Question

Assunto: Divisão de P(x) por ax + b, a diferente de 0.

Quero o resto da divisão utilizando a técnica de P ( - b / a)


4/27 . x^3 + 4/9 . x^2 + 2/3 . x + 1 dividido por 2x - 3


Resposta: r = 7/2


x^3 - 3x - 1/27 dividido 3x - 1


resposta: r = - 1

Para a primeira questão fiz P (3/2) e para segunda, P(1 / 3) mas não bate.

Answer 1

Vejamos o úso da técnica:

a)

$P(x)=\frac{4}{27}x^3+\frac{4}{9}x^2+\frac{2}{3}x+1\\ \\ P(\frac{3}{2})=\frac{4}{27}(\frac{3}{2})^3+\frac{4}{9}(\frac{3}{2})^2+\frac{2}{3}.(\frac{3}{2})+1\\ \\ P(\frac{3}{2})=\frac{4}{27}(\frac{27}{8})^3+\frac{4}{9}(\frac{9}{4})^2+\frac{2}{3}.(\frac{3}{2})+1\\ \\ P(\frac{3}{2})=\frac{4}{27}(\frac{27}{8})+\frac{4}{9}(\frac{9}{4})+\frac{2}{3}.(\frac{3}{2})+1\\ \\ P(\frac{3}{2})=\frac{1}{2}+1+1+1=\frac{7}{2}$

b)

$P(x)=x^3 - 3x - \frac{1}{27}\\ \\ P(\frac{1}{3})=(\frac{1}{3})^3 - 3(\frac{1}{3}) - \frac{1}{27}\\ \\ P(\frac{1}{3})=\frac{1}{27} - 1 - \frac{1}{27}=-1$

Tudo certo. :-)

Answer 2

Explicação passo-a-passo:

1)

$\sf P\left(\dfrac{3}{2}\right)=\dfrac{4}{27}\cdot\left(\dfrac{3}{2}\right)^3+\dfrac{4}{9}\cdot\left(\dfrac{3}{2}\right)^2+\dfrac{2}{3}\cdot\dfrac{3}{2}+1$

$\sf P\left(\dfrac{3}{2}\right)=\dfrac{4}{27}\cdot\dfrac{27}{8}+\dfrac{4}{9}\cdot\dfrac{9}{4}+1+1$

$\sf P\left(\dfrac{3}{2}\right)=\dfrac{1}{2}+1+1+1$

$\sf P\left(\dfrac{3}{2}\right)=\dfrac{1}{2}+3$

$\sf P\left(\dfrac{3}{2}\right)=\dfrac{1+6}{2}$

$\sf \red{P\left(\dfrac{3}{2}\right)=\dfrac{7}{2}}$

2)

$\sf P\left(\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^3-3\cdot\dfrac{1}{3}-\dfrac{1}{27}$

$\sf P\left(\dfrac{1}{3}\right)=\dfrac{1}{27}-1-\dfrac{1}{27}$

$\sf P\left(\dfrac{1}{3}\right)=\dfrac{1-27-1}{27}$

$\sf P\left(\dfrac{1}{3}\right)=\dfrac{-27}{27}$

$\sf \red{P\left(\dfrac{1}{3}\right)=-1}$