How do you simplify #\frac { b ^ { 2} + 14b + 18} { b ^ { 2} + 15b +56}#?
1 Answer
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
Multiplying both sides by
#b+38 = A(b+8)+B(b+7)#
Putting
#31 = A#
Putting
#30 = -B#
So
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)#