How do you simplify #\frac { 6n - 18} { 5n ^ { 2} - 8n - 21}#?

1 Answer
Tony B
May 2, 2017

#6/(5n+7)#

Explanation:

Lets see if we can figure this out with logic.

It is always worth seeing if we can factor 'things' out for cancelling. So lets have a play!

#color(blue)("Consider the "5n^2-8n-21)#

Whole numbers factors of 21 is 3 and 7 so we must have something like: #(?n+-3)(?n+-7)#
As the 21 is negative then the two constants must be opposite signs

The 5 from #5x^2# is a prime number so we can only have #1xxx5x=5x^2#

Thus we have 1 of 2 possibilities

Possibility 1#->(5n+-3)(n+-7)#
Possibility 2#->(5n+-7)(n+-3)#
.................................................................................................
Consider #(5n+3)(n-7)-> 5n^2-35n+3n-21color(red)(" Fail")#
Consider #(5n-3)(n+7)-> 5n^2+35n-3n-21color(red)(" Fail")#

Consider #(5n+7)(n-3)-> 5n^2-15n+7n-21#
#" "->5n^2-8n-21color(red)(" Works")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all back together")#

#(6n-18)/((5n+7)(n-3))#

Notice that 6 is a common factor for the numerator giving:

#(6(n-3))/((5n+7)(n-3)) larr" we now have something we can cancel"#

#6/(5n+7)xx(n-3)/(n-3)#

But #(n-3)/(n-3) # is the same as 1 giving:

#6/(5n+7)#

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