Question

Qual é o desenvolvimento do binômio(3x – 1)4 (3x - 1) elevado a quarta potência

Answer 1

O binômio de grau 4 dará origem a um polinômio, também, de grau 4, ou seja, um polinômio na forma:

$\boxed{ax^4~+~bx^3~+~cx^2~+~dx~+~e}$

Utilizando expressão do termo geral do binômio (abaixo), podemos montar o polinômio gerado a partir de (x+b)ⁿ.

$\boxed{T_{p+1}~=~\dbinom{n}{p}\cdot x^{n-p}\cdot b^p}$

Assim, teremos:

$(3x-1)^4~=\\\\\\=~\dbinom{4}{0}\cdot (3x)^{4-0}\cdot(-1)^0~+~\dbinom{4}{1}\cdot (3x)^{4-1}\cdot(-1)^1~+~\dbinom{4}{2}\cdot (3x)^{4-2}\cdot(-1)^2\\\\~\\~~~~~+~\dbinom{4}{3}\cdot (3x)^{4-3}\cdot(-1)^3~+~\dbinom{4}{4}\cdot (3x)^{4-4}\cdot(-1)^4\\\\\\\\=~\frac{4!}{0!\cdot(4-0)!}\cdot(3x)^4\cdot1~+~\frac{4!}{1!\cdot(4-1)!}\cdot(3x)^3\cdot(-1)~+~\frac{4!}{2!\cdot(4-2)!}\cdot(3x)^2\cdot1\\\\\\~~~+~\frac{4!}{3!\cdot(4-3)!}\cdot(3x)^1\cdot(-1)~+~\frac{4!}{4!\cdot(4-4)!}\cdot(3x)^0\cdot(-1)$

$=~\frac{24}{1\cdot24}\cdot81x^4\cdot1~+~\frac{24}{1\cdot6}\cdot27x^3\cdot(-1)~+~\frac{24}{2\cdot2}\cdot9x^2\cdot1\\\\\\~~~+~\frac{24}{6\cdot1}\cdot3x\cdot(-1)~+~\frac{24}{24\cdot1}\cdot1\cdot(-1)\\\\\\\\=~1\cdot81x^4\cdot1~+~4\cdot27x^3\cdot(-1)~+~6\cdot9x^2\cdot1\\\\\\~~~+~4\cdot3x\cdot(-1)~+~1\cdot1\cdot(-1)\\\\\\\\=~\boxed{81x^4~-~108x^3~+~54x^2~-~12x~-~1}$