sat-dog

The correct answer is $-419$. It's given that the expression $\left(3x-23\right)\left(19x+6\right)$ is equivalent to the expression $a{x}^2+bx+c$, where $a$, $b$, and $c$ are constants. Applying the distributive property to the given expression, $\left(3x-23\right)\left(19x+6\right)$, yields $\left(3x\right)\left(19x\right)+\left(3x\right)\left(6\right)-\left(23\right)\left(19x\right)-\left(23\right)\left(6\right)$, which can be rewritten as $57x^2+18x-437x-138$. Combining like terms yields $57x^2-419x-138$. Since this expression is equivalent to $a{x}^2+bx+c$, it follows that the value of $b$ is $-419$.