Barely Sufficient Equations

$\Rightarrow \quad { x }^{ 4 }+6{ x }^{ 2 }+25\\ \Rightarrow \quad { x }^{ 4 }+10{ x }^{ 2 }+25-4{ x }^{ 2 }\\ \Rightarrow \quad { ({ x }^{ 2 }+5) }^{ 2 }-{ (2x) }^{ 2 }\\ \Rightarrow \quad ({ x }^{ 2 }+2x+5)({ x }^{ 2 }-2x+5)\\$

Now $f\left( x \right)$ can be any one of the above factors so we have to check for the common factor by dividing it by other polynomial.

Hence, on dividing ${ x }^{ 2 }-2x+5$ by $3{ x }^{ 4 }+4{ x }^{ 2 }+28x+5$ we get remainder 0.

$\therefore \quad f\left( 1 \right)=1-2+5=\boxed { 4 } .$