Question 216079
{{{(x^2-6x+9)/(x-3)}}}


Step 1.  Look at the numerator {{{x^2-6x+9}}}.


Step 2.  Need to integers m and n whose sum is -6 and product is 9.


Step 3.  The two integers are -3 and -3.  We can factor through grouping where -6x=-3x-3x.  We can also recognize that the trinomial is a perfect square, that is  {{{x^2-6x+9=(x-3)^2}}}.  Nevertheless the following steps, will show the technique of grouping.


{{{x^2-6x+9=(x^2-3x)+(-3x+9)=x(x-3)-3(x-3)}}}


Factor (x-3) from both groups


{{{x^2-6x+9=(x-3)^2}}}


Step 4.  Now substitute the expression in Step 3 with {{{(x-3)^2}}}


{{{(x^2-6x+9)/(x-3)=(x-3)^2/(x-3)}}}


Note we can cancel the factor (x-3) in the numerator and denominator.


{{{(x^2-6x+9)/(x-3)=x-3}}}


Step 4.  ANSWER:  {{{(x^2-6x+9)/(x-3)=x-3}}}


I hope the above steps were helpful. 


For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


And good luck in your studies!


Respectfully,
Dr J