### Instruction: Let $b$ and $c$ be real numbers.  If the polynomial $x^2+bx+c$ has exactly one real root and $b=c+1$, find the value of the product of all possible values of $c$.### Response: I know that a quadratic polynomial has exactly one real root when its discriminant, which is the part under the square root in the quadratic formula, is equal to zero.
So, I need to find the values of $b$ and $c$ that make $b^2-4c=0$.
I also know that $b=c+1$, so I can substitute that into the equation and get $(c+1)^2-4c=0$.
Expanding the square, I get $c^2+2c+1-4c=0$, which simplifies to $c^2-2c+1=0$.
This is a perfect square, so I can factor it as $(c-1)^2=0$.
This means that $c=1$ is the only possible value that satisfies the conditions.
Therefore, the product of all possible values of $c$ is just $1$.