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Question Number 215656 by MathematicalUser2357 last updated on 13/Jan/25

$$3x^3-2x^2-12x+8=0$$

Answered by Wuji last updated on 13/Jan/25

$$(\mathrm{x}-2)({3\mathrm{x}}^2+4\mathrm{x}-4)=0$$$$\mathrm{x}=2,{3\mathrm{x}}^2+4\mathrm{x}-4=0$$$$\mathrm{let}\mathrm{y}={3\mathrm{x}}^2+4\mathrm{x}-4$$$${\mathrm{y}}^{\prime }=6\mathrm{x}+4$$$$\mathrm{from}{\mathrm{y}}^{\prime }=\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}$$$$6\mathrm{x}+4=\pm \sqrt{4^2-4\left(3\right)(-4)}$$$$6\mathrm{x}+4=\pm \sqrt{16+48}$$$$6\mathrm{x}+4=\pm \sqrt{64}$$$$6\mathrm{x}+4=\pm 8$$$$6\mathrm{x}=\pm 8-4$$$$6\mathrm{x}=8-4,-8-4$$$$6\mathrm{x}=4,-12$$$$\mathrm{x}=\frac{4}{6},\frac{-12}{6}$$$$\mathrm{x}=\frac{2}{3}{}^{\prime }-2$$$$\therefore \mathrm{x}=2,\frac{2}{3},-2$$

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