c^2-2c+1-9f^2

Ashley:

Knowing that I don't see an equal sign on this polynomial, it is meant that it must be treated as an expression, so I assume the objective with this expression is to factor it.

If factoring is the objective in this exercise, consider there are two areas of interest in this expression:

=> First, *c*^{2} - 2*c* + 1 is a perfect square trinomial, so it is factored this way:

*c*^{2} - 2*c* + 1 = *c*^{2} - *
c* - *c* + 1

= *c*(*c* - 1) - 1(*c* - 1)

= (*c* - 1)(*c* - 1)

= (*c* - 1)^{2}

Note: The above trinomial was factored by grouping.

Second, the term 9*f *^{2} can be written as (3*f*)^{2} knowing that both 9 and *f*^{ 2} are squares.

With both areas of interest being covered, the original expression can be rewritten as (*c* - 1)^{2} - (3*f
*)^{2}.

Notice that the rewritten expression has the following notation: *x*^{2} -
*y*^{2}, which is a difference of squares. In this case, *x* = (*c* - 1)^{2} and
*y* = (3*f*)^{2}.

Since, for any difference of squares, *x*^{2} - *y*^{2} = (*x* -
*y*)(*x* + *y*), the expression

(*c* - 1)^{2} - (3*f*)^{2} = [(*c* - 1) - 3*f*] [(*c* - 1) + 3*f*].

The use of brackets was needed to distinguish one factor from the other.

## Comments

Ashley- what are you supposed to do with this equation? i.e. does it ask you to solve for f, factor, graph, etc. ?

I don't know where to start with this either. What was the prompt for all of the problems in the section this was under?