Question

Jessica foi chamada no quadro para resolver a seguinte divisão do polinômio P(x) = x⁴ + 2x³ - x² + 1 por Q(x) = 2x - 3

Qual é o quociente correto achando por Jessica? ​

Answer 1

Resposta:

x⁴ + 2x³ - x² + 1     |    2x-3

                                  x³/2 +7x²/4+ 17x/8+51/16 é o quociente

-x⁴+3x³/2

+7x³/2-x²+1

-7x²/2+21x²/4

+17x²/4 +1

-17x²/4+51x/8

+51x/8+1

-51x/8+153/16

153/16 +1 =153/16+16/16 =169/16 é o resto

Answer 2

Olá, bom dia ◉‿◉.

Organizando a divisão:

$\begin{array}{r|c}x {}^{4} + 2x {}^{3} -x {}^{2} + 1&2x - 3 \\ \end{array}$

Primeiro ciclo:

Para começar, pegue o primeiro número do divisor e divida pelo primeiro número do dividendo:

$\frac{x {}^{4} }{2x} = \boxed{ \frac{x{}^{3} }{ 2 } }$

Colocando o resultado no quociente:

$\begin{array}{r|c}x {}^{4} + 2x {}^{3} -x {}^{2} + 1&2x - 3 \\ & \frac{x {}^{3} }{2} \end{array}$

Agora multiplique o quociente pelos termos do dividendo:

$\frac{x {}^{3} }{2} . \frac{2x - 3}{1} \\ \frac{x {}^{3}.(2x - 3) }{2} \\ \frac{2x {}^{4} - 3x {}^{3} }{2} \\ \frac{2x {}^{4} }{2} - \frac{3x {}^{3} }{2} \\ \boxed{ x {}^{4} - \frac{3x {}^{3} }{2}}$

Para finalizar o primeiro ciclo, passe o resultado para o outro lado da divisão com os sinais trocados.

$\begin{array}{r|c} \cancel{x {}^{4}} + 2x {}^{3} -x {}^{2} + 1&2x - 3 \\\cancel{- x {}^{4}} + \frac{3x {}^{3} }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & \frac{x {}^{3}}{2} \\ \frac{7x {}^{3} }{2}-x^{2}+1 \: \end{array}$

Segundo ciclo:

Para resolver os outros ciclos, vamos usar a mesma lógica dos anteriores:

$\frac{ \frac{7x {}^{3} }{2} }{2x} = \frac{7x {}^{3} }{2} . \frac{1}{2x} = \frac{7x {}^{3} }{4x} = \boxed{ \frac{7x {}^{2} }{4} }$

$\begin{array}{r|c} \cancel{x {}^{4}} + 2x {}^{3} -x {}^{2} + 1&2x - 3 \\ \cancel{- x {}^{4}} + \frac{3x {}^{3} }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & \frac{x {}^{3} }{2} + \frac{7x {}^{2} }{4} \\ \frac{7x {}^{3} }{2} - x {}^{2} + 1 \: \end{array}$

$\frac{7x {}^{2} }{4} . \frac{2x - 3}{1} \\ \frac{7x {}^{2} .(2x - 3)}{4} \\ \frac{14x {}^{3} - 21 {x}^{2} }{4} \\ \frac{14x {}^{3} }{4} - \frac{21x {}^{2} }{4} \\ \boxed{\frac{7x {}^{3} }{2} - \frac{21x {}^{2} }{4}}$

$\begin{array}{r|c} \cancel{x {}^{4}} + 2x {}^{3} -x {}^{2} + 1&2x - 3 \\ \cancel{- x {}^{4}} + \frac{3x {}^{3} }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & \frac{x {}^{3} }{2} + \frac{7x {}^{2} }{4} \\ \cancel{\frac{7x {}^{3} }{2}} - x {}^{2} + 1 \: \\ \cancel{- \frac{7x {}^{3} }{2} }+ \frac{21x {}^{2} }{4} \: \: \: \: \: \\ \frac{17x {}^{2} }{4} + 1 \end{array}$

$\frac{ \frac{17x {}^{2} }{4} }{2x} = \frac{17x {}^{2} }{4} . \frac{1}{2x} = \frac{17x {}^{2} }{8x} = \boxed{ \frac{17x}{8} }$

$\begin{array}{r|c} \cancel{x {}^{4}} + 2x {}^{3} -x {}^{2} + 1&2x - 3 \\ \cancel{- x {}^{4}} + \frac{3x {}^{3} }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: &\frac{x {}^{3} }{2} + \frac{7x {}^{2} }{4} + \frac{17x}{8} \\ \cancel{\frac{7x {}^{3} }{2}} - x {}^{2} + 1 \: \\ \cancel{- \frac{7x {}^{3} }{2} }+ \frac{21x {}^{2} }{4} \: \: \: \: \: \: \\ \frac{17x {}^{2} }{4} + 1 \end{array}$

$\frac{17x}{8} . \frac{2x - 3}{1} \\ \frac{17x.(2x - 3)}{8} \\ \frac{34x {}^{2} - 51x }{8} \\ \frac{34x {}^{2} }{8} - \frac{51x}{8} \\ \boxed{\frac{17x {}^{2} }{4} - \frac{51x}{8} }$

$\begin{array}{r|c} \cancel{x {}^{4}} + 2x {}^{3} -x {}^{2} + 1&2x - 3 \\ \cancel{- x {}^{4}} + \frac{3x {}^{3} }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: &\frac{x {}^{3}}{2} + \frac{7x {}^{2} }{4} + \frac{17x}{8} \\ \cancel{\frac{7x {}^{3} }{2}} - x {}^{2} + 1 \: \\ \cancel{- \frac{7x {}^{3} }{2} }+ \frac{21x {}^{2} }{4} \: \: \: \: \: \: \\ \cancel{\frac{17x {}^{2} }{4}} + 1 \\ \cancel{ - \frac{17x {}^{2} }{4}} + \frac{51x}{8}\\ \frac{51x}{8} + 1 \end{array}$

$\frac{ \frac{51x}{8} }{2x} = \frac{51x}{8} . \frac{1}{2x} =\frac{51x}{16x} = \boxed{ \frac{51}{16} }$

$\begin{array}{r|c} \cancel{x {}^{4}} + 2x {}^{3} -x {}^{2} + 1&2x - 3 \\ \cancel{- x {}^{4}} + \frac{3x {}^{3} }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & \frac{x {}^{3}}{2} + \frac{7x {}^{2} }{4} + \frac{17x}{8} + \frac{51}{16} \\ \cancel{\frac{7x {}^{3} }{2}} - x {}^{2} + 1 \: \\ \cancel{- \frac{7x {}^{3} }{2} }+ \frac{21x {}^{2} }{4} \: \: \: \: \: \: \\ \cancel{\frac{17x {}^{2} }{4}} + 1 \\ \cancel{ - \frac{17x {}^{2} }{4}} + \frac{51x}{8}\\ \frac{51x}{8} + 1 \end{array}$

$\frac{51}{16} . \frac{2x - 3}{1} \\ \frac{51.(2x - 3)}{16} \\ \frac{102x - 153}{16} \\ \frac{102x}{16} - \frac{153}{16} \\ \boxed{\frac{51x}{8} - \frac{153}{16}}$

$\begin{array}{r|c} \cancel{x {}^{4}} + 2x {}^{3} -x {}^{2} + 1&2x - 3 \\ \cancel{- x {}^{4}} + \frac{3x {}^{3} }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & \frac{x {}^{3}}{2} + \frac{7x {}^{2} }{4} + \frac{17x}{8} + \frac{51}{16} \\ \cancel{\frac{7x {}^{3} }{2}} - x {}^{2} + 1 \: \\ \cancel{- \frac{7x {}^{3} }{2} }+ \frac{21x {}^{2} }{4} \: \: \: \: \: \: \\ \cancel{\frac{17x {}^{2} }{4}} + 1 \\ \cancel{ - \frac{17x {}^{2} }{4}} + \frac{51x}{8}\\ \cancel{\frac{51x}{8}} + 1 \\ \cancel{- \frac{51x}{8}} + \frac{153}{16} \\ \frac{169}{16} \rightarrow resto \end{array}$

RESPOSTA:

$\boxed{ \frac{ {x}^{3} }{2} + \frac{7x {}^{2} }{4} + \frac{17x}{8} + \frac{51}{16}}$