Oh, okay!
]]>That is two quadratics for the numerators also. Because
That is why I went right to 2 quadratics in the numerators.
]]>Shouldn't the partial fraction we expect be of the form:
Because in the "fit" for the coefficients a,b,c,d,e,f it gets taken care of all by itself.
]]>to tell that our numerators would be quadratics
One way is to say the numerator is an 8th degree poly, a quadratic times by a six degree
poly is an 8th degree poly.
Now what you do is form a 6 x 6 set of simultaneous linear equations. This is done by substituting x = -3,-2,-1,0,1,2 in the above. That wipes out the x and you are left with:
Which is exactly what we expected.
]]>I will provide something in a few minutes.
]]>Where I am stuck is with regards to where to go with these 2 pieces of knowledge.
]]>The way I would do it would be by solving a set of simultaneous linear equations.
You would need two things:
1) To know the form for A and B beforehand.
2) The ability to solve a 6 x 6 set of linear eqautions.
Which of course gives a massive expansion, if it's of interest:
Which, when you add it all together gives:
Which, of course, is:
So I had satisfied myself with the denominators and I tried using my standard partial fractions method and said
But then I got:
And gave up, because I knew that I must have been on the wrong track. And, well, that's when I asked you lol
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