Question

The coefficient of x³ in the expansion (2x-1)³​

Answer 1

`\qquad \qquad \textit{binomial theorem expansion} \\\\ \qquad \qquad (2x-1)^(3)~\hspace{4em} \begin{array}{cccl} term&coefficient&value\\ \cline{1-3}&\\ 1&+1&(2x)^(3 )(-1)^0\\ 2&+3&(2x)^(2)(-1)^1\\ 3&+3&(2x)^(1)(-1)^2\\ 4&+1&(2x)^(0)(-1)^3 \end{array} \\\\\\ 1(8x^3)+3(-4x^2)+3(2x)+1(-1)\implies \text{\LARGE 8}x^3-12x^2+6x-1`

Answer 2

Answer: 8

Explanation:

Since the expression is raised to the 3rd power, the highest exponent x will have will be 3. So in order to get the coefficient you multiply 2x by itself 3 times, getting (2x)³ = (2 * 2 * 2)(x * x * x) = (8)(x³) = 8x³