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Question Number 78568 by TawaTawa last updated on 18/Jan/20

$$\mathrm{Find}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}$$$${\mathrm{bx}}^3-(3\mathrm{b}+2){\mathrm{x}}^2-2(5\mathrm{b}-3)\mathrm{x}+20=0$$

Answered by key of knowledge last updated on 18/Jan/20

$$={\mathrm{bx}}^3-{3\mathrm{bx}}^2-10\mathrm{bx}-{2\mathrm{x}}^2+6\mathrm{x}+20=$$$$\mathrm{bx}({\mathrm{x}}^2-3\mathrm{x}-10)-2({\mathrm{x}}^2-3\mathrm{x}-10)=$$$$(\mathrm{bx}-2)(\mathrm{x}-5)(\mathrm{x}+2)=0$$$$\mathrm{x}=5,-2,\frac{2}{\mathrm{b}}(\mathrm{b}\ne 0)$$

Commented by TawaTawa last updated on 18/Jan/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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