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OpenStudy (anonymous):
In partial fraction,can we factorize the numerator(2x+8/(x+5)(x+3))?
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OpenStudy (lgbasallote):
uhmm you can..but it wouldnt make any sense? lol
OpenStudy (anonymous):
no,we factorize denominator alwayssssssss
OpenStudy (anonymous):
what so you mean?
like my teacher said before we begin we need to factorise.but i don't know which one
OpenStudy (anonymous):
*do
OpenStudy (anonymous):
@nitz Always denominator?
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OpenStudy (anonymous):
see the resolution of f(x)/g(x) into partial fractions depends mainly upon the nature of the factors of g(x)
OpenStudy (anonymous):
ie the denominator,,,,,
OpenStudy (anonymous):
oh!!! ok..ok I see @nitz can you solve this for me pls.
OpenStudy (anonymous):
ya sure
OpenStudy (anonymous):
\[A/(X+5) + B/(X+3) = (2X+8)/(X+5)(X+3) \]
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OpenStudy (anonymous):
2x+8/(x+5)(x+3)=A/(x+5)+B/(x+3)
2x+8/(x+5)(x+3)=(A(x+3)+B(x+5))/(x+5)(x+3)
OpenStudy (anonymous):
cancel the common factors ie (x+3)(x+5) on both sides assuming them to be non zero
OpenStudy (anonymous):
we are left with
2x+8=A(x+3)+B(x+5)
OpenStudy (anonymous):
put x=-3,we get
2(-3)+8=A(-3+3)+B(-3+5)
2=2B
B=1
OpenStudy (anonymous):
SImilarly put x=-5
YOU WILL GET A =1
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OpenStudy (anonymous):
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