SOLUTION: {{{7/(x^3+9*x^2+5*x+45)}}}={{{A/(x^2+5)}}}+{{{B/(x+9)}}} What is B? I tried the following approach: 7=A*(x+9)+B*{{{(X^2+5)}}} 7=x*(A+Bx)+9A+5B 7=9A+5B & x(A+Bx)=0 I cou

SOLUTION: {{{7/(x^3+9x^2+5x+45)}}}={{{A/(x^2+5)}}}+{{{B/(x+9)}}} What is B? I tried the following approach: 7=A(x+9)+B{{{(X^2+5)}}} 7=x*(A+Bx)+9A+5B 7=9A+5B & x(A+Bx)=0 I cou

Question 95435: =+
What is B?
I tried the following approach:
7=A*(x+9)+B*
7=x*(A+Bx)+9A+5B
7=9A+5B & x(A+Bx)=0
I could only say B=(7-9A)/5 & either x=0 or x=-A/B
I'm not sure how to move from here to the next in order to find the solution.
Could you help? The question was given by teacher on the board, not from a textbook.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Since the denominator of the 1st fraction is x^2+5
you need Ax+B as its numerator and C as the numerator of the 2nd fraction
------------------
Then multiplying thru by (x^2+5)(x+9) you would get:
7 = (Ax+B)(x+9) + C(x^2+5)
7 = Ax^2+9Ax+Bx+9B+Cx^2+5C
7 = (A+C)x^2+(9A+B)x+(9B+5C)
---------------
Let x=0, then 0A+9B+5C = 7
Let x=1, then 10A+10B+6C=7
Let x=-1, then -8A+8B+6C=7
------------------
Solving that with a TI-83 Matrix facility you get
A = -0.8139...
B = 0.73255...
C = 0.08139...
=============
That may or may not be what you are looking for but
you cannot arrive at an answer with A as the numerator over
x^2+5 and B as the numerator over x+9
================
Cheers,
Stan H.

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