O resto de um polinômio será sempre um grau a menos que o divisor do polinômio [tex]\displaystyle \sf P(x) = q(x)\cdot \underbrace{\sf (x-1)\cdot (x-2)\cdot (x-3)}_{\displaystyle \text{divsor de grau 3}}+r(x) \\\\\\ \text{Se o divisor \'e de grau 3, ent\~ao o resto ser\'a no m\'aximo de grau 2} : \\\\ Da{\'i}}: \\\\ P(x) = q(x)\cdot (x-1)\cdot (x-2)\cdot (x-3)+ax^2+bx+c \\\\\ P(1) = 2 \to a.1^2+b.1+c= 2 \\\\ P(2)= 1 \to a.2^2+b.2+c=1 \\\\ P(3) = -4 \to a.3^2+3.b+c = -4 \\\\ temos : \\\\ (I) \ \ a+b+c = 2 \\\\ (II) \ 4a+2b+c=1 \\\\ (III) 9a+3b+c =-4[/tex]
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O resto de um polinômio será sempre um grau a menos que o divisor do polinômio
[tex]\displaystyle \sf P(x) = q(x)\cdot \underbrace{\sf (x-1)\cdot (x-2)\cdot (x-3)}_{\displaystyle \text{divsor de grau 3}}+r(x) \\\\\\ \text{Se o divisor \'e de grau 3, ent\~ao o resto ser\'a no m\'aximo de grau 2} : \\\\ Da{\'i}}: \\\\ P(x) = q(x)\cdot (x-1)\cdot (x-2)\cdot (x-3)+ax^2+bx+c \\\\\ P(1) = 2 \to a.1^2+b.1+c= 2 \\\\ P(2)= 1 \to a.2^2+b.2+c=1 \\\\ P(3) = -4 \to a.3^2+3.b+c = -4 \\\\ temos : \\\\ (I) \ \ a+b+c = 2 \\\\ (II) \ 4a+2b+c=1 \\\\ (III) 9a+3b+c =-4[/tex]
[tex]\displaystyle \sf \text{Fa\c camos}: \\\\ (III)-(II) : \\\\ 9a+3b+c-4a-2b-c = -4-1 \\\\ 5a+b = -5 \\\\\\ (II)-(I) : \\\\ 4a+2b+c - a-b-c = 1-2\\\\ 3a+b=-1 \\\\ Da{\'i}}: \\\\ 5a+b=-5 \\ \underline{3a+b=-1\ \ \ \ \ \ -} \\\\ 5a+b-3-b=-5-(-1) = -4 \\\\ 2a = -4\to \boxed{\sf a =-2} \\\\ b = -1-3a\to b=-1-3(-2) \to \boxed{\sf b=5 } \\\\ a+b+c=2 \to -2+5+c=1 \to \boxed{\sf c = -1}[/tex]
Portanto o resto da divisão de P(x) por (x-1)(x-2)(x-3) é :
[tex]\Large\boxed{\sf \ -2x^2+5x-1 \ }\checkmark[/tex]