Solving &#x222B;<!-- ∫ --> ( 5 x 2

Audrina Jackson 2022-07-14 Answered
Solving ( 5 x 2 + 3 x 2 x 3 + 2 x 2 ) via partial fractions
So, first of all, we must factorize the denominator:
x 3 + 2 x 2 = ( x + 2 ) x 2
Great. So now we write three fractions:
A x 2 + B x + C x + 2
Eventually we conclude that
A ( x + 2 ) + B ( x + 2 ) ( x ) + C ( x 2 ) = 5 x 2 + 3 x 2
So now we look at what happens when x=−2:
C=12
When x=0:
A=−1And now we are missing B, but we can just pick an arbitrary number for x like... 1:
B=−1
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Answers (2)

Zachery Conway
Answered 2022-07-15 Author has 7 answers
You have an algebra error. When you put x=−2, the constraint equation collapses to 4C=12, so C=3.
Your method for determining B is correct – the constraining identity must hold for any x. Putting x=1 is a fine choice as the arithmetic is easier. However the incorrect value for C will affect the value you determine for B.
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Jamison Rios
Answered 2022-07-16 Author has 6 answers
I think this is the easiest way.
( B + C ) x 2 + ( A + 2 B ) x + 2 A = 5 x 2 + 3 x 2
2 A = 2 A = 1 A + 2 B = 1 + 2 B = 3 B = 2 B + C = 2 + C = 5 C = 3
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