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

When a polynomial $P(x)$ is divided by $(x-k)$, the remainder equals $P(k)$.

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$P(x) = ax^2 + bx + 5$

Dividing by $(x-1)$, the remainder is $P(1)=3$.

$P(1) = a(1)^2 + b(1) + 5 = 3$

$\Rightarrow a + b + 5 = 3$

$\Rightarrow a + b = -2 \quad \rightarrow$ (1)

Dividing by $(x+1) = (x-(-1))$, the remainder is $P(-1)=2$.

$P(-1) = a(-1)^2 + b(-1) + 5 = 2$

$\Rightarrow a - b + 5 = 2$

$\Rightarrow a - b = -3 \quad \rightarrow$ (2)

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Adding equations (1) and (2):

$2a = -5$

$a = -\frac{5}{2}$

Substituting into equation (1):

$-\frac{5}{2} + b = -2$

$b = \frac{1}{2}$

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Calculating $2b - 4a$:

$=2(\frac{1}{2}) - 4(-\frac{5}{2})$

$= 1 + 10$

$= 11$