Question

Encontre o polinomio p (x), de grau 3, tal que : p (x+1)-p (x)=x2

Answer 1

Sendo
$p(x) = ax^3 + bx^2 + cx + d$

Entao
$p(x+1) = a(x+1)^3 + b(x+1)^2 + c(x+1) + d \\ = a(x^3+3x^2+3x+1) + b(x^2 + 2x + 1) + c(x+1) + d \\ = ax^3 + (3a+b)x^2 + (3a+2b+c)x + (a+b+c+d)$

Como p(x+1)-p(x) = x^2
$ax^3 + (3a+b)x^2 + (3a+2b+c)x + (a+b+c+d) - ax^3 - bx^2 - cx - d = 0x^3 + 1x^2 + 0x + 0 \\ \implies 3ax^2 + (3a+2b)x + (a+b+c) = 0x^3 + 1x^2 + 0x + 0 \\ \\ \implies \\ \\ 3a = 1 \\ 3a+2b=0 \\ a+b+c = 0 \\ \\ \implies \\ \\ a= \frac{1}{3} \\ b=- \frac{1}{2} \\ c=\frac{1}{6}$

Portanto

$\boxed{\boxed{ p(x) = \frac{1}{3} x^3 - \frac{1}{2} x^2 + \frac{1}{6} x + d}}$

Sendo "d" qualquer real (as exigências enunciadas não restringem o valor de "d")