How do you simplify #(x^2-12x+36)/(4x-24)#?

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Answer 1 · 2015-10-21T10:01:27

#((x-6))/4#

Look for potential of parts of the numerator also occurring in the denominator so that you can cancel out.

Consider the numerator: #x^2-12x+36#

To get anything at all that starts to looks like the denominator we have to factorise. Lets try it!

Try: # (x+6)(x-6)#

That does not work. Why is that?
Look at the constants: #(-6) times (+6) # gives #-36#. That is wrong so we need to change something.

Try: #(x-6)(x-6) = x^2 -12x +36# Now we have found the factors.

Substitute this into our original equation giving:

#((x-6)(x-6))/(4x-24)#

We will have found what we want if we factor out 4 from the denominator giving:

#((x-6)(x-6))/(4(x-6))#

Write as:

#((x-6))/4 times ((x-6))/((x-6))#

But:

#((x-6))/((x-6)) = 1#

giving:

#((x-6))/4#

Answer 2 · 2015-10-21T10:50:28

The answer is #(x-6)/4#.

The numerator #(x^2-12x+36)# is in the form #a^2-2ab+b^2#, where #a=x and b=6#.

Rewrite the numerator.

#((x^2)-2(x)(6)+(6^2))/(4x-24)#

Apply the square of a difference #(a-b)^2=a^2-2ab+b^2#.

#(x-6)^2/(4x-24)#

Factor #4# out of the denominator.

#(x-6)^2/(4(x-6))#

Rewrite the numerator.

#((x-6)(x-6))/(4(x-6))#

Cancel #(x-6)# from the numerator and denominator.

#((cancel(x-6))(x-6))/(4cancel((x-6))##=#

#(x-6)/4#