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Solve the polynomial equation x^3-27=0 by factoring and using the principle of zero products



Answer :

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[tex]x^3-27=0 \\ x^3-3^3=0 \\ (x-3)(x^2+3x+3^2)=0 \\ (x-3)(x^2+3x+9)=0 \\ x-3=0 \ \lor \ x^2+3x+9=0 \\ \\ 1. \\ x-3=0 \\ x=3[/tex]

[tex]2. \\ x^2+3x+9=0 \\ a=1 \\ b=3 \\ c=9 \\ b^2-4ac=3^2-4 \times 1 \times 9=9-36=-27 \\ \sqrt{b^2-4ac}=\sqrt{-27}=\sqrt{9 \times 3 \times (-1)}=\pm 3\sqrt{3}i \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-3 \pm 3\sqrt{3}i}{2 \times 1}=\frac{-3 \pm 3\sqrt{3}i}{2}=-\frac{3}{2} \pm \frac{3\sqrt{3}}{2}i \\ x=-\frac{3}{2}-\frac{3\sqrt{3}}{2}i \ \lor \ x=-\frac{3}{2}+\frac{3\sqrt{3}}{2}i[/tex]

[tex]\boxed{x=3 \hbox{ or } x=-\frac{3}{2}-\frac{3\sqrt{3}}{2}i \hbox{ or } x=-\frac{3}{2}+\frac{3\sqrt{3}}{2}i}[/tex]