-A)
    B-2A+2C=-24
    -2B-7C = 35
    solve for A and B using (2-A) for C
    B-2A+2(2-A) = -24
    2B+7(2-A) = -35

    B - 4 A = -28
    2B - 7A = -49

    2 B - 8 A = -56
    2 B - 7A = -49
    subtract
    -A = -7
    so
    A = 7
    then B = 4(7) - 28 = 0
    C = 2-A = -5
    7, 0 , -5 indeed work

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