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Question Number 14421 by chux last updated on 31/May/17

$$\mathrm{express}\frac{{2\mathrm{x}}^2-\mathrm{x}+2}{{(\mathrm{x}+2)}^2(1-2\mathrm{x})}\mathrm{as}\mathrm{partial}$$$$\mathrm{fraction}$$

Answered by RasheedSindhi last updated on 01/Jun/17

$$\mathrm{Let}$$$$\frac{{2\mathrm{x}}^2-\mathrm{x}+2}{{(\mathrm{x}+2)}^2(1-2\mathrm{x})}=\frac{{\mathrm{A}}_1}{\mathrm{x}+2}+\frac{{\mathrm{A}}_2}{{(\mathrm{x}+2)}^2}+\frac{\mathrm{B}}{1-2\mathrm{x}}$$$${2\mathrm{x}}^2-\mathrm{x}+2={\mathrm{A}}_1(\mathrm{x}+2)(1-2\mathrm{x})+{\mathrm{A}}_2(1-2\mathrm{x})+\mathrm{B}{(\mathrm{x}+2)}^2$$$$\mathrm{This}\mathrm{is}\mathrm{an}\mathrm{identity}\mathrm{i}-\mathrm{e}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{equal}$$$$\mathrm{for}\mathrm{all}\mathrm{values}\mathrm{of}\mathrm{x}.$$$$\mathrm{Let}\mathrm{x}=-2$$$$2{(-2)}^2-(-2)+2={\mathrm{A}}_2(1-2(-2))$$$${\mathrm{A}}_2=\frac{12}{5}$$$$\mathrm{Let}\mathrm{x}=1/2$$$$2{\left(1/2\right)}^2-\left(1/2\right)+2=\mathrm{B}{(\frac{1}{2}+2)}^2$$$$\mathrm{B}\left(\frac{25}{4}\right)=\frac{1}{2}-\frac{1}{2}+2$$$$\mathrm{B}=2\left(\frac{4}{25}\right)=\frac{8}{25}$$$$\mathrm{To}\mathrm{determine}{\mathrm{A}}_1:$$$${2\mathrm{x}}^2-\mathrm{x}+2={\mathrm{A}}_1(\mathrm{x}+2)(1-2\mathrm{x})+{\mathrm{A}}_2(1-2\mathrm{x})+\mathrm{B}{(\mathrm{x}+2)}^2$$$$\mathrm{Substitute}{\mathrm{A}}_2=\frac{12}{5}\mathrm{\&}\mathrm{B}=\frac{8}{25}\mathrm{in}\mathrm{above}.$$$${2\mathrm{x}}^2-\mathrm{x}+2={\mathrm{A}}_1(\mathrm{x}+2)(1-2\mathrm{x})+\left(\frac{12}{5}\right)(1-2\mathrm{x})+\left(\frac{8}{25}\right){(\mathrm{x}+2)}^2$$$${25\mathrm{x}}^2-25\mathrm{x}+50={25\mathrm{A}}_1(-{2\mathrm{x}}^2-3\mathrm{x}+2)$$$$+60(1+2\mathrm{x})+8({\mathrm{x}}^2+4\mathrm{x}+4)$$$${25\mathrm{x}}^2-25\mathrm{x}+50=-{50\mathrm{A}}_1{\mathrm{x}}^2-{75\mathrm{A}}_1\mathrm{x}+{50\mathrm{A}}_1+60+120\mathrm{x}+{8\mathrm{x}}^2+32\mathrm{x}+32$$$${25\mathrm{x}}^2-25\mathrm{x}+50=(-{50\mathrm{A}}_1+8){\mathrm{x}}^2-{75\mathrm{A}}_1\mathrm{x}+{50\mathrm{A}}_1+60+120\mathrm{x}+32\mathrm{x}+32$$$$$$$$$$$$\subset \oplus \cap \mathrm{\top }!\cap \cup \in ...$$

Commented by Tinkutara last updated on 31/May/17

$$\mathrm{Sorry},\mathrm{but}{\mathrm{A}}_1=\frac{-21}{25},{\mathrm{A}}_2=\frac{12}{5}\mathrm{and}$$$$\mathrm{B}=\frac{8}{25}\left(\mathrm{By}\mathrm{calculator}\right)$$

Commented by chux last updated on 01/Jun/17

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}$$

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