How do you simplify #(12x^6)/(6x^ -2)#?

1 Answer
Jan 5, 2016

Answer:

#(12x^6)/(2x^(-2)) = 2x^8#

Explanation given in detail. The method should help you deal with any question of this type.

Explanation:

Splitting this into 2 parts; numbers and variables

#color(blue)("Part 1: The numbers") color(white)(".....") 12/6#

Divide top and bottom by 6 as #12 -: 6# gives a whole number answer.

#(12 -: 6)/(6 -:6) = 2/1 = 2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Part 2 : The variables") color(white)(".....") (x^6)/(x^-2)#

'..............................................................................................
#color(brown)("To explain what is happening: ")#

Suppose we had a different variable, say #z#.

If this is written as #z^(-2)# then it is another way of writing #1/(z^2)#
If this is written as #1/(z^(-2))# then it is another way of writing #z^2#

So to summarise: If written this way in the numerator it is actually the denominator. On the other hand, if it is written this way in the denominator it is actually the numerator.

#color(green)("You 'move it' to the other side of the dividing line.")#
'...........................................................................................

Write #x^6/(x^(-2)) " as " x^6 xx x^2#

But # x^6 xx x^2# is the same value as # x^((6+2)) = x^8#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together")#

#(12x^6)/(2x^(-2)) = 2x^8#

'///////////////////////////////////////////////////////////////////////////////////////'
Another way of dealing with #1/x^(-2)#

Multiply by 1 in the form of# color(white)(..)x^2/x^2# giving:

#1/x^(-2) xx x^2/x^2#

#=(1xx x^2)/(x^(-2)xxx^(2))#

#= x^2/(x^((-2+2)))#

#x^2/x^0#

But #x^0=1# giving

#x^2/1 = x^2#