Question 209704
You don't need to use the quadratic formula.



{{{x^2-2ax+a^2-b^2}}} Start with the given expression.



{{{x^2-2ax+a^2}}} We're going to ignore the last term (for now) and focus on factoring the first three terms.



Looking at the expression {{{x^2-2ax+a^2}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-2}}}, and the last coefficient is {{{1}}}.



Now multiply the first coefficient {{{1}}} by the last coefficient {{{1}}} to get {{{(1)(1)=1}}}.



Now the question is: what two whole numbers multiply to {{{1}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-2}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{1}}} (the previous product).



Factors of {{{1}}}:

1

-1



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{1}}}.

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


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-2}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>1+1=2</font></td></tr><tr><td  align="center"><font color=red>-1</font></td><td  align="center"><font color=red>-1</font></td><td  align="center"><font color=red>-1+(-1)=-2</font></td></tr></table>



From the table, we can see that the two numbers {{{-1}}} and {{{-1}}} add to {{{-2}}} (the middle coefficient).



So the two numbers {{{-1}}} and {{{-1}}} both multiply to {{{1}}} <font size=4><b>and</b></font> add to {{{-2}}}



Now replace the middle term {{{-2ax}}} with {{{-ax-ax}}}. Remember, {{{-1}}} and {{{-1}}} add to {{{-2}}}. So this shows us that {{{-ax-ax=-2ax}}}.



{{{x^2+highlight(-ax-ax)+a^2}}} Replace the second term {{{-2ax}}} with {{{-ax-ax}}}.



{{{(x^2-ax)+(-ax+a^2)}}} Group the terms into two pairs.



{{{x(x-a)+(-ax+a^2)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x-a)-a(x-a)}}} Factor out the GCF {{{-a}}} from the second group.



{{{(x-a)(x-a)}}} Combine like terms.



{{{(x-a)^2}}} Condense.



So {{{x^2-2ax+a^2}}} factors to {{{(x-a)^2}}}



----------------------------------



So {{{x^2-2ax+a^2-b^2}}} becomes {{{(x-a)^2-b^2}}}




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



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.



{{{(x-a)^2-b^2=(x-a+b)(x-a-b)}}} Plug in {{{A=x-a}}} and {{{B=b}}}.



So this shows us that {{{(x-a)^2-b^2}}} factors to {{{(x-a+b)(x-a-b)}}}.



In other words {{{(x-a)^2-b^2=(x-a+b)(x-a-b)}}}.



============================================================


Answer:


So {{{x^2-2ax+a^2-b^2}}} completely factors to {{{(x-a+b)(x-a-b)}}}



In other words, {{{x^2-2ax+a^2-b^2=(x-a+b)(x-a-b)}}}