Here's the question you clicked on:
janirish
factor x2 - 14x + 49
try to do the same process, as what we do on x^2-6x+9, earlier.
\[\Huge{x^2-14x+99}\]\[\Large{\implies x^2-7x-7x+49}\]\[\Large{\implies x(x-7)-7(x-7)=??}\]
i know you can do it janirish, do not be afraid to try, :)
break the midterm x^2-7x-7x+49 x(x-7)-7(x-7) (x-7)(x-7)
why we ask such question
@sami-21 plz do not give full answers http://openstudy.com/code-of-conduct