Answer
Two different values are $34$ and $29$.
$2(n+1)(3n+14)$ and $ (3n+4)(2n+7)$.
Work Step by Step
The given expression is
$6n^2+\_n+28$
The standard form of the trinomial is
$ax^2+bx+c$
Where, $a=6,c=28$ and $b=?$.
Multiply $a$ and $c$.
$ac=6\times 28=168$
Two possible factors are $6\times 28$ and $21\times 8$.
So the values of $b$ are $6+28=34$ and $21+8=29$.
The first trinomial is
$\Rightarrow 6n^2+34n+28$
Factor.
$\Rightarrow 2(n+1)(3n+14)$.
The second trinomial is
$\Rightarrow 6n^2+29n+28$
Factor.
$\Rightarrow (3n+4)(2n+7)$.
