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

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

Polynomial $$ax^2 + bx + 5$$ leaves a remainder 3 when divided by $$x - 1$$. Thus, as per the remainder theorem, when x = 1, the remainder is 3. 

f(1) = 3 => $$a+b+5=3$$ => $$a+b=-2\rightarrow1$$

Polynomial $$ax^2 + bx + 5$$ leaves a remainder 2 when divided by $$x + 1$$. Thus, as per the remainder theorem, when x = -1, the remainder is 2.

f(-1) = 3 => $$a-b+5=2$$ => $$a-b=-3\rightarrow2$$

Adding eq. 1 and eq. 2 - 

=> $$2a=-5$$ => $$a=-2.5$$

Subtracting eq. 2 from eq. 1 -

=> $$2b=1$$ => $$b=0.5$$

Therefore the value of $$2b-4a=2(0.5)-4(-2.5)=1+10=11$$