Answer:
Let f(x) = 4[tex]x^{3}[/tex] + 16[tex]x^{2}[/tex] + 19x + 10.
Comparing (2x + 5) with (x - a), a = [tex]\frac{-5}{2}[/tex]
Now,
Using remainder theorem,
f(a) = Remainder(R)
or, R = 4[tex](-2.5)^{3}[/tex] + 16 [tex](-2.5)^{2}[/tex] + 19(-2.5) + 10
or, R = -62.5 + 100 - 47.5 + 10
or, R = 0
Therefore, the result is zero.
