Take out a common factor of y^2y2 first.
y^2(y^2-14y color(red)(+49)) " "rarr color(red)(+49)y2(y2−14y+49) →+49 a clue!
You should recognize 4949 as being a perfect square!
In (x+y)^2 = x^2 +2xy +y^2 = ax^2+bx + c(x+y)2=x2+2xy+y2=ax2+bx+c there is always a special relationship between the coefficient of the second term (b) and the third term (c)
(bdiv 2) " and then squared gives " c(b÷2) and then squared gives c.
If this relationship exists in the trinomial you will know you have a square of a binomial.
Let's look closer..... (14div2)^2 = 7^2=49!!(14÷2)2=72=49!!
:. y^2color(red)((y^2-14y+49)) = y^2color(red)((y-7)^2)
If you missed that, you can try to find factors of 49 which add to 14. This will give you factors of 7 and 7 anyway:
y^2(y-7)(y-7)
=y^2(y-7)^2