# Algebra

posted by .

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 -

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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