Question

How do I complete the square and factor the resulting trinomial?

Answer 1

To complete the square we have to divided the linear coefficient by 2, and then squared it. In this case

`\begin{gathered} y^2-5y+(-(5)/(2))^2 \\ =y^2-5y+(25)/(4) \end{gathered}`

to factor it we have to remember that this will always be the squared term plus what we add (before squaring). Then

`\begin{gathered} y^2-5y+(-(5)/(2))^2=y^2-5y+(25)/(4) \\ =(y-(5)/(2))^2 \end{gathered}`