Answer:
[tex]y^2+28y+196[/tex]
Step-by-step explanation:
We are given that
[tex]y^2+28y+..[/tex]
We have to fill in the number that makes the polynomial a perfect square quadratic.
We know that
[tex](a+b)^2=a^2+2\times a\times b+b^2[/tex]
Using the formula
Expand the expression
[tex](y)^2+2\times y\times 14+(14)^2[/tex]
[tex](y+14)^2[/tex]
[tex]=y^2+28y+196[/tex]
