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Question Number 163000 by tounghoungko last updated on 03/Jan/22

Answered by Rasheed.Sindhi last updated on 03/Jan/22

y=f(x)=2x^3 −x^2 −3x  zeroes of f(x)  2x^3 −x^2 −3x=0  x(2x^2 −x−3)=0  x{2x^2 −3x+2x−3}=0  x{x(2x−3)+1(2x−3)}=0  x(x+1)(2x−3)=0  x=0 ∣ x=−1 ∣ x=(3/2)

$${y}={f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{3}{x} \\ $$$${zeroes}\:{of}\:{f}\left({x}\right) \\ $$$$\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{3}{x}=\mathrm{0} \\ $$$${x}\left(\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{3}\right)=\mathrm{0} \\ $$$${x}\left\{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}{x}−\mathrm{3}\right\}=\mathrm{0} \\ $$$${x}\left\{{x}\left(\mathrm{2}{x}−\mathrm{3}\right)+\mathrm{1}\left(\mathrm{2}{x}−\mathrm{3}\right)\right\}=\mathrm{0} \\ $$$${x}\left({x}+\mathrm{1}\right)\left(\mathrm{2}{x}−\mathrm{3}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\mid\:{x}=−\mathrm{1}\:\mid\:{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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