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Lecture 11

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MTH 162 Lecture 11: 6.3 Partial Fractions Notes
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University of Miami

Mathematics

MTH 162

Pachero

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MTH 162
Calculus II
6.3 Partial Fractions Notes
L. Sterling
Abstract
Provide a generalization to each of the key terms listed in this section.
Rational Functions
p(x)
▯ Rational functions are functions that are in the common form of R(x) = q(x) which shows
where both p(x) and q(x) are polynomial functions while it’s only q(x) can’t be a zero
polynomial.
– This would generally mean the following:
q (x) 6= 0
▯ When you are trying to ﬁgure out the domain, besides that fact that q (x) 6= 0, then you ﬁnd
out that it [the domain] will be all real numbers.
▯ If you have any rational function that is in the general form of R(x) = q(x) in its lowest
form, then you’ll have R will have a vertical asymptote for each and every value of x for
which q(x) = 0.
Steps to Partial Fractions
▯ Partial fractions deals with writing ration functions as the sum of the simpler fractions and
these are the steps to partial fractions:
– These are the steps when the rational function is actually while it’s numerator’s degree
is less than the denominator’s degree:
▯ Factor q(x), which is the denominator.
▯ Create a sum of fractions with the all of the numerators are all diﬀerent variables
with the denominators are actually each of those factors.
▯ Create fractions with factor’s increasing powers in there are any repeated factors.
▯ Multiply by q(x), which is the denominator.
▯ Set the numerator, which is p(x), to 0.
▯ Solve for all variables.
– These are the steps when the rational function is actually while it’s numerator’s degree
is greater than the denominator’s degree:
▯ Perform any long division that can come in the form of the following:
p(x) = f (x) + r (x)
q (x) q (x)
r(x)
▯ Find the remainder’s, which is q(x) partial function.
1 Partial Fraction Example: Decomposing
Decompose the following into partial functions:
x
R(x) =
(x + 1)(x ▯ 4)
x
(x+1)(x▯4)
x a0 a1
(x▯4)(x+1)= x▯4+ x+1
x(x▯4)(x+1) a0(x▯4)(x+1) a1(x▯4)(x+1)
(x▯4)(x+1) = x▯4 + x+1
x = a0(x + 1) + a1(x ▯ 4)
(▯1) = a0((▯1) + 1) + a1((▯1) ▯ 4)
▯1 = a (0) + a (▯5)
0 1
▯1 = ▯5a 1
▯1 ▯5a
▯5 = ▯51
1
a1= 5
x = a0(x + 1) + a1(x ▯ 4)
4 = a0(1 + 4) + a1(4 ▯ 4)
4 = a0(5) + a1(0)
4 = 5a
0
4= 5a0
5 4
a0= 5

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