Answer
$A=-\frac{7}{2}$
Work Step by Step
We are given:
$f(x)=2x^{3}+\mathrm{A}x^{2}+4x-5$ and $f(2)=5$
We solve for $A$:
$f(2)=2(2)^{3}+A(2)^{2}+4(2)-5$
$f(2)=16+4A+8-5$
$f(2)=4A+19$
Since $f(2)=5$:
$5=4A+19$
$-14=4A$
$A=\frac{-14}{4}$
$A=-\frac{7}{2}$
