Question 597877
{{{c^2-1}}} Start with the given expression.



{{{(c)^2-(1)^2}}} Rewrite {{{1}}} as {{{(1)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=c}}} and {{{B=1}}}.



So let's use the difference of squares formula {{{A^2-B^2=(A+B)(A-B)}}} to factor the expression:



{{{A^2-B^2=(A+B)(A-B)}}} Start with the difference of squares formula.



{{{(c)^2-(1)^2=(c+1)(c-1)}}} Plug in {{{A=c}}} and {{{B=1}}}.



So this shows us that {{{c^2-1}}} factors to {{{(c+1)(c-1)}}}.



In other words {{{c^2-1=(c+1)(c-1)}}}.



Hopefully this helps. If not, let me know.