Question

Dividindo-se o polinômio P(x) por x-1, o resto é 1 e dividindo-se por (x-2) e (x-3), o resto é 5.
Determine o resto da divisão de P(x) por (x-1)(x-2)(x-3).​

Answer 1

$\displaystyle \sf P(x)=q_1(x)\cdot(x-1)+1 \to P(1) = 1 \\\\ P(x)=q_2(x)\cdot (x-2)+5 \to P(2)=5 \\\\ P(x)=q_3(x)\cdot(x-3)+5\to P(3)=5 \\\\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\\\ P(x) = Q(x)\cdot (x-1)\cdot(x-2)\cdot(x-3)+ax^2+bx+c$

Daí :

$\displaystyle \sf x=1 \ \therefore a+b+c= 1 \\\\ x=2 \ \therefore 4a+2b+c=5 \\\\ x=3 \ \therefore 9a+3b+c = 5 \\\\\\ \left\{ \begin{array}{I} \sf (I) \ \ a+b+c=1\\\\ \sf (II)\ 4a+2b+c=5\\\\ \sf (III)\ 9a+3b+c=5 \end{array} \right \\\\\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\\\\\ \left\{ \begin{array}{I}\sf (III)-(II) \to 5a +b=0 \\\\ \sf (II)-(I) \to 3a+b=4 \end{array} \right$

$\displaystyle \sf \underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\\\\\ \left\{\sf \begin{array}{I} \sf 5a-3a +b-b = 0-4 \\\\ \sf 2a = - 4 \to \boxed{\sf a = -2} \end{array} \right$

$\sf 5a+b=0 \to \boxed{\sf b = 10 }\\\\ c = 1-a-b \to c = 1+2-10 \to \boxed{\sf c = - 7 }$

Portanto o resto de P(x) por (x-1)(x-2)(x-3) vale :

$\displaystyle \huge\boxed{\sf -2x^2+10x-7 \ }\checkmark$