Question

Considere 1, 2 e 3 raízes de um polinômio do terceiro grau P(x) em que P(0) = 1, logo, P(10) vale?

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{P(x) = [(x - 1).(x - 2).(x - 3)].k}$

$\mathsf{P(x) = [(x - 1).(x^2 - 3x - 2x + 6)].k}$

$\mathsf{P(x) = [(x - 1).(x^2 - 5x + 6)].k}$

$\mathsf{P(x) = [x^3 - 5x^2 + 6x - x^2 + 5x - 6].k}$

$\mathsf{P(x) = [x^3 - 6x^2 + 11x - 6].k}$

$\mathsf{P(0) = [(0)^3 - 6.(0)^2 + 11(0) - 6].k}$

$\mathsf{P(0) = [0 - 0 + 0 - 6].k}$

$\mathsf{P(0) = -6k}$

$\mathsf{1 = -6k}$

$\mathsf{k = -\dfrac{1}{6}}$

$\mathsf{P(10) = [(10)^3 - 6(10)^2 + 11(10) - 6].\left(-\dfrac{1}{6}\right)}$

$\mathsf{P(10) = [1000 - 600 + 110 - 6].\left(-\dfrac{1}{6}\right)}$

$\mathsf{P(10) = \left(-\dfrac{504}{6}\right)}$

$\boxed{\boxed{\mathsf{P(10) = -84}}}$