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February 1, 2015

February 1, 2015

Posted by **Anonymous** on Friday, April 18, 2008 at 6:29pm.

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 -
**Megan**, Friday, April 18, 2008 at 6:49pmMy answers are

A=2-C

B=(35-7C)/(-2)

- Algebra -
**Damon**, Friday, April 18, 2008 at 8:36pmI 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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