How do you simplify #(r+6)/(4^2-r-6)# and state the restrictions?

1 Answer
Apr 7, 2015

I assume your question is #(r + 6)/(r^2 - r - 6)#

Let's Factorise the #color(red)(DENOMINATOR)# first.

The Denominator is #color(red)(r^2 - r - 6)#

We can use Splitting the Middle Term technique to factorise this.

It is in the form #ax^2 + bx + c# where #a=1, b=-1, c= -6#

To split the middle term, we need to think of two numbers #N_1 and N_2# such that:
#N_1*N_2 = a*c and N_1+N_2 = b#
#N_1*N_2 = (1)*(-6) and N_1+N_2 = -1#
#N_1*N_2 = -6 and N_1+N_2 = -1#

After Trial and Error, we get #N_1 = 2 and N_2 = -3#
#(2)*(-3) = -6# and #(2) + (-3) = -1#

So we can write the denominator as
#color(red)(r^2 +2r -3r - 6)#
# = r*(r+2) - 3*(r+2)#
# = (r+2)*(r-3)#

The Denominator can be written as #color(red)((r-2)*(r+3))#

The expression we have been given is
#(r + 6)/(r^2 - r - 6)#

After the denominator was factorised, the Expression can now be written as :

#((r+6))/((r+2)*(r-3))#

#(r + 6)/(r^2 - r - 6)# = #((r+6))/((r+2)*(r-3))#