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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
	$y=f(x)=2x^3 −x^2 −3x zeroes of f(x) 2x^3 −x^2 −3x=0 x(2x^2 −x−3)=0 x{2x^2 −3x+2x−3}=0 x{x(2x−3)+1(2x−3)}=0 x(x+1)(2x−3)=0 x=0 ∣ x=−1 ∣ x=(3/2)$
	$y = f (x) = 2 x^{3} - x^{2} - 3 x$ $z e r o e s o f f (x)$ $2 x^{3} - x^{2} - 3 x = 0$ $x (2 x^{2} - x - 3) = 0$ $x {2 x^{2} - 3 x + 2 x - 3} = 0$ $x {x (2 x - 3) + 1 (2 x - 3)} = 0$ $x (x + 1) (2 x - 3) = 0$ $x = 0 ∣ x = - 1 ∣ x = \frac{3}{2}$
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