Question

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

Answer 1

`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`


`2. \\ x^2+3x+9=0 \\ a=1 \\ b=3 \\ c=9 \\ b^2-4ac=3^2-4 * 1 * 9=9-36=-27 \\ √(b^2-4ac)=√(-27)=√(9 * 3 * (-1))=\pm 3√(3)i \\ x=(-b \pm √(b^2-4ac))/(2a)=(-3 \pm 3√(3)i)/(2 * 1)=(-3 \pm 3√(3)i)/(2)=-(3)/(2) \pm (3√(3))/(2)i \\ x=-(3)/(2)-(3√(3))/(2)i \ \lor \ x=-(3)/(2)+(3√(3))/(2)i`


`\boxed{x=3 \hbox{ or } x=-(3)/(2)-(3√(3))/(2)i \hbox{ or } x=-(3)/(2)+(3√(3))/(2)i}`