Answer
(a) $2 x^{2}+5x+11,\ \ \ 29$
(b) $ -2 x^{2}+5x-3,\ \ \ \ -1$
(c) $10x^3+8x^2+35x+28,\ \ \ 210$
(d) $\dfrac{5 x+4}{ 2 x^{2}+7}m\ \ \ \ \dfrac{14}{ 15}$
Work Step by Step
Given $$f(x)=5 x+4 ;\ \ g(x)=2 x^{2}+7$$
(a) The sum is given by
\begin{align*}
f(x)+g(x)&=5 x+4 +2 x^{2}+7\\
&=2 x^{2}+5x+11
\end{align*}
The sum at $x=2$ is $ 2(2)^2+5(2)+11=29$
(b) The difference is given by
\begin{align*}
f(x)-g(x)&=5 x+4 -2 x^{2}-7\\
&=-2 x^{2}+5x-3
\end{align*}
The difference at $x=2$ is $ -2(2)^2+5(2)-3=-1$
(c) The product is given by
\begin{align*}
f(x)*g(x)&=(5 x+4)( 2 x^{2}+7)\\
&= 10x^3+8x^2+35x+28
\end{align*}
The product at $x=2$ , $10(2)^3+8(2)^2+35(2)+28 = 210$
(d) The quotient
\begin{align*}
\frac{f(x)}{g(x)}&=\frac{5 x+4}{ 2 x^{2}+7}
\end{align*}
The quotient at $x=2$ , $\dfrac{5 (2)+4}{ 2 (2)^{2}+7}=\frac{14}{15}$
