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Question Number 78568 by TawaTawa last updated on 18/Jan/20
Find the roots of the equation   bx^3  − (3b + 2)x^2  − 2(5b − 3)x + 20  =  0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\mathrm{bx}^{\mathrm{3}} \:−\:\left(\mathrm{3b}\:+\:\mathrm{2}\right)\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{5b}\:−\:\mathrm{3}\right)\mathrm{x}\:+\:\mathrm{20}\:\:=\:\:\mathrm{0} \\ $$
Answered by key of knowledge last updated on 18/Jan/20
=bx^3 −3bx^2 −10bx−2x^2 +6x+20=  bx(x^2 −3x−10)−2(x^2 −3x−10)=  (bx−2)(x−5)(x+2)=0  x=5 , −2 , (2/b)(b≠0)
$$=\mathrm{bx}^{\mathrm{3}} −\mathrm{3bx}^{\mathrm{2}} −\mathrm{10bx}−\mathrm{2x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{20}= \\ $$$$\mathrm{bx}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{10}\right)−\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{10}\right)= \\ $$$$\left(\mathrm{bx}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{5}\right)\left(\mathrm{x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{5}\:,\:−\mathrm{2}\:,\:\frac{\mathrm{2}}{\mathrm{b}}\left(\mathrm{b}\neq\mathrm{0}\right) \\ $$
Commented by TawaTawa last updated on 18/Jan/20
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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