Respuestas
Respuesta dada por:
4
Hola!!☺☺
Recordemos que:
![\boxed{ \mathtt{ \boldsymbol{D = dq + r}}}\\ \\ Donde \\ \\ D = Dividendo \\ \\ d = divisor \\ \\ q = cociente \\ \\ r = residuo \boxed{ \mathtt{ \boldsymbol{D = dq + r}}}\\ \\ Donde \\ \\ D = Dividendo \\ \\ d = divisor \\ \\ q = cociente \\ \\ r = residuo](https://tex.z-dn.net/?f=+%5Cboxed%7B+%5Cmathtt%7B+%5Cboldsymbol%7BD+%3D+dq+%2B+r%7D%7D%7D%5C%5C++%5C%5C+Donde+%5C%5C++%5C%5C+D+%3D+Dividendo+%5C%5C++%5C%5C+d+%3D+divisor+%5C%5C++%5C%5C+q+%3D+cociente+%5C%5C++%5C%5C+r+%3D+residuo)
Entonces
![d = {x}^{2} + 1 \\ \\ q = x + 2 \\ \\ r = x - 3 d = {x}^{2} + 1 \\ \\ q = x + 2 \\ \\ r = x - 3](https://tex.z-dn.net/?f=d+%3D++%7Bx%7D%5E%7B2%7D++%2B+1+%5C%5C++%5C%5C+q+%3D+x+%2B+2+%5C%5C++%5C%5C+r+%3D+x+-+3)
![D = dq + r \\ \\ D = ( {x}^{2} + 1)(x + 2) + (x - 3) \\ \\ D = ( {x}^{3} + 2 {x}^{2} + x + 2) + (x - 3) \\ \\ D = {x}^{3} + 2 {x}^{2} + 2x - 1 D = dq + r \\ \\ D = ( {x}^{2} + 1)(x + 2) + (x - 3) \\ \\ D = ( {x}^{3} + 2 {x}^{2} + x + 2) + (x - 3) \\ \\ D = {x}^{3} + 2 {x}^{2} + 2x - 1](https://tex.z-dn.net/?f=D+%3D+dq+%2B+r+%5C%5C++%5C%5C+D+%3D+%28+%7Bx%7D%5E%7B2%7D++%2B+1%29%28x+%2B+2%29+%2B+%28x+-+3%29+%5C%5C++%5C%5C+D+%3D+%28+%7Bx%7D%5E%7B3%7D++%2B+2+%7Bx%7D%5E%7B2%7D++%2B+x+%2B+2%29+%2B+%28x+-+3%29+%5C%5C++%5C%5C+D+%3D++%7Bx%7D%5E%7B3%7D++%2B+2+%7Bx%7D%5E%7B2%7D++%2B+2x+-+1)
Recordemos que:
Entonces
Preguntas similares
hace 6 años
hace 6 años
hace 9 años
hace 9 años