Temos :
[tex]\displaystyle \text {P(x)} = \text{a.x}^3+\text{b.x}^2+\text{c.x}+\text d \\\\ \text {Q(x)}=\text {e.x}^3+\text{f.x}^2+\text{g.x}+\text h \\\\ \text{P(x)+Q(x)} = 2\text x^3+3\text x^2-3\text x-4 \\\\ \text{isto {\'e}} : \\\\ \text{(a + e).x}^3+\text{(b + f).x}^2+\text{(c + g).x}+\text h +\text d = 2\text x^3+3\text x^2-3\text x-4 \\\\ \underline {\text{Com isso, sabemos que ser{\'a} poss{\'i}vel se}}:[/tex]
[tex]\displaystyle \text{a + e} = 2\to \text a = 2-\text e\\\\ \text{b + f}=3\to \text b=3-\text f \\\\ \text{c + g} = -3 \to \text c = -3-\text g\\\\ \text {h + d}=-4 \to \text d=-4-\text h[/tex]
[tex]\text{Fa{\c c}amos P}(\sqrt{2}) =3 : \\\\ \text a(\sqrt{2})^3+\text b.(\sqrt{2})^2+\text c.\sqrt{2}+\text d = 3 \\\\ 2\sqrt{2}.\text a+2.\text b+\text c.\sqrt{2}+\text d = 3[/tex]
[tex]\text{Fa{\c c}amos Q}(\sqrt{2}) : \\\\ \text e(\sqrt{2})^3+\text f.(\sqrt{2})^2+\text g.\sqrt{2}+\text h \\\\ 2\sqrt{2}.\text e+2.\text f+\text g.\sqrt{2}+\text h[/tex]
vamos usar a informação :
[tex]\displaystyle \text a = 2-\text e\\\\ \text b=3-\text f \\\\ \text c = -3-\text g\\\\ \text d=-4-\text h[/tex]
vamos multiplicar dos dois lados até ficar parecida com os coeficientes de [tex]\text P(\sqrt{2})[/tex], assim :
[tex]\displaystyle 2\sqrt{2}.\text a = 2.2\sqrt{2}-2.\sqrt{2}.\text e\\\\ 2.\text b=2.3-2.\text f \\\\ \sqrt{2}.\text c = -3.\sqrt{2}-\sqrt{2}.\text g\\\\ \text d=-4-\text h[/tex]
somando ambos os lados :
[tex]\displaystyle 2\sqrt{2}.\text a+2.\text b+\text c.\sqrt{2}+\text d = 4\sqrt{2}-2\sqrt{2}.\text e+6-2\text f-3\sqrt{2}-\sqrt{2}.\text g-4-\text h \\\\ \text{Por{\'e}m sabemos que o lado esquerdo {\'e} P}(\sqrt{2)}=3 \ , \text{da{\'i}} : \\\\ 3 = 4\sqrt{2}-3\sqrt{2}+6-4-(2\sqrt{2}.\text e+2\text f+\sqrt{2}\text g+\text h) \\\\ \text{Po{\'em} o que est{\'a} entre parenteses {\'e} justamente Q(}\sqrt{2)} \ , \text{Da{\'i}}: \\\\ 3=\sqrt{2}+2-\text {Q(}\sqrt{2}) \\\\ \text{Q}(\sqrt{2})=\sqrt{2}+2-3[/tex]
[tex]\huge\boxed{\text{Q}(\sqrt{2})=\sqrt{2}-1\ }\checkmark[/tex]
