Answer:
[tex]a^{4}-3\cdot a^{2}+9 = \left(a^{2}-\frac{3}{2}-i\,\frac{\sqrt{27}}{2}\right)\cdot \left(a^{2}-\frac{3}{2}+i\,\frac{\sqrt{27}}{2}\right)[/tex]
Step-by-step explanation:
Let [tex]a^{4}-3\cdot a^{2}+9[/tex], if we apply the algebraic substitution [tex]u = a^{2}[/tex], we can reduce this quartic equation into a quadratic equation:
[tex]u^{2}-3\cdot u + 9[/tex] (1)
The roots of this quadratic equation is determined by the Quadratic Formula:
[tex]u_{1,2} = \frac{3\pm\sqrt{(-3)^{2}-4\cdot (1)\cdot (9)}}{2\cdot (1)}[/tex]
[tex]u_{1,2} = \frac{3}{2}\pm i\,\frac{1}{2}\cdot \sqrt{27}[/tex]
The solutions of this polynomial are: [tex]u_{1} = \frac{3}{2}+i\,\frac{\sqrt{27}}{2}[/tex], [tex]u_{2} = \frac{3}{2}-i\,\frac{\sqrt{27}}{2}[/tex]. Then, the factorized form is:
[tex]\left(u-\frac{3}{2}-i\,\frac{\sqrt{27}}{2} \right)\cdot \left(u-\frac{3}{2}+i\,\frac{\sqrt{27}}{2} \right)[/tex]
[tex]\left(a^{2}-\frac{3}{2}-i\,\frac{\sqrt{27}}{2} \right)\cdot \left(a^{2}-\frac{3}{2}+i\,\frac{\sqrt{27}}{2} \right)[/tex]
