Răspuns :
[tex] \it E(x) = x^3-2x^2-x+2=x^2(x-2)-(x-2) =(x-2)(x^2-1)=
\\ \\
=(x-2)(x-1)(x+1).[/tex]
b)
[tex] \it E(-3) = (-3-2)(-3-1)(-3+1) = -5\cdot(-4)\cdot(-2)=-40 [/tex]
c)
[tex] \it E(\sqrt2) = (\sqrt2)^3-2(\sqrt2)^2-\sqrt2+2=2\sqrt2-2\cdot2-\sqrt2+2=\sqrt2-2
\\ \\
E(-\sqrt2) = (-\sqrt2)^3-2(-\sqrt2)^2-(-\sqrt2)+2=
\\ \\
=-2\sqrt2-2\cdot2+\sqrt2+2=-\sqrt2-2=-(\sqrt2+2)
\\ \\
E(\sqrt2)\cdot E(-\sqrt2) =- (\sqrt2-2)(\sqrt2+2)=-[(\sqrt2)^2-2^2] =
\\ \\
=-(2-4)=-(-2)=2 \in\mathbb{N} [/tex]
Am anexat inca o rezolvare ---
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