I need to find $b$ and $c$ such that $f(x)$ is divisible by $(x - 1)^2.$ One way to do that is to use polynomial long division and make sure the remainder is zero. I think the second way is faster and easier, so I'll use that. To find $f(1),$ I just plug in $x = 1$ into $f(x):$ $f(1) = 1^3 + b + c = 1 + b + c.$ Now I have two equations and two unknowns: $1 + b + c = 0$ and $3 + b = 0.$ I can solve for $b$ from the second equation: $b = -3.$ Then I can plug in $b = -3$ into the first equation and solve for $c:$ $1 - 3 + c = 0,$ so $c = 2.$ Therefore, the ordered pair $(b,c)$ is $(-3,2).$ 
Let $f(x) = x^3 + bx + c.$  If $(x - 1)^2$ is a factor of $f(x),$ then find the ordered pair $(b,c).$