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Question Number 87556 by M±th+et£s last updated on 05/Apr/20

Commented by jagoll last updated on 05/Apr/20

x^3 −2x^2 +7x−1 = a(x−2)^2 + b(x−3)(x−2)^2 +c(x−3)^2 (x−2)^2 + d(x−3)^3 +e(x−3)^3 (x−2)

$$\mathrm{x}^{\mathrm{3}} −\mathrm{2x}^{\mathrm{2}} +\mathrm{7x}−\mathrm{1}\:=\:\mathrm{a}\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} + \\ $$$$\mathrm{b}\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{c}\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} \left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} + \\ $$$$\mathrm{d}\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{3}} +\mathrm{e}\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{3}} \left(\mathrm{x}−\mathrm{2}\right) \\ $$

Answered by $@ty@m123 last updated on 05/Apr/20

Let ((x^3 −2x^2 +7x−1)/((x−3)^3 (x−2)^2 ))=(A/((x−3)^3 ))+(B/((x−3)^2 )) +(C/((x−3)))+(D/((x−2)^2 ))+(E/(x−2)) ⇒A(x−2)^2 +B(x−3)(x−2)^2 +C(x−3)^2 (x−2)^2 +D(x−3)^2 +E(x−3)^3 (x−2)= x^3 −2x^2 +7x−1 ....(i) Put x=0 in (i) ⇒4A+12B+36C+9D+54E=−1 ...(ii) Put x=1 in (i) A−2B+4C+4D−8E=5 ....(iii) Put x=2 in (i) D=8−8+14−1=13 Put x=3 in (i) A=27−18+20=29 Now do it yourself: ■ Put values of A & D in (ii) &(iii) and form equation (iv) &(v) ■ Put x=4 in (i) and form equation (vi) ■ Solve (iv), (v) &(vi) for B, C &E using Cramer′s Rule.

$${Let}\:\frac{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{3}} \left({x}−\mathrm{2}\right)^{\mathrm{2}} }=\frac{{A}}{\left({x}−\mathrm{3}\right)^{\mathrm{3}} }+\frac{{B}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{C}}{\left({x}−\mathrm{3}\right)}+\frac{{D}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }+\frac{{E}}{{x}−\mathrm{2}} \\ $$$$\Rightarrow{A}\left({x}−\mathrm{2}\right)^{\mathrm{2}} +{B}\left({x}−\mathrm{3}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} +{C}\left({x}−\mathrm{3}\right)^{\mathrm{2}} \left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{D}\left({x}−\mathrm{3}\right)^{\mathrm{2}} +{E}\left({x}−\mathrm{3}\right)^{\mathrm{3}} \left({x}−\mathrm{2}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{1}\:....\left({i}\right) \\ $$$${Put}\:{x}=\mathrm{0}\:{in}\:\left({i}\right) \\ $$$$\Rightarrow\mathrm{4}{A}+\mathrm{12}{B}+\mathrm{36}{C}+\mathrm{9}{D}+\mathrm{54}{E}=−\mathrm{1}\:...\left({ii}\right) \\ $$$${Put}\:{x}=\mathrm{1}\:{in}\:\left({i}\right) \\ $$$${A}−\mathrm{2}{B}+\mathrm{4}{C}+\mathrm{4}{D}−\mathrm{8}{E}=\mathrm{5}\:....\left({iii}\right) \\ $$$${Put}\:{x}=\mathrm{2}\:{in}\:\left({i}\right) \\ $$$${D}=\mathrm{8}−\mathrm{8}+\mathrm{14}−\mathrm{1}=\mathrm{13} \\ $$$${Put}\:{x}=\mathrm{3}\:{in}\:\left({i}\right) \\ $$$${A}=\mathrm{27}−\mathrm{18}+\mathrm{20}=\mathrm{29} \\ $$$${Now}\:{do}\:{it}\:{yourself}: \\ $$$$\blacksquare\:{Put}\:{values}\:{of}\:{A}\:\&\:{D}\:{in}\:\left({ii}\right)\:\&\left({iii}\right) \\ $$$$\:{and}\:{form}\:{equation}\:\left({iv}\right)\:\&\left({v}\right) \\ $$$$\blacksquare\:{Put}\:{x}=\mathrm{4}\:{in}\:\left({i}\right)\:{and}\:{form}\:{equation}\:\left({vi}\right) \\ $$$$\blacksquare\:{Solve}\:\left({iv}\right),\:\left({v}\right)\:\&\left({vi}\right)\:{for}\:{B},\:{C}\:\&{E}\:{using} \\ $$$${Cramer}'{s}\:{Rule}. \\ $$

Commented by peter frank last updated on 05/Apr/20

thank you

$${thank}\:{you} \\ $$

Commented by M±th+et£s last updated on 05/Apr/20

god bless you sir

$${god}\:{bless}\:{you}\:{sir} \\ $$

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