Posted by Anonymous on Friday, April 18, 2008 at 6:29pm.
Is this right?
Decompose into partial fractions.
(2x^224x+35)/(x^2+2x7)(x2)=
(Ax+B)/(x^2+2x7)+C/(x2)
=Ax^2+Bx2Ax2B+Cx^2+2xC7C
=(A+C)x^2+Bx+2(Ax+CxB)7C
A+C=2
2Ax+Bx+2Cx=24
2B7C=35
A=2C
B=(357C)/(2)

Algebra  Megan, Friday, April 18, 2008 at 6:49pm
My answers are
A=2C
B=(357C)/(2)

Algebra  Damon, Friday, April 18, 2008 at 8:36pm
I agree to here:
Ax^2+Bx2Ax2B+Cx^2+2xC7C
but then
(A+C)x^2 + (B2A+2C) x +(2B7C)
so
A+C=2 so C=(2A)
B2A+2C=24
2B7C = 35
solve for A and B using (2A) for C
B2A+2(2A) = 24
2B+7(2A) = 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 = 2A = 5
7, 0 , 5 indeed work
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