How do you simplify #\frac { b ^ { 2} + 14b + 18} { b ^ { 2} + 15b +56}#?

1 Answer
Jun 14, 2018

#(b^2+14b+18)/(b^2+15b+56) = 1-31/(b+7)+30/(b+8)#

Explanation:

Given:

#(b^2+14b+18)/(b^2+15b+56)#

One form of simplifying involves separating out the polynomial part of the rational function, which we can do as follows:

#(b^2+14b+18)/(b^2+15b+56) = ((b^2+15b+56)-(b+38))/(b^2+15b+56)#

#color(white)((b^2+14b+18)/(b^2+15b+56)) = (b^2+15b+56)/(b^2+15b+56)-(b+38)/(b^2+15b+56)#

#color(white)((b^2+14b+18)/(b^2+15b+56)) = 1-(b+38)/(b^2+15b+56)#

Further, we can note that:

#b^2+15b+56 = (b+7)(b+8)#

Hence we can split the remaining rational expression into partial fractions:

#(b+38)/(b^2+15b+56) = A/(b+7)+B/(b+8)#

for some #A, B# to be determined.

Multiplying both sides by #b^2+15b+56# this becomes:

#b+38 = A(b+8)+B(b+7)#

Putting #b=-7#, we find#

#31 = A#

Putting #b=-8#, we find:

#30 = -B#

So #B=-30#

So:

#(b+38)/(b^2+15b+56) = 31/(b+7)-30/(b+8)#

and:

#(b^2+14b+18)/(b^2+15b+56) = 1-(b+38)/(b^2+15b+56) = 1-31/(b+7)+30/(b+8)#