2)
[tex] \bf 2x^2+3\sqrt{5}x+5~~\begin{cases}5=&\sqrt{5}\cdot \sqrt{5}\\3\sqrt{5}=&2\sqrt{5}+\sqrt{5}\end{cases}\implies \left( 2x+\sqrt{5} \right)\left(x+\sqrt{5} \right) [/tex]
3)
let's keep in mind that
1² = 1
1³ = 1
1⁴ = 1
1¹⁰⁰⁰⁰⁰⁰⁰⁰ = 1
[tex] \bf x^4-5x^2+4\implies (x^2)^2-5x^2+4~~\begin{cases}-5x^2=&-4x^2-x^2\\4=&-4\cdot -1\end{cases}\\\\\\(x^2-4)(x^2-1)\implies \stackrel{\textit{difference of squares on both}}{(x^2-2^2)(x^2-1^2)}\\\\\\(x-2)(x+2)(x-1)(x+1) [/tex]
