If the polynomial \(ax^2 + bx + 5\) leaves a remainder \(3\) when divided by \(x - 1\), and a remainder \(2\) when divided by \(x + 1\), then \(2b - 4a\) equals

The value of $2b-4a$ is $11$. Evaluating the polynomial at $1$ and $-1$ gives $a+b=-2$ and $a-b=-3$, so $a=-\frac52$ and $b=\frac12$. Reversing the subtraction that finds $b$ changes $2b$ to $-1$ and gives $9$.