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(5x²+17x+15)/((x+2)²(x-3))≡

A/(x+2)+B/(x+2)²+C/(x-3).

Therefore:

A(x+2)(x-3)+B(x-3)+C(x+2)²=

Ax²-Ax-6A+Bx-3B+Cx²+4Cx+4C=

(A+C)x²+(-A+B+4C)x+(-6A-3B+4C)≡

       5x²+                17x+                  15.

So, A+C=5, C=5-A; -A+B+4C=17; -6A-3B+4C=15.

4C=17+A-B=15+6A+3B, 2=5A+4B, B=(2-5A)/4.

We now have B and C in terms of A, so substitute:

-A+(2-5A)/4+20-4A=17, 

-4A+2-5A+80-16A=68.

14=25A, A=14/25; C=5-A=111/25; B=(2-5A)/4=-1/5.

We can write these as decimals: A=0.56, B=-0.2, C=4.44.

Partial fractions:

0.56/(x+2)-0.2/(x+2)²+4.44/(x-3).

ago by Top Rated User (660k points)

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