Answer:
[tex]a) f(x)=3x^{3}-6x^{2}-15x+30[/tex]
Step-by-step explanation:
1) In this question we've been given "a", the leading coefficient. and two roots:
[tex]x_{1}=\sqrt{5}\:x_{2}=2[/tex]
2) There's a theorem, called the Irrational Theorem Root that states:
If one root is in this form [tex]x'=\sqrt{a}+b[/tex] then its conjugate [tex]x''=\sqrt{a}-b[/tex]. is also a root of this polynomial.
Therefore
[tex]x_3=-\sqrt{5}[/tex]
3) So, applying this Theorem we can rewrite the equation, by factoring. Remembering that x is the root. Since the question wants it in this expanded form then:
[tex]f(x)=3(x-\sqrt{5})(x+\sqrt{5})(x-2)\Rightarrow 3x^{3}-6x^{2}-15x+30[/tex]
