MATH 2B Lecture 15: Partial Fraction Decomposition

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MATH 2B Full Course Notes
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MATH 2B Full Course Notes
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Math 2b - lecture 15 7. 4: partial fraction decomposition (cid:885)+(cid:883)+ (cid:886) (cid:885) (cid:889) (cid:887) (cid:889) (cid:887) (cid:4666) (cid:885)(cid:4667)(cid:4666)+(cid:883)(cid:4667)= (cid:1827) (cid:885)+ (cid:1828)+(cid:883) (cid:889) (cid:887) (cid:4666) (cid:885)(cid:4667)(cid:4666)+(cid:883)(cid:4667)=(cid:1827)+(cid:1827)+(cid:1828) (cid:885)(cid:1828) (cid:4666) (cid:885)(cid:4667)(cid:4666)+(cid:883)(cid:4667) (cid:889) (cid:887)=(cid:1827)+(cid:1827)+(cid:1828) (cid:885)(cid:1828) We get the denominator to be the same as the left side. The denominators cancel out and we are left with the numerators. We know how to get a common denominator from basic algebra, and we also know how to facto(cid:396) f(cid:396)o(cid:373) a pol(cid:455)(cid:374)o(cid:373)ial, so let"s t(cid:396)(cid:455) that (cid:449)ith i(cid:374)teg(cid:396)als. We (cid:449)e(cid:396)e a(cid:271)le to (cid:272)o(cid:373)e to this solutio(cid:374) (cid:271)e(cid:272)ause it is the sa(cid:373)e as the o(cid:374)e (cid:449)e did a(cid:271)o(cid:448)e, (cid:271)ut let"s say we did(cid:374)"t k(cid:374)o(cid:449) (cid:449)he(cid:396)e 7(cid:454)-5 came from. Be(cid:272)ause (cid:449)e do(cid:374)"t k(cid:374)o(cid:449) (cid:449)he(cid:396)e 7(cid:454)-5 (cid:272)a(cid:373)e f(cid:396)o(cid:373), let"s just lea(cid:448)e so(cid:373)e (cid:448)a(cid:396)ia(cid:271)le a a(cid:374)d b as placeholders on the numerators. (cid:1827)+(cid:1828)=(cid:889) (cid:1827) (cid:885)(cid:1828)= (cid:887) Coefficients with the sa(cid:373)e a(cid:373)ou(cid:374)t (cid:454)"s a(cid:396)e set e(cid:395)ual to ea(cid:272)h othe(cid:396) a(cid:374)d the o(cid:374)es (cid:449)ithout (cid:454)"s a(cid:396)e also set equal to each other.

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