Question

help with partial fractions

Answer 1

\[\frac{9x^2+19x+56}{x(x-8)(x-7)}=\frac{A}{x}+\frac{B}{x-8}+\frac{C}{x-7}\]
=\[\frac{A(x-8)(x-7)+Bx(x-7)+Cx(x-8)}{x(x-8)(x-7)}\]
since the denominators are the same then the numerators are the same
\[9x^2+19x+56=A(x-8)(x-7)+Bx(x-7)+Cx(x-8)\]
\[9x^2+19x+56=A(x^2-15x+56)+B(x^2-7x)+C(x^2-8x)\]

Answer 2

\[9x^2+19x+56=Ax^2+Bx^2+Cx^2-15Ax-7Bx-8Cx+56A\]
\[9x^2+19x+56=x^2(A+B+C)+x(-15A-7B-8C)+56A\]

Answer 3

so A+B+C=9
-15A-7B-8C=19
56A=56

Answer 4

so 56A=56 implies A=1
so we have 1+B+C=9 and -15-7B-8C=19

Answer 5

B+C=8 and -7B-8C=34
B=8-C -7(8-C)-8C=34
-56+7C-8C=34
-C=90
C=-90
B=8-(-90)=8+90=98

Answer 6

A=1, B=98, C=-90
A/x+B/(x-8)+C/(x-7)=1/x+98/(x-8)-90/(x-7)

Answer 7

i didn't do on scratch paper
there may be a mistake above but if you look at what i was trying to do you should be able to figure how to do this problem and get the right answer
you can also check my if you want by combining the partial fractions

Answer 8

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2000=1605
2001=1728
2002=1776
2003=1903
2004=2072
2005=2050
2006=2141
2007=2150
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